How To Find Angular Acceleration From Angular Velocity: Problem And Examples

January 21, 2024January 17, 2022 by AKSHITA MAPARI

angular acceleration from angular velocity 0

In the field of physics and mechanics, understanding rotational motion is essential. One important concept in rotational motion is angular acceleration, which relates to the rate of change of angular velocity. In this blog post, we will explore how to find angular acceleration from angular velocity, step by step. We will also cover special cases and related concepts, providing clear explanations and examples along the way.

How to Calculate Angular Acceleration from Angular Velocity

angular acceleration from angular velocity 1

The Mathematical Formula

angular acceleration from angular velocity 3

The formula to calculate angular acceleration $(alpha$ ) from angular velocity $(omega$ ) is given by:

$alpha = frac{{Delta omega}}{{Delta t}}$

Here, $Delta omega$ represents the change in angular velocity, and $Delta t$ represents the change in time.

Step-by-Step Process

To find the angular acceleration from angular velocity, follow these steps:

Identify the initial angular velocity $(omega_i$ ) and the final angular velocity $(omega_f$ ).
Determine the time interval $(Delta t$ ) during which the change in angular velocity occurs.
Calculate the change in angular velocity $(Delta omega$ ) by subtracting the initial angular velocity from the final angular velocity: $Delta omega = omega_f - omega_i$ .
Divide the change in angular velocity by the change in time to obtain the angular acceleration: $alpha = frac{{Delta omega}}{{Delta t}}$ .

Worked out Example

Image by User:Cdang – Wikimedia Commons, Wikimedia Commons, Licensed under CC BY-SA 4.0.

Let’s work through an example to better understand how to find angular acceleration from angular velocity.

Example: A wheel starts with an initial angular velocity of 2 rad/s and accelerates uniformly to a final angular velocity of 8 rad/s over a time interval of 4 seconds. Calculate the angular acceleration.

Solution:
Given:
Initial angular velocity $(omega_i$ ) = 2 rad/s
Final angular velocity $(omega_f$ ) = 8 rad/s
Time interval $(Delta t$ ) = 4 s

To find the angular acceleration $(alpha$ ), we can use the formula: $alpha = frac{{Delta omega}}{{Delta t}}$ .

First, calculate the change in angular velocity $(Delta omega$ ):
$Delta omega = omega_f - omega_i = 8 , text{rad/s} - 2 , text{rad/s} = 6 , text{rad/s}$ .

Next, divide the change in angular velocity by the change in time:
$alpha = frac{{Delta omega}}{{Delta t}} = frac{{6 , text{rad/s}}}{{4 , text{s}}} = 1.5 , text{rad/s}^2$ .

Therefore, the angular acceleration of the wheel is $1.5 , text{rad/s}^2$ .

Special Cases in Finding Angular Acceleration

How to Find Angular Acceleration without Time

In some cases, the time interval $(Delta t$ ) may not be given. However, it is still possible to find the angular acceleration using other known quantities.

If the initial angular velocity $(omega_i$ ), final angular velocity $(omega_f$ ), and the change in angular displacement $(Delta theta$ ) are known, the angular acceleration can be found using the following formula:

$alpha = frac{{(omega_f)^2 - (omega_i)^2}}{{2 cdot Delta theta}}$

Where $Delta theta$ represents the change in angular displacement.

How to Find Angular Acceleration from Angular Velocity and Radius

In situations where the angular velocity $(omega$ ) and radius $(r$ ) are known instead of time, the angular acceleration can be determined using the following formula:

$alpha = frac{{omega^2}}{{r}}$

Where $r$ represents the radius.

How to Find Angular Acceleration from Angular Velocity and Time

If the angular velocity $(omega$ ) and tangential acceleration $(a_t$ ) are given, the angular acceleration can be calculated using the formula:

$alpha = frac{{a_t}}{{r}}$

Where $r$ is the radius.

Related Concepts

How to Find Tangential Acceleration from Angular Velocity

To find the tangential acceleration $(a_t$ ) from angular velocity $(omega$ ) and radius $(r$ ), you can use the formula:

$a_t = omega cdot r$

This formula relates the linear velocity $(v$ ) to the angular velocity $(omega$ ) and radius $(r$ ), as tangential acceleration is the rate of change of linear velocity.

How to Find Linear Acceleration from Angular Velocity

Linear acceleration $(a$ ) can be determined from angular velocity $(omega$ ) and radius $(r$ ) using the formula:

$a = alpha cdot r$

Where $alpha$ represents the angular acceleration.

How to Calculate Centripetal Acceleration from Angular Velocity

Centripetal acceleration $(a_c$ ) can be calculated using the formula:

$a_c = frac{{v^2}}{{r}} = omega^2 cdot r$

Here, $v$ represents the linear velocity and $r$ is the radius.

Understanding how to find angular acceleration from angular velocity is crucial for analyzing rotational motion. By following the steps outlined in this blog post, you can calculate angular acceleration accurately. Remember to consider special cases and related concepts to gain a comprehensive understanding of this topic.

How can you find the constant angular acceleration from the given angular velocity in a motion?

To find the constant angular acceleration from the given angular velocity, you can follow the steps mentioned in the article Finding constant angular acceleration in motion. First, determine the final angular velocity and initial angular velocity. Then, calculate the change in angular velocity and the change in time. Finally, divide the change in angular velocity by the change in time to obtain the constant angular acceleration. This method helps in quantifying the change in angular velocity over a specific period of time, enabling a deeper understanding of the motion’s behavior.

Numerical Problems on how to find angular acceleration from angular velocity

Problem 1:

An object is rotating with an angular velocity of 4 rad/s. It accelerates uniformly at a rate of 2 rad/s^2 for a time of 5 seconds. Find the final angular velocity of the object.

Solution:
Given:
Initial angular velocity, $omega_{i} = 4$ rad/s
Angular acceleration, $alpha = 2$ rad/s $^2$
Time, $t = 5$ s

The final angular velocity can be calculated using the formula:
$[omega_{f} = omega_{i} + alpha t]$

Substituting the given values, we have:
$[omega_{f} = 4 + 2 times 5 = 14 text{ rad/s}]$

Therefore, the final angular velocity of the object is 14 rad/s.

Problem 2:

A wheel starts from rest and rotates with an angular acceleration of 3 rad/s^2. If it rotates for a time of 10 seconds, find the final angular velocity of the wheel.

Solution:
Given:
Initial angular velocity, $omega_{i} = 0$ rad/s (as the wheel starts from rest)
Angular acceleration, $alpha = 3$ rad/s $^2$
Time, $t = 10$ s

The final angular velocity can be calculated using the formula:
$[omega_{f} = omega_{i} + alpha t]$

Substituting the given values, we have:
$[omega_{f} = 0 + 3 times 10 = 30 text{ rad/s}]$

Therefore, the final angular velocity of the wheel is 30 rad/s.

