How to Find Velocity from Displacement: A Simple Guide

In physics, velocity represents the rate of change of displacement over time. It is a crucial concept in understanding the motion of objects. In this blog post, we will explore different methods to find velocity from displacement. We will cover various scenarios, including situations where time, displacement vector, displacement function, and acceleration are given. So, let’s dive in and learn how to calculate velocity from displacement!

How to Calculate Velocity from Displacement and Time

The Formula for Velocity

To calculate velocity from displacement and time, we use the formula:

$v = \frac{\Delta x}{\Delta t}$

Where:
– $v$ represents velocity,
– $\Delta x$ represents the change in displacement, and
– $\Delta t$ represents the change in time.

Step-by-Step Guide to Calculate Velocity

Determine the initial displacement ( $x_1$ ) and final displacement ( $x_2$ ).
Calculate the change in displacement ( $\Delta x = x_2 - x_1$ ).
Determine the initial time ( $t_1$ ) and final time ( $t_2$ ).
Calculate the change in time ( $\Delta t = t_2 - t_1$ ).
Substitute the values of $\Delta x$ and $\Delta t$ into the velocity formula: $v = \frac{\Delta x}{\Delta t}$ .
Calculate the value of velocity.

Worked Out Examples

Let’s work through a couple of examples to solidify our understanding.

Example 1:
Suppose a car travels from point A to point B, with an initial displacement of 10 meters and a final displacement of 50 meters. The time taken to travel from point A to point B is 5 seconds. Calculate the velocity.

Solution:
– $x_1 = 10m$
– $x_2 = 50m$
– $\Delta x = 50m - 10m = 40m$
– $t_1 = 0s$
– $t_2 = 5s$
– $\Delta t = 5s - 0s = 5s$

Substituting the values into the velocity formula:
$v = \frac{40m}{5s} = 8m/s$

Therefore, the velocity of the car is 8 m/s.

Example 2:
Consider a person running on a circular track. The person completes one lap, covering a displacement of 400 meters in 60 seconds. Calculate the velocity.

Solution:
– $x_1 = 0m$ (starting point)
– $x_2 = 400m$ (ending point)
– $\Delta x = 400m - 0m = 400m$
– $t_1 = 0s$
– $t_2 = 60s$
– $\Delta t = 60s - 0s = 60s$

Substituting the values into the velocity formula:
$v = \frac{400m}{60s} = 6.67m/s$

Therefore, the velocity of the person running is approximately 6.67 m/s.

How to Determine Velocity from Displacement without Time

Understanding the Concept of Instantaneous Velocity

In some scenarios, we may only have information about displacement and not the time. To determine the velocity from displacement alone, we need to understand the concept of instantaneous velocity.

Instantaneous velocity is defined as the velocity of an object at a specific point in time. It represents the object’s velocity at an infinitesimally small time interval. It can be calculated by taking the derivative of the displacement function with respect to time.

Calculating Instantaneous Velocity from Displacement

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

To calculate the instantaneous velocity from displacement, follow these steps:

Determine the displacement function $x(t)$ .
Differentiate the displacement function with respect to time ( $\frac{dx}{dt}$ ).
Substitute the value of time at a specific point to get the instantaneous velocity.

Practical Examples

Let’s consider a couple of practical examples to illustrate the calculation of instantaneous velocity.

Example 1:
A motorboat starts from rest and accelerates uniformly for 10 seconds. The displacement function of the motorboat is given by $x(t) = 2t^2$ (where $x$ is in meters and $t$ is in seconds). Find the instantaneous velocity of the motorboat at $t = 5s$ .

Solution:
– Given displacement function: $x(t) = 2t^2$
– Differentiating the displacement function: $\frac{dx}{dt} = 4t$
– Substituting $t = 5s$ into $\frac{dx}{dt}$ :
$\frac{dx}{dt} = 4(5) = 20m/s$

Therefore, the instantaneous velocity of the motorboat at $t = 5s$ is 20 m/s.

Example 2:
A particle moves along a straight line. Its displacement as a function of time is given by $x(t) = t^3 - 2t^2 + 5$ (where $x$ is in meters and $t$ is in seconds). Find the instantaneous velocity of the particle at $t = 2s$ .

Solution:
– Given displacement function: $x(t) = t^3 - 2t^2 + 5$
– Differentiating the displacement function: $\frac{dx}{dt} = 3t^2 - 4t$
– Substituting $t = 2s$ into $\frac{dx}{dt}$ :
$\frac{dx}{dt} = 3(2)^2 - 4(2) = 8m/s$

Therefore, the instantaneous velocity of the particle at $t = 2s$ is 8 m/s.

How to Find Velocity from Displacement Vector

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

Understanding Displacement Vectors

In physics, displacement can be represented as a vector quantity. A displacement vector includes both magnitude (length) and direction. To find velocity from a displacement vector, we need to consider the change in displacement and the time taken.

Calculating Velocity from Displacement Vector

To calculate velocity from a displacement vector, follow these steps:

Determine the initial displacement vector ( $\vec{r}_1$ ) and final displacement vector ( $\vec{r}_2$ ).
Calculate the change in displacement vector ( $\Delta \vec{r} = \vec{r}_2 - \vec{r}_1$ ).
Determine the time taken ( $\Delta t$ ).
Substitute the values of $\Delta \vec{r}$ and $\Delta t$ into the formula: $\vec{v} = \frac{\Delta \vec{r}}{\Delta t}$ .
Calculate the value of velocity vector.

Examples for Better Understanding

Let’s consider a couple of examples to illustrate the calculation of velocity from a displacement vector.

Example 1:
A car initially at position $\vec{r}_1 = 2\hat{i} + 3\hat{j}$ moves to a final position $\vec{r}_2 = 8\hat{i} + 10\hat{j}$ . The time taken to move from the initial position to the final position is 4 seconds. Calculate the velocity.

Solution:
– Initial displacement vector: $\vec{r}_1 = 2\hat{i} + 3\hat{j}$
– Final displacement vector: $\vec{r}_2 = 8\hat{i} + 10\hat{j}$
– Change in displacement vector: $\Delta \vec{r} = \vec{r}_2 - \vec{r}_1 = (8\hat{i} + 10\hat{j}) - (2\hat{i} + 3\hat{j}) = 6\hat{i} + 7\hat{j}$
– Time taken: $\Delta t = 4s$

Substituting the values into the velocity formula:
$\vec{v} = \frac{6\hat{i} + 7\hat{j}}{4s} = \frac{3}{2}\hat{i} + \frac{7}{4}\hat{j}$

Therefore, the velocity of the car is $\frac{3}{2}\hat{i} + \frac{7}{4}\hat{j}$ .

Example 2:
Consider a particle moving along a curved path. The initial displacement vector is $\vec{r}_1 = 2\hat{i} - 3\hat{j}$ , and the final displacement vector is $\vec{r}_2 = -4\hat{i} + 5\hat{j}$ . The time taken to move from the initial position to the final position is 3 seconds. Calculate the velocity.

Solution:
– Initial displacement vector: $\vec{r}_1 = 2\hat{i} - 3\hat{j}$
– Final displacement vector: $\vec{r}_2 = -4\hat{i} + 5\hat{j}$
– Change in displacement vector: $\Delta \vec{r} = \vec{r}_2 - \vec{r}_1 = (-4\hat{i} + 5\hat{j}) - (2\hat{i} - 3\hat{j}) = -6\hat{i} + 8\hat{j}$
– Time taken: $\Delta t = 3s$

Substituting the values into the velocity formula:
$\vec{v} = \frac{-6\hat{i} + 8\hat{j}}{3s} = -2\hat{i} + \frac{8}{3}\hat{j}$

Therefore, the velocity of the particle is $-2\hat{i} + \frac{8}{3}\hat{j}$ .

How to Measure Velocity from Displacement Function

Understanding Displacement Function

In some cases, we may have a displacement function that describes the motion of an object. The displacement function provides the position of an object at any given time. To measure velocity from a displacement function, we need to differentiate the function with respect to time.

Calculating Velocity from Displacement Function

To calculate velocity from a displacement function, follow these steps:

Determine the displacement function $x(t)$ .
Differentiate the displacement function with respect to time ( $\frac{dx}{dt}$ ).
The derivative gives the velocity function $v(t)$ .

Worked Out Examples

Let’s work through a couple of examples to understand how to measure velocity from a displacement function.

Example 1:
Consider an object moving along a straight line. Its displacement function is given by $x(t) = 2t^2 - 4t + 7$ , where $x$ is in meters and $t$ is in seconds. Calculate the velocity from the displacement function.

Solution:
– Given displacement function: $x(t) = 2t^2 - 4t + 7$
– Differentiating the displacement function: $\frac{dx}{dt} = 4t - 4$
– The derivative gives the velocity function: $v(t) = 4t - 4$

Therefore, the velocity from the displacement function is $v(t) = 4t - 4$ .

Example 2:
Suppose an object is moving in a circular path. Its displacement function is given by $x(t) = R\cos(t)$ , where $R$ represents the radius of the circular path. Calculate the velocity from the displacement function.

Solution:
– Given displacement function: $x(t) = R\cos(t)$
– Differentiating the displacement function: $\frac{dx}{dt} = -R\sin(t)$
– The derivative gives the velocity function: $v(t) = -R\sin(t)$

Therefore, the velocity from the displacement function is $v(t) = -R\sin(t)$ .