Problem 3:

Image by EnEdC – Wikimedia Commons, Wikimedia Commons, Licensed under CC BY-SA 3.0.

angular acceleration from angular velocity 2

A fan blade initially rotates with an angular velocity of 8 rad/s. It decelerates uniformly at a rate of 4 rad/s^2 for a time of 2 seconds. Find the final angular velocity of the fan blade.

Solution:
Given:
Initial angular velocity, $omega_{i} = 8$ rad/s
Angular acceleration, $alpha = -4$ rad/s $^2$ (negative sign indicates deceleration)
Time, $t = 2$ s

The final angular velocity can be calculated using the formula:
$[omega_{f} = omega_{i} + alpha t]$

Substituting the given values, we have:
$[omega_{f} = 8 - 4 times 2 = 0 text{ rad/s}]$

Therefore, the final angular velocity of the fan blade is 0 rad/s.

Also Read:

How to Find Final Velocity with Acceleration and Distance: A Comprehensive Guide

June 25, 2024January 16, 2022 by SAKSHI KM

Summary

Determining the final velocity of an object given its acceleration and distance traveled is a fundamental concept in physics. This comprehensive guide will walk you through the step-by-step process, providing detailed explanations, formulas, examples, and problem-solving techniques to help you master this essential skill.

Understanding the Kinematic Equation

The key formula used to find the final velocity of an object, given its acceleration and distance, is the kinematic equation:

$v^2 = u^2 + 2as$

Where:
– $v$ is the final velocity (in m/s)
– $u$ is the initial velocity (in m/s)
– $a$ is the acceleration (in m/s²)
– $s$ is the displacement or distance traveled (in m)

This equation is derived from the basic principles of motion and can be used to solve a variety of problems in physics.

Applying the Formula: Examples and Explanations

Example 1: Finding Final Velocity with Known Initial Velocity and Acceleration

Suppose an object starts with an initial velocity of 5 m/s and accelerates at a constant rate of 2 m/s² over a distance of 10 meters. To find the final velocity, we can plug the values into the formula:

$v^2 = 5^2 + 2 \times 2 \times 10$
$v^2 = 25 + 40$
$v^2 = 65$
$v = \sqrt{65} \approx 8.06 \text{ m/s}$

In this example, the object’s final velocity is approximately 8.06 m/s.

Example 2: Finding Final Velocity with Different Units

Consider a car that accelerates from 30 km/h to a final velocity over a distance of 200 meters. If the acceleration is 2 m/s², we need to convert the initial velocity to meters per second (30 km/h = 8.33 m/s) and then use the formula:

$v^2 = 8.33^2 + 2 \times 2 \times 200$
$v^2 = 69.44 + 800$
$v^2 = 869.44$
$v = \sqrt{869.44} \approx 29.47 \text{ m/s}$

In this case, the car’s final velocity is approximately 29.47 m/s.

Example 3: Finding Final Velocity with Time

If you know the time over which the acceleration occurs, you can use the formula:

$v = u + at$

For example, if an object accelerates from 10 m/s at a rate of 3 m/s² over 5 seconds:

$v = 10 + 3 \times 5$
$v = 10 + 15$
$v = 25 \text{ m/s}$

In this scenario, the object’s final velocity is 25 m/s.

Solving Numerical Problems

To further solidify your understanding, let’s work through some numerical problems:

An object starts with an initial velocity of 8 m/s and accelerates at a rate of 4 m/s² over a distance of 20 meters. Find the final velocity.
A car accelerates from 20 km/h to a final velocity over a distance of 150 meters. If the acceleration is 3 m/s², calculate the final velocity.
An object is initially moving at 15 m/s and undergoes an acceleration of 2.5 m/s² for 8 seconds. Determine the final velocity.

Solutions:

Plugging the values into the formula:
$v^2 = 8^2 + 2 \times 4 \times 20$
$v^2 = 64 + 160$
$v^2 = 224$
$v = \sqrt{224} \approx 14.97 \text{ m/s}$
First, convert the initial velocity to meters per second:
$20 \text{ km/h} = 5.56 \text{ m/s}$
Then, use the formula:
$v^2 = 5.56^2 + 2 \times 3 \times 150$
$v^2 = 30.94 + 900$
$v^2 = 930.94$
$v = \sqrt{930.94} \approx 30.51 \text{ m/s}$
Using the formula $v = u + at$:
$v = 15 + 2.5 \times 8$
$v = 15 + 20$
$v = 35 \text{ m/s}$

Key Takeaways

The formula $v^2 = u^2 + 2as$ is used when you know the acceleration and distance.
The formula $v = u + at$ is used when you know the acceleration and time.
Ensure that the units of velocity, acceleration, and distance are consistent (e.g., meters per second for velocity and acceleration, and meters for distance).
Practice solving a variety of problems to develop a strong understanding of the concepts.
Refer to the provided examples and solutions to guide your problem-solving process.

References

Is Horizontal Velocity Constant: Why, When, How, Problem Examples

January 21, 2024January 16, 2022 by Manish Naik

The article discusses about is horizontal velocity constant or why horizontal velocity constant in projectile motion, along with its problem examples.

After the applied force sets the parabolic path in projectile motion, only the gravity force acts downward to a projectile in the vertical direction. So an object moves downward with varying vertical velocity but constant horizontal velocity.

When is Horizontal Velocity Constant?

The horizontal velocity during projectile motion is constant.

As per definition, the projectile motion concerns a single gravity force that provokes a net force toward the earth’s center. Since no horizontal force acts in projectile motion, its horizontal velocity stays constant and progresses towards the earth’s center due to gravity force.

To understand the concept of downward acceleration with constant horizontal velocity, consider an example of a boy throwing a ball in a straight direction with high velocity. Let’s say the gravity is negligible in this case, then how does the ball move in the absence of gravity?

Newton’s law says, “an object in motion continues to move in motion unless some force acts on it“. That means; a ball moves in a straight path with a constant velocity, which agrees with the law of inertia.

What is Inertia? — **Law of Inertia in Projectile Motion**
(credit: shutterstock)

Now in the same case, what if gravity exists? The gravity force influences the ball’s vertical motion, causing a vertical acceleration, agreeing with the free-falling object accelerating at a rate of acceleration due to gravity g. Because of gravity force, the ball will drop vertically below its straight path after parabolic trajectory as per the attribute of the projectile motion.

Suppose that boy applies a force of 20N to throw the ball. The ball achieves a velocity of 10m/s and persists in moving with the same velocity horizontally. At the same time, the vertical velocity achieves the velocity at 9.8m/s each second due to gravity force. That means the ball has a vertical acceleration of about 9.8 m/s but no horizontal acceleration.

**Horizontal Velocity Constant**
(credit: shutterstock)

The downward acceleration results in a lowered trajectory from the straight line the projectile ball travels if the gravity force is negligible. The ball possesses its constant horizontal velocity in the existence of gravity which acts to drive the exact vertical velocity as before.

However, the gravity or the vertical force works perpendicular to the horizontal velocity of the projectile. Since the perpendicular components are independent of each other, the gravity force cannot impact the ball’s horizontal velocity. That’s why any projectile moves with downward acceleration but constant horizontal velocity.

Why is Horizontal Velocity Constant?

The horizontal velocity of the projectile is constant because of its inertia.

If there is no force, an object’s state will not modify as per the law of inertia. A projectile’s tendency to remain in a motion with constant velocity resulted in the horizontal motion of the projectile. Therefore, the projectile drives with a stable horizontal velocity.

**Why is Horizontal Velocity Constant** (credit: shutterstock)

What Is The Initial Horizontal Velocity Of A Projectile: How to Find, Examples

January 21, 2024January 16, 2022 by Raghavi Acharya

In this post on physics, let us study the concept of what is the initial horizontal velocity of a projectile and its examples.

Whenever any projectile is released from its state of rest, it gains some velocity that helps it to move forward. The starting velocity that is gained by the projectile on its horizontal path is termed as initial horizontal velocity and can be found using the formulas that are mentioned below,

v_ix = v_x – + 1/2a_xt

v_ix = v_fx – a_xt

v_ix² = v_fx² – 2a_xx

v_ix = v_x /cosθ

Before studying in detail about what is the initial horizontal velocity of a projectile, let us know about projectile motion.

Projectile Motion: Concept and features

Projectile motion is one of the essential types of motion that is an essential part of mechanical physics.

When any object is thrown into open space, it travels along a trajectory that makes a shape of the parabolic curve under the influence of only gravitational force. This projectile motion is a particular case because the motion takes place in both the x and y component of a coordinate axis.

what is the horizontal velocity of a projectile

It’s time to know the main focus of this article: the initial horizontal velocity of a projectile.

What is the initial horizontal velocity of a projectile?

The word-initial refers to the beginning or starting point; displacement, velocity, etc.

Here the initial horizontal velocity of a projectile indicates the velocity gained by the projectile as soon as it launches or releases from the state of rest into the earth’s atmospheric surface. A projectile gains this initial velocity along the horizontal straight path. Hence it is given the name initial horizontal velocity of projectile motion.

v_ixt = v_x – + 1/2a_xt

v_ix = v_fx – a_xt

v_ix² = v_fx² – 2a_xx

v_ix = v_x /cosΘ

Now take a peek into how to find the initial horizontal velocity of any projectile.

How to find the initial horizontal velocity of a projectile?

We can easily and quickly measure the value of initial horizontal velocity by noticing the formula to calculate a projectile’s horizontal velocity.

We have already known the formulas used to calculate horizontal velocity; consider those formulas and rearrange them so that you get you to get initial horizontal velocity as the main component to be measured. See below; we have arranged for them to get V_ix.

Let us know these formulas elaborately by knowing the different approaches of what is the initial horizontal velocity of a projectile.

Different approaches to find what is the initial horizontal velocity of a projectile

The different approaches of what is the initial horizontal velocity of a projectile are as given as follows,

The first approach is to use the kinematics equation of mechanical physics to find out the initial horizontal velocity of a projectile.
The second approach is to look into the formulas used to measure the value of the horizontal velocity of a projectile.

We have to make some minor modifications to the initial velocity as the main component to be measured in both these approaches.

How does the initial velocity of a projectile affect its range?

There is an influence of the initial velocity of any projectile in some motion on its range.

We have to note down the three main factors that tell about the influence of initial velocity on the range. The main three factors are given below.

If the initial velocity V_ix is more then, even the range of the projectile that is under the same motion will reach a greater height., In contrast, if the initial velocity is less, even the range will be limited.
Depending on whether the initial velocity is greater or lesser, the angle of the horizontal component of velocity varies.
Suppose we take any two projectiles launched at different angles, having different horizontal velocities into consideration. In that case, the range may be identical for the projectiles, but it reaches different maximum heights.

After knowing the influence of V_ix on its range, let us study the method to find initial horizontal velocity in projectile motion.

How to find initial velocity in projectile motion?

One can find initial velocity in any projectile motion by using the below formulas,

Here if you are finding on different components of projectile motion, take into account the x or y component in place of velocity in the following formulas,

v_ixt = v_x – + 1/2a_xt

v_ix = v_fx – a_xt

v_ix² = v_fx² – 2a_xx

Therefore, it will become easy to calculate the initial velocity by knowing these formulas. By solving some fundamental problems on what is the initial horizontal velocity of a projectile, we can study it better.

Problems based on the initial horizontal velocity of a projectile

Here are some problems based on the initial velocity of a projectile to better understand the concept.

Problem 1

A pack of garbage is thrown into the bin. The horizontal velocity is 25m/s, and the launching angle of the pack of garbage is 12°. Now find out the value of the initial horizontal velocity?

Solution: Let us first note down the given data,

The velocity along the horizontal path = V_x = 25m/s and θ = 12°

Now we can use one of the formulas as mentioned in the above concepts to measure initial horizontal velocity

V_ix = V_x / Cos θ

V_ix = (25) /Cos (12°)

V_ix = 29.65m/s

Therefore, the required initial horizontal velocity is 29.65m/s

Problem 2

A single coin is thrown into the earth’s atmosphere. The horizontal velocity will be almost equal to 29m/s, and the angle of the coin when it is thrown is 20°. Now measure the initial horizontal velocity of the coin?

Solution: Let us first note down the given data,

The velocity along the horizontal path = V_x = 29m/s and θ = 20°

Now we can use one of the formulas as mentioned in the above concepts to measure initial horizontal velocity

V_ix = V_x / Cos θ

V_ix = (29)/ Cos (20°)

V_ix = 30.88m/s

Therefore, the required initial horizontal velocity is 30.88m/s

What are the different examples of the initial horizontal velocity of a projectile in real life?

There are many examples that we find around us that will be a great example of the initial horizontal velocity of any projectile.

Every motion of a projectile occurs with gaining of velocity known to be initial velocity. When thrown above apart from a straight angle, all the material makes a parabolic curve while reaching the ground. So, we can consider the below-mentioned examples for initial horizontal velocity.

Long jump
Running of horses
Discus throw
Throwing of basketball into the net
Fireworks

Long jump

Before taking a long jump, the athlete warms up their body then begins the motion. Here the component required for a long jump includes initial velocity. If the initial velocity is more than that, the long jump taken by the athlete will be higher.

what is the initial horizontal velocity of a projectile — Image Credit: Pixabay free images

Running of horses

We have seen the running of horses may be in real or in any wildlife channels. If you notice carefully, while running, horses’ legs make a trajectory path. At the beginning of motion, the velocity present to initiate is known to be initial horizontal velocity.

Discus throw

Even in the game of discus throw, while throwing a disc, the velocity possessed by the disc along the horizontal path is initial horizontal velocity. If the velocity is more, even the disc will reach a greater height, increasing the chance of winning the game.

Throwing of basketball into the net

Even if we observe the basketball game, we can identify the parabolic path made by the player while making sure the ball enters the net. It is an example of projectile motion and even an example for initial horizontal velocity because the velocity as soon as the ball is thrown possessed by it.

Fireworks

We make sure that we lit fireworks and enjoy ourselves during the festival or any celebration. The velocity gained by the firework as soon as it lits termed to be initial horizontal velocity that influences the range of the fireworks to reach a certain point in the atmosphere. If the V is more, then, obviously, the fireworks reach maximum height.

Therefore, we can notice the projectile motion mainly in sports.

What factors tell initial velocity to affect the maximum height of a projectile?

Initial velocity directly influences the point of maximum height in a projectile motion.

If the speed of initial velocity is more, then the angle at which it is launched will be steeper. If this angle is increased, then there is a possibility that the maximum height in a projectile motion will increase. Therefore, we can infer that the change in initial velocity affects the maximum height of the projectile.

In this way, initial velocity and maximum height act based on each other.

Why does increasing initial velocity increase range?

The increasing initial horizontal velocity tends to increase the range of a projectile.

The term increasing initial horizontal velocity signifies that the position of a projectile with the time increase makes the projectile in motion move to a different height. These phenomena directly make the projectile a more extended height, increasing the projectile‘s range.

So, there is a direct effect of initial horizontal velocity on the range and maximum height of any projectile.

To study more:

What is the horizontal velocity of a projectile

Frequently Asked Questions | FAQs

What are the three main things affected by the initial horizontal velocity of a projectile motion?

The main factors that are affected by the initial horizontal velocity are

Projectile angle
Maximum height
Range of a projectile motion

What are the best examples of the initial horizontal velocity of a projectile motion?

The best examples of initial horizontal velocity are as below,

Waterfall
Jumping from a certain height or any vehicle
Baseball game
Launching of rockets
Falling of meteoroids.
Throwing any waste into the dustbin.

Also Read:

Mastering Cryogenic Velocity Computations: A Comprehensive Guide

June 25, 2024January 15, 2022 by Core SME lambdageeks.com

Cryogenics, the study of materials and processes at extremely low temperatures, is a critical field in various industries, from aerospace to energy. Accurately computing velocity in cryogenic systems is essential for optimizing performance, ensuring safety, and advancing scientific research. In this comprehensive guide, we will delve into the intricacies of computing velocity in cryogenics, exploring … Read more

Mastering the Determination of Velocity in Molecular Orbitals: A Comprehensive Guide

June 25, 2024January 15, 2022 by Core SME lambdageeks.com

how to determine velocity in molecular orbitals

Determining the velocity of electrons within molecular orbitals is a crucial aspect of understanding the behavior and properties of molecules. This comprehensive guide will delve into the theoretical background, calculation methods, quantifiable data, and practical applications of determining velocity in molecular orbitals, providing a valuable resource for physics students and researchers. Theoretical Background Molecular Orbital … Read more

How To Find Acceleration In Velocity Time Graph: Problems And Examples

February 17, 2024January 14, 2022 by AKSHITA MAPARI

How to Find Acceleration in Velocity-Time Graph

Understanding the Basics of Velocity-Time Graph

When studying the motion of an object, it is often helpful to analyze its velocity as a function of time. A velocity-time graph, also known as a V-T graph, provides a graphical representation of an object’s velocity over a specific time interval. The graph consists of a horizontal time axis and a vertical velocity axis. By examining the shape and characteristics of the graph, we can gain valuable insights into the object’s motion.

In a velocity-time graph, the slope of the line represents the object’s acceleration. Acceleration is the rate at which an object’s velocity changes over time. It indicates how quickly the object is speeding up or slowing down. When the velocity-time graph has a straight line, the acceleration is constant. A steeper slope indicates a higher acceleration, while a shallower slope represents a lower acceleration.

Importance of Acceleration in Velocity-Time Graph

Acceleration is a fundamental concept in physics that helps us understand the changes in an object’s motion. By examining the acceleration in a velocity-time graph, we can determine whether an object is accelerating, decelerating, or moving at a constant velocity. This information is crucial for analyzing various physical phenomena, including the motion of vehicles, projectiles, and celestial bodies.

Calculating the acceleration in a velocity-time graph allows us to quantify the rate at which an object’s velocity is changing. This information helps us predict an object’s future motion, determine the forces acting upon it, and analyze its overall dynamics.

Steps to Calculate Acceleration in Velocity-Time Graph

Image by Pradana Aumars – Wikimedia Commons, Wikimedia Commons, Licensed under CC0.

To calculate acceleration using a velocity-time graph, follow these steps:

1. Identifying Key Points on the Graph

Examine the velocity-time graph and identify two key points: the initial velocity (v_i) and the final velocity (v_f). The initial velocity represents the object’s velocity at the beginning of the time interval, while the final velocity represents its velocity at the end of the interval.

2. Using the Slope Formula to Calculate Acceleration

The slope of the velocity-time graph represents the object’s acceleration. To calculate the slope, we use the formula:

$text{slope} = frac{text{change in velocity}}{text{change in time}} = frac{v_f - v_i}{t}$

where v_f is the final velocity, v_i is the initial velocity, and t is the time interval.

3. Interpreting the Results

Once you have calculated the acceleration using the slope formula, you can interpret the results. A positive acceleration indicates that the object is speeding up, while a negative acceleration indicates that it is slowing down. If the acceleration is zero, the object is moving at a constant velocity.

Worked Out Examples

Let’s work through a few examples to solidify our understanding of calculating acceleration in velocity-time graphs.

Example of a Positive Acceleration

Suppose we have a velocity-time graph that shows a straight line with a positive slope. The initial velocity (v_i) is 10 m/s, the final velocity (v_f) is 30 m/s, and the time interval (t) is 5 seconds.

Using the slope formula, we can calculate the acceleration:

$text{slope} = frac{v_f - v_i}{t} = frac{30 , text{m/s} - 10 , text{m/s}}{5 , text{s}} = frac{20 , text{m/s}}{5 , text{s}} = 4 , text{m/s}^2$

Therefore, the object has a positive acceleration of 4 m/s^2, indicating that it is speeding up.

Example of a Negative Acceleration

Now let’s consider a velocity-time graph with a negative slope. The initial velocity (v_i) is 20 m/s, the final velocity (v_f) is 10 m/s, and the time interval (t) is 2 seconds.

Using the slope formula:

$text{slope} = frac{v_f - v_i}{t} = frac{10 , text{m/s} - 20 , text{m/s}}{2 , text{s}} = frac{-10 , text{m/s}}{2 , text{s}} = -5 , text{m/s}^2$

In this case, the object has a negative acceleration of -5 m/s^2, indicating that it is slowing down.

Example of Zero Acceleration

Finally, let’s examine a velocity-time graph with a horizontal line. This line represents a constant velocity. If the initial velocity (v_i) is 15 m/s, the final velocity (v_f) is also 15 m/s, and the time interval (t) is 4 seconds, the acceleration will be zero.

Using the slope formula:

$text{slope} = frac{v_f - v_i}{t} = frac{15 , text{m/s} - 15 , text{m/s}}{4 , text{s}} = frac{0 , text{m/s}}{4 , text{s}} = 0 , text{m/s}^2$

The object has a zero acceleration, indicating that its velocity remains constant.

Common Misconceptions in Calculating Acceleration in Velocity-Time Graph

While calculating acceleration in a velocity-time graph, it is crucial to avoid common misconceptions. Let’s address a few of them:

Misconception about the Slope

Some people may mistakenly assume that the slope of a velocity-time graph represents the velocity instead of the acceleration. Remember, the slope represents the change in velocity over time, which gives us the acceleration.

Misconception about the Area under the Graph

Another misconception is that the area under a velocity-time graph directly provides the acceleration. However, the area under the graph represents the displacement or distance traveled, not the acceleration.

Misconception about the Time Interval

People sometimes assume that the time interval for calculating acceleration should be the entire duration of the graph. However, the time interval should only include the specific segment for which you want to calculate the acceleration.

Understanding how to find acceleration in a velocity-time graph is essential for analyzing an object’s motion. By identifying key points on the graph and using the slope formula, we can determine the acceleration and gain valuable insights into the object’s dynamics. Remember to be aware of common misconceptions to ensure accurate calculations.

How can we find acceleration and velocity using velocity-time graphs and constant acceleration?

To understand the concept of finding acceleration and velocity using velocity-time graphs and constant acceleration, we can explore the article on How to Find Acceleration in Velocity-Time Graph. This article provides a comprehensive explanation of the steps involved in determining acceleration from a velocity-time graph. Additionally, to understand the connection between acceleration and velocity, you can refer to the article on Finding Velocity with Constant Acceleration. This article elucidates the process of calculating velocity when the acceleration remains constant. By integrating these two concepts, we can gain a deeper understanding of the relationship between velocity and acceleration in dynamic systems.

Numerical Problems on how to find acceleration in velocity time graph

Problem 1:

Image by MikeRun – Wikimedia Commons, Wikimedia Commons, Licensed under CC BY-SA 4.0.

A car accelerates uniformly from rest to a velocity of 40 m/s in 10 seconds. Find the acceleration of the car.

Solution:

Given:
Initial velocity, $u = 0 , text{m/s}$
Final velocity, $v = 40 , text{m/s}$
Time taken, $t = 10 , text{s}$

The acceleration ( $a$ ) can be found using the formula:

$a = frac{{v - u}}{{t}}$

Substituting the given values, we have:

$a = frac{{40 - 0}}{{10}}$

Simplifying the equation, we get:

$a = frac{{40}}{{10}}$

Therefore, the acceleration of the car is $4 , text{m/s}^2$ .

Problem 2:

A train starts from rest and accelerates uniformly. It covers a distance of 200 m in 10 seconds. Find the acceleration of the train.

Solution:

Given:
Initial velocity, $u = 0 , text{m/s}$
Distance covered, $s = 200 , text{m}$
Time taken, $t = 10 , text{s}$

The acceleration ( $a$ ) can be found using the formula:

$a = frac{{2(s - ut)}}{{t^2}}$

Substituting the given values, we have:

$a = frac{{2(200 - 0 cdot 10)}}{{10^2}}$

Simplifying the equation, we get:

$a = frac{{2 cdot 200}}{{100}}$

Therefore, the acceleration of the train is $4 , text{m/s}^2$ .

Problem 3:

A particle is moving with a constant acceleration. Its initial velocity is 10 m/s and it covers a distance of 100 m in 5 seconds. Find the acceleration of the particle.

Solution:

Given:
Initial velocity, $u = 10 , text{m/s}$
Distance covered, $s = 100 , text{m}$
Time taken, $t = 5 , text{s}$

The acceleration ( $a$ ) can be found using the formula:

$a = frac{{2(s - ut)}}{{t^2}}$

Substituting the given values, we have:

$a = frac{{2(100 - 10 cdot 5)}}{{5^2}}$

Simplifying the equation, we get:

$a = frac{{2 cdot 50}}{{25}}$

Therefore, the acceleration of the particle is $4 , text{m/s}^2$ .

Also Read:

What Is The Horizontal Velocity Of A Projectile: Different Aspects And Facts

January 21, 2024January 14, 2022 by Raghavi Acharya

The horizontal velocity is considered to occur along the x-axis. In this post, let us study about what is the horizontal velocity of a projectile.

The horizontal velocity(v_x) of any projectile is the primary component of the body undergoing projectile motion. It is generally used to know the velocity value at which the particle travels along the horizontal path in any projectile motion. It is even considered the initial path covered by the body in motion.

To know the different aspects and facts of the horizontal velocity of a projectile, let us move further in this post.

What is the horizontal velocity of a projectile at its highest point?

As soon as the projectile reaches its highest point, it is attracted by gravitation force and moves downwards.

Suppose the particle in projectile motion reaches its highest point along the horizontal path. In that case, the horizontal velocity will usually be zero because as soon as it reaches the peak point, it is made to fall vertically downwards by the influence of the force of gravity. This horizontal velocity at its highest point does not encounter any influence from vertical velocity.

So, from this, we can infer that horizontal velocity will become zero whenever it reaches its highest point in a projectile motion. Now let us look into the fact of what is the horizontal velocity of a projectile and its formula.

The horizontal velocity of a projectile formula

We have studied earlier how to find velocity in a motion; similarly, we can consider one of the methods of calculating velocity by taking into account displacement and time to calculate horizontal velocity.

The following formula helps measure the horizontal velocity of the projectile; we can either use the kinematics equation or only the initial velocity and launch angle to measure the horizontal velocity.
The kinematics equation includes displacement, acceleration, time, initial and final velocities.

V_x = V_i Cos θ

x = v_ixt + 1/2a_xt²

v_fx = v_ix + a_xt

v_fx² = v_ix² + 2a_xx

These are some crucial formulas that help measure the horizontal velocity. This concept helps in knowing what the horizontal velocity of the projectile is

The horizontal velocity of a projectile at its maximum height

As soon as the particle in projectile motion along the horizontal path reaches its maximum height, it will be constant.

The horizontal velocity in a projectile is usually measured in V_x. It will be constant until it reaches h_max; only the vertical velocity will be zero at the point of h_max, and later it undergoes variation due to gravity.

Now to study in detail the calculation of maximum height.

How to find the horizontal velocity at maximum height?

The maximum height of a projectile along the horizontal path gains horizontal velocity, and it is calculated using the formula as shown below,

h_max = (V_Y)² / 2g

Here

h_max refers to a maximum height

V_Y refers to vertical velocity

g refers to gravitational force

The above equation depicts the calculation of maximum height at the launching of the projectile. It only comes under the influence of the vertical vector component of the velocity at the beginning.

The horizontal velocity of a projectile launched at an angle

When thrown into open-air, every particle undergoes projectile motion and makes a certain angle while moving down towards the ground.

We know that the nature of the path in a projectile motion is parabolic in shape. When thrown, it moves with initial constant horizontal velocity and makes an angle θ as it reaches its maximum height. This angle is considered the launch angle of a projectile along the horizontal path.

Image: Horizontal velocity and angle of projectile

At this point of the post, let us go in-depth to know more facts about what is the horizontal velocity of a projectile at the peak of travel.

What is the horizontal velocity of a projectile at the peak of travel?

The peak of travel in a horizontal component of a projectile also refers to the maximum height attained by a projectile.

As we have studied in the above concepts, the velocity component along the vertical displacement will be zero at the maximum height. It becomes a horizontal component entirely at the point of h_max in the parabolic path. We can infer that as the velocity will be only along the horizontal path, the horizontal velocity will be zero at the travel peak.

To know more about the vertical velocity of a projectile motion, let us study the next concept.

What is the effect on the horizontal velocity in motion as it falls?

As soon as the projectile falls, it will be no more in the state of horizontal displacement, and hence there will be the absence of horizontal velocity.

There will be the absence of influence of gravitational force along the horizontal path; hence the horizontal velocity will maintain its constant velocity.
As the velocity will be constant, there will be not much change in acceleration. When it falls after reaching h_max, the horizontal part is now entirely vertical, and at this point, the influence of gravity acts on it.
Both the components will be independent of one another.

After knowing the components of the horizontal velocity of a projectile, let us study the variation in horizontal velocity at different launch angles.

What is the horizontal velocity of a projectile thrown at an angle of 60?

We know how to measure horizontal velocity with having the values of displacement, initial velocity, time and launch angle. But in this particular question, they have mentioned only the launch angle.

In the earlier concepts, we have come across that the horizontal velocity of a projectile will be along with the horizontal displacement of a projectile thrown at a certain angle and forms a trajectory; it is represented in the horizontal x-axis of velocity.

We can calculate horizontal velocity b=of a projectile that is thrown at an angle of 60 by considering the below equation,

V_x = V_i Cos θ

Here we can consider the value of θ as 60, focussing the horizontal direction then we can substitute the value of the angle in the above equation as shown,

V_x = V_i Cos 60

From trigonometric values, we can know the values of cos60 as ½ or 0.5

V_x = 1/2 V_i

Therefore, from the above substitution, we can conclude that when a projectile is thrown in the horizontal direction at an angle of 60, the obtained horizontal velocity will be half of the total projectile velocity.

Let us about more changes in horizontal velocity.

Can the horizontal velocity of a projectile change?

The horizontal velocity of any projectile motion does not change.

In any projectile motion, the horizontal component of velocity will remain constant until it reaches its certain height and only in the vertical direction it undergoes variation.

Therefore, we can infer that the velocity of a projectile in a horizontal direction does not change.

How does the horizontal velocity of a projectile become zero?

The horizontal velocity cannot be zero but will be constant throughout the motion.

When the initiation of a projectile motion takes place, it reaches a particular point called maximum height; till that point, the velocity remains constant as there will be no other force, and here the vertical velocity component will be zero.

So, from these facts, we can come to the theory that only the vertical component of velocity in a projectile motion will be zero, and anyway, the horizontal component of velocity will always be constant; that is, it maintains the value of initial velocity.

Why will the horizontal component of a projectile velocity be constant?

The horizontal component of the velocity that is considered in the projectile motion can be considered shortly as horizontal velocity.

The horizontal velocity is constant throughout the projectile motion because no other external force acts on the body in motion. It is the reason why gravity only influences the vertical component and not on the horizontal that makes the horizontal velocity component remain constant.

What do you mean by maximum height in projectile motion?

The maximum height in a projectile motion is nothing but the highest point a projectile reaches during the movement.

Usually, we observe that when any material or object is thrown above, it goes up to a certain height and then travels back towards us due to the influence of gravitation.
The particular height a body reaches will be considered as the maximum height. At this point, there will be zero velocity, and the vertical motion of the body begins. Here, the initial velocity is applied to take the body along some height of horizontal path, known as the projectile range.

What is the relationship between a projectile motion’s horizontal and vertical velocity?

Both components of a projectile motion never depend on each other.

We have already known that any projectile motion consists of a horizontal and vertical velocity.
These two components are considered because projectile motion takes place in two-dimension.
The horizontal velocity component of the projectile motion has differed entirely in terms of nature and forces from the vertical component of the velocity vector of the body undergoing projectile motion.
As the name suggests, both the velocities occur along respective x and y coordinates.
Since they occur in different directions, they are independent of each other.
There is no specific relationship between horizontal velocity and vertical velocity.

Define the relation between horizontal velocity and maximum height?

There is no particular relationship between horizontal velocity and maximum height, but there is a specific relation between horizontal displacement and maximum height.

The displacement along the horizontal path simultaneously includes horizontal velocity. The displacement in the horizontal path is also termed its range; even this range influences the horizontal velocity. It is given a special formula as given below,

h/R = tanθ/4

Here, there is an increase in angle, even the maximum height increases.

What do you mean by a range of a projectile?

The range of a projectile is usually influenced by horizontal velocity or speed.

The range is a unique name given to the distance travelled by a projectile along the horizontal direction. There will be no effect of acceleration along this path since, in a projectile, the only force of action that occurs is the gravitational force. This range is calculated by an essential formula that is mentioned below,

R = v₀²sin2θ/g

R indicates the horizontal range (m) of a projectile
V_o indicates the initial velocity (m/s) of a projectile
g indicates acceleration due to gravity that is experienced by a projectile
θ indicates the angle of the initial velocity from the horizontal plane

What is the relation between horizontal velocity and the range of a projectile?

The relation between horizontal velocity and the range of a projectile is determined by the horizontal path along which the particle moves in trajectory.

The maximum height of any projectile motion is a specific point at which some change in velocity occurs and where the vertical velocity of the projectile will always be zero. From this position, the projectile will take a curve to move downwards. We can notice that the horizontal velocity will remain the same until the body reaches a maximum height where the horizontal component will be zero and influences its range.

How do you find horizontal velocity with maximum height?

Horizontal velocity can be measured by considering various components such as maximum height, range and launch angle.

To know more about how to measure horizontal velocity and to know what is the horizontal velocity of the projectile, we have one formula that is useful in finding the horizontal velocity.

H_max = (V_i)² Sin2θ_i/g

Here initial velocity and the launched angle is taken into consideration.

How will the horizontal component of velocity for a projectile be affected by the vertical component Quizlet?

The vertical component of projectile velocity won’t affect its horizontal component.

Both horizontal and vertical velocity are the two main components of projectile velocity and will act perpendicularly to one another. At the same time, it’s in the perpendicular direction; both the components are not under the influence of one another.

What happens to the horizontal velocity in motion as it moves up?

As soon as the projectile starts to move up, it reaches a certain height, takes diversion, and begins to move vertically.

There will be no acceleration component that acts on the horizontal path that tends to make the horizontal velocity maintain its constant phase throughout its displacement. After reaching its max. Height and an acceleration amount of 9.8m/s act on the body attracted by gravity.
From this observation, we can imply that only the vertical velocity varies by an acceleration of 9.8m/s, whereas on the horizontal path, its velocity remains constant throughout its range.

Why does the horizontal velocity do not change during its flight?

The horizontal velocity in projectile motion does not change during its flight because no gravitational force acts on it.

We have studied in our earlier classes that for any change in velocity, there must be some gravitational force that acts on it. Keeping this concept in mind, if we see the nature of horizontal velocity, we know that the acceleration component is absent along the horizontal path; due to this, no gravity acts on it, leading to the absence of horizontal forces. All these factors contribute to maintaining the constant phase of horizontal velocity during its flight.

Why is horizontal velocity constant in a projectile motion?

The horizontal velocity is constant in a projectile motion due to the absence of change in acceleration along its horizontal path.

We can consider both horizontal velocity and the velocity at the initiation of a projectile motion identical. As the name suggests, the projectile travels through a horizontal straight path. We can notice the horizontal speed and velocity. This velocity will maintain a constant phase until it reaches a certain height, and the acceleration due to gravity only occurs on the vertical path and not on the horizontal.
We can sum up it since the acceleration is constant or absent throughout the horizontal displacement, and the horizontal velocity will be constant.

Why does horizontal velocity does not affect vertical velocity?

When anybody or object is thrown above, it gains velocity to move further.

In a projectile motion, as soon as trajectory occurs, the beginning initial velocity along the projectile’s parabolic path is considered horizontal velocity. At some point, the object reaches a certain height called maximum height. It is made to travel downwards as it gets attracted by the gravitational force. It is the one and the only force that acts in the projectile motion of any particle.
As soon as the particle moves below, a vertical displacement and velocity occur. Here both the horizontal and vertical velocities are opposite each other, and hence there will be no effect on horizontal velocity due to vertical velocity.

Mention the relationship between launched angle, maximum height and range of horizontal velocity?

The angle of maximum height increases as there will be an increase in horizontal range.

We have already studied no direct relation between horizontal velocity and range. If the projectile thrown at a certain angle reaches its maximum height, let us assume that the length will be comparably maximum, then even the range will be maximum. But as soon as the projectile moves vertically, the horizontal velocity will be zero.

Also Read:

Is Velocity Zero At The Highest Point: When, How, Why, Problems And Facts

January 21, 2024January 13, 2022 by AKSHITA MAPARI

In this article, we will talk about zero velocity, is velocity zero at the highest point, reason, and various facts.

The object accelerates in the upward direction with kinetic energy, its kinetic energy is converted into potential energy and has the highest potential energy once nil kinetic energy remains with the object and thus the velocity of the object becomes zero.

What is zero velocity?

The velocity is the rate of a shift in the position in a given time.

The velocity of the object is zero if the object tends to remain fixed at one position during a certain time or an object returns back to its initial position after displacement.

Position-Time Graph for Zero Velocity

On the position-time graph, the velocity of the object is zero if the slope of a graph is parallel to the x-axis.

The following graph explains the same.

is velocity zero at the highest point — **Position-Time Graph for an object at rest**

The above graph shows at the position of the object remain unchanged all the time. Hence the velocity of the object is zero.

It is also true when the object returns back to the position from where it had started its journey. It is clearly indicated in the below graph.

The above graph shows that the object was displaced from x₁ to x₂, but then returns back to x₁ after a certain time duration. Hence, the change in a position now becomes zero and therefore the velocity is also equal to zero.

Why is Velocity Zero at the Highest Point?

As the object goes higher and higher above the ground surface, it stores the energy in the form of potential energy.

Once all the kinetic energy of the object is converted into the potential energy, the object does not accelerate further but is held there for a few seconds until exerts a force due to gravity which pulls the object towards itself.

The object then returns down towards the ground on converting the potential energy that it has gained into kinetic energy.

Zero Horizontal Velocity of an Object in a projectile motion

An object in a projectile motion moves in two dimensions, x-direction and y-direction.

Initially, when the object starts its journey from its origin, we have both horizontal velocity as well as vertical velocity. Once it attains the highest point where all of its kinetic energy is converted into the potential energy it stops accelerating for a while and continues to move in the horizontal direction rather than the vertical route, and then due to gravity accelerates down.

What happens to the object when its velocity becomes zero at the highest point?

The object is held at rest for the smallest duration of time when its velocity becomes zero.

When the object reaches the highest point in its flight, the kinetic energy of the object becomes zero and has the highest potential energy that it can have.

The variation of the energy of the object is shown in the below diagram.

It is seen that the ball has the maximum kinetic energy on kicking which goes on decreasing as the ball attains height. The decreasing kinetic energy is converted into the potential energy and is the maximum when the ball reaches the highest point.

The velocity of the ball is the maximum during its initial flight.

At the highest point K.E. =0; that is

1/2mv²=0

Since mass is conserved, it is evident that the velocity of the object is zero at this point.

v²=0

The energy of the object is conserved during its flight. The initial kinetic energy is converted to potential energy until the kinetic energy of the object has vanished, and then the potential energy is converted into kinetic energy.

The velocity of the object in a flight above the surface due to gravity is

When the ball has the maximum potential energy it tends to occupy the space at its highest point once all its kinetic energy is converted into potential energy. It is only because the gravitational pull of the Earth exerted on the object, pulls the object down to the surface. Hence, the object observes the freefall and accelerates downward.

Problem1: Calculate the kinetic energy and the potential energy of the object of mass 5 kg hanging above the ground at a height of 100 meters.

Solution: The height of the object from the ground is h=100m.

The mass of the object m=5kg

The potential energy of the object is given by the equation

V=mgh

=5kg x 9.8m/s² x100=4900

=4.9k Joules

The potential energy associated with the object is 4.9kJoules.

The velocity of the objectis zero. Hence, the kinetic energy of the object is

K.E.=1/2mv²=1/2m x 0=0

The kinetic energy of the object is zero.

Problem 2: Consider the same object tied at a height of 100 meters from above the ground that is slightly drifted by the high wind speed. Find the change in potential energy of the object if it displaces its position 10cm in 2 seconds.

Given: Displacement x=10 cms=0.10m, time t=2 seconds

v= Δx/Δt = 0.1/2 = 0.05m/s

The kinetic energy of the object now is

K.E.= 1/2mv²= 1/2 x 5 x (0.05)²

=1/2 x 5 x 0.0025=0.00625 Joules

The potential energy of the object at a height of 100 meters was 4.9kJoules.

When the object sets into motion, the potential energy of the object is converted into kinetic energy.

Thus the potential energy associated with the object after the object displaces 10cm from the above height will be

P.E. = P.E._initial – K.E. = (4.9k-0.00625)J = 4.89k Joules

Read more on the Potential Energy.

Frequently Asked Questions

Is the velocity of the object falling down from the height is negative?

The velocity of the object due to gravity varying with height is given by the formula

As the difference between the final and the initial height becomes negative due to decreasing height, the velocity of the object falling down is negative.

Is the momentum of the object at the highest point conserved?

The object has momentum associated with it when it is moving with a velocity.

When the object is at the highest point above the surface, the object has no kinetic energy associated with it as the velocity of the object is zero; hence the momentum of the object is also zero.

What is a free fall?

The object after attaining the highest point in its flight and gaining the highest potential energy tends to return back to the surface.

The rapid fall and free movement of the object due to the effect of the gravitational force exerted by the Earth is called the free fall of the object.

Why acceleration of the object at the highest point is zero?

The acceleration will be zero if there is no variation in the velocity of the object.

At the highest point, the object has no kinetic energy where all of its kinetic energy is converted into the potential energy, and therefore the velocity is zero and the object does not accelerate.

Also Read:

17 Constant Velocity Example: Detailed Explanation And Facts

January 21, 2024January 13, 2022 by AKSHITA MAPARI

In this article, we will discuss various examples of constant velocity with detailed explanations, and facts.

The following is a list of constant velocity examples:-

Object in a circular motion

An object moving in a circular path of radius ‘r’ elapses a distance of 2πr on each round with angular velocity ω.

Consider an object in a circular motion that covers a distance ‘s’ in a certain time interval ‘t’. Let ‘θ’ be an angle made by the particle displacing from its initial position.

**The motion of an object in a circular motion**

The linear velocity of the particle is a change in the position of the object in time t.

v=s / t

‘s’ is a displacement of a particle which is an arc length and can be calculated as a product of angle made by the particle on displacement and the radius of the circle.

s= θr

On substituting this in the above equation, we have

v=rθ/ t

Since the angular velocity is equal to the change in angle with respect to time; we can rewrite the equation as

v=rω

Where ω is an angular velocity

The angular velocity of the object is constant if the linear velocity of the object remains constant.

Motion of an object due to Centripetal Force

The object in a centripetal motion exerts the centripetal force which is acting towards the center of the circular path and the linear velocity of the object remains perpendicular to the centripetal force.

The centripetal force exerted on the object in a circular motion is given by the equation

F=mv²/r

Where r is the radius of the circular path

V is a linear velocity

M is a mass of the particle

If there is an external force acting on the object that keeps the object in the circular motion the

ma=mv²/r

a=v²/r

v²=ar

Hence, the velocity of the object is constant if the angular acceleration of the object is constant.

Revolution of a moon around the Earth

The velocity of a moon revolving around the Earth is almost at a constant rate. The moon completes one revolution around the Earth in 27.3 days which is equal to T=27.3 x 24 x 60 x 60=23,58,720 seconds

The distance of the moon from the center of the Earth is 3,84,000 km

The distance covered by the moon in one revolution is equal to the circumference of the circular path. Hence, the velocity of the motion is V= 2π r/T

V= 2π x 384000 x 1000/2358720=1022 m/s

The velocity of a moon orbiting around the Earth is 1022m/s and is constant.

A person walking on the street at a constant speed

A man walking on the street at his constant speed will cover an equal distance in an equal interval of time. This can be an example of a constant velocity.

Drawing water from a well with the help of a pulley

While drawing water from a well using a pulley, the force is applied downward but the reaction force is in the upward direction. The velocity of the bucket lifting up depends upon the length of the rope stretched on every pull.

The length of the rope that is stretched depends upon the movement of the arm and the length of the hand. Hence the velocity of the bucket and the angular acceleration of the pulley will be constant.

A ray of light

A light travels in a straight line at a constant speed of 3 x 10⁸ m/s. A light shows various other phenomena in nature like scattering, dispersion, reflection, refraction, total internal refraction, interference, diffraction, etc.

The speed of light is equal to the product of the wavelength of the light and the frequency of the electromagnetic wave as c=fλ.

Speed of the object in a vacuum

If the object falls in the vacuum, it accelerates at a constant speed and experiences a free fall. All the objects in the vacuum will move at the same speed irrespective of their shape, size, density, or weight. The velocity of the object in a motion in a vacuum is constant.

Sound wave

The sound wave travels at a speed of 332m/s at normal temperature and pressure conditions. The speed of sound is determined by the distance traveled by the sound waves in a certain time duration.

The velocity of a sound wave varies depending upon the density of different mediums and sound waves travel at a constant rate.

Clock

The minute hand, an hour hand, and a second hand on the clock move at a constant speed. The point at the center where all the hands are attached resembles an instantaneous central point.

A clock measures angle of 360 degrees and each minute on the clock is equal to 1 degree. A second-hand travels 360 degrees in one minute, hence the speed of the second hand is

A minute hand covers 360 degrees in 1 hour, hence the velocity of the minute hand is

An hour hand displaces 30 degrees in 1 hour, therefore the velocity of an hour hand is

A car traveling on a road at constant speed

A car moving at a constant speed will elapse equal distance in an equal duration of time hence is an example of constant velocity. The velocity of a car is measured as the ratio distance covered by the car from its initial position to reach a certain distance in time ‘t’.

A ball moving on a plane surface

A ball can travel at the same speed unless exerts an external force that makes increases or decreases the speed of the ball to displace its position.

Fan

A fan rotating at a constant speed gives a constant angular velocity until the speed of the fan is changed.

Hourglass

A sand-filled hourglass is dropped down from the hole at a constant velocity.

The hourglass is designed such that the frictional forces due to sand and glass are canceled and the constant pressure is exerted on the hole that makes the sand drop down from the hole. The sand filled in glass drops at a constant rate and is hence used as a timer.

Train

A train is an example of a constant velocity, which elapses the same distance in a certain interval of time.

Electric vehicles without gears

Electric vehicles work on electric energy. A vehicle without gear will move at the same speed and in the front direction only.

Photon

A photon being a light particle is easily carried away and moves with the speed of light. The velocity of a photon is constant.

Birds Flying

The velocity of the birds is mostly constant while they are flying. Most of the birds are observed to sway at constant velocity. Therefore, we can estimate the expected date and time of birds venturing from the far locations.

What is constant velocity?

The velocity of the object is defined as the rate of change of position of an object in a fixed interval of time.

If the distance elapsed by the object in a given time interval will be constant for every time interval then the velocity of the object will be constant.

Hence,

x₂-x₁=Constant

For the velocity to be constant, the change in the position of the object has to be constant along with time.

Read more on Constant Velocity.

Position-Time Graph for constant velocity

The displacement of the object in time is represented in the following position-time graph.

constant velocity example — **Position-Time Graph**

The slope of the position-time graph gives the velocity of the object between the two time intervals while displacing from one position to another. The slope of the graph is linear and is constant throughout the slope.

Frequently Asked Questions

Q1. If an object is traveling at a constant velocity of 12m/s, then calculate the distance covered by the object after 1 minute.

Solution: 1 minute=60 seconds

The distance covered by the object in 60 seconds traveling at a speed of 12m/s will be

= 12m/s\times 60s=720m

Hence, the distance traveled by the object in 1 minute is 720m.

How does a graph of velocity v/s time will looks like for a constant velocity?

An object with constant velocity will travel in a straight line covering an equal distance in a given time interval.

Since the velocity of the object will be constant all the time, the acceleration of the object which is a slope of the graph will be zero. This implies that the slope of the graph will be a straight line.

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