How to Find Velocity from Displacement and Acceleration

Understanding the Role of Acceleration in Velocity

Acceleration plays a crucial role in determining the change in velocity over time. If we have information about displacement and acceleration, we can find velocity by integrating the acceleration function with respect to time.

Calculating Velocity from Displacement and Acceleration

To calculate velocity from displacement and acceleration, follow these steps:

Determine the displacement function $x(t)$ .
Differentiate the displacement function to get the velocity function $v(t)$ .
Determine the acceleration function $a(t)$ .
Integrate the acceleration function with respect to time to get the velocity function.

Practical Examples

Let’s consider a couple of practical examples to understand how to find velocity from displacement and acceleration.

Example 1:A motorboat starts from rest and experiences a constant acceleration given by , where is in seconds. Find the velocity of the motorboat after 5 seconds.

Solution:
– Given acceleration function: $a(t) = 2t$
– Integrating the acceleration function to get the velocity function:
$\int a(t) \,dt = \int 2t \,dt = t^2 + C$
– Substituting $t = 5s$ into the velocity function:
$v(t) = (5s)^2 + C$

Therefore, the velocity of the motorboat after 5 seconds is $v(5s) = 25s^2 + C$ .

Example 2:
Consider an object moving along a straight line. Its acceleration function is given by $a(t) = 3t^2 - 6t + 4$ , where $t$ is in seconds. Find the velocity of the object at $t = 2s$ .

Solution:
– Given acceleration function: $a(t) = 3t^2 - 6t + 4$
– Integrating the acceleration function to get the velocity function:
$\int a(t) \,dt = \int (3t^2 - 6t + 4) \,dt$
$v(t) = t^3 - 3t^2 + 4t + C$
– Substituting $t = 2s$ into the velocity function:
$v(2s) = (2s)^3 - 3(2s)^2 + 4(2s) + C$

Therefore, the velocity of the object at $t = 2s$ is $v(2s) = 8s^3 - 12s^2 + 8s + C$ .

By following the methods explained above, you can find the velocity from displacement, even when acceleration is involved.

Calculating velocity from displacement is an essential skill in physics. Whether you have information about time, displacement vectors, displacement functions, or acceleration, there are specific techniques to determine velocity accurately. By utilizing formulas, differentiation, integration, and working through examples, you can confidently find velocity from displacement in various scenarios. Keep practicing and exploring different situations to strengthen your understanding of this fundamental concept in physics. Happy calculating!

Numerical Problems on how to find velocity from displacement

Problem 1:

A particle moves along the x-axis with a displacement given by the equation: $x(t) = 2t^2 + 3t + 1$ , where $x$ is the displacement and $t$ is the time. Find the velocity of the particle at time $t = 2$ seconds.

Solution:
To find the velocity, we need to differentiate the displacement equation with respect to time. Let’s take the derivative of $x(t)$ .

$\begin{align}v(t) &= \frac{dx}{dt} \&= \frac{d}{dt}(2t^2 + 3t + 1) \&= 4t + 3\end{align}$

Now, substitute $t = 2$ into the velocity equation to find the velocity at $t = 2$ seconds.

$\begin{align}v(2) &= 4(2) + 3 \&= 8 + 3 \&= 11 \, \text{m/s}\end{align}$

Therefore, the velocity of the particle at $t = 2$ seconds is $11$ m/s.

Problem 2:

A car accelerates uniformly from rest and covers a distance of 120 meters in 10 seconds. Find the velocity of the car at the end of the 10-second interval.

Solution:
When a car accelerates uniformly, the equation for displacement is given by:

$\begin{align}x(t) &= \frac{1}{2}at^2\end{align}$

where $x$ is the displacement, $a$ is the acceleration, and $t$ is the time.

In this case, we are given the displacement $x = 120$ meters and the time $t = 10$ seconds. Since the car starts from rest, the initial velocity is $u = 0$ .

We can rearrange the equation to solve for acceleration:

$\begin{align}a &= \frac{2x}{t^2}\end{align}$

Substituting the given values:

$\begin{align}a &= \frac{2(120)}{(10)^2} \&= \frac{240}{100} \&= 2.4 \, \text{m/s}^2\end{align}$

Now, let’s find the final velocity using the equation:

$\begin{align}v &= u + at\end{align}$

Substituting $u = 0$ , $a = 2.4$ m/s^2, and $t = 10$ s:

$\begin{align}v &= 0 + (2.4)(10) \&= 24 \, \text{m/s}\end{align}$

Therefore, the velocity of the car at the end of the 10-second interval is $24$ m/s.

Problem 3:

A ball is thrown vertically upwards with an initial velocity of 20 m/s. The equation for the height of the ball above the ground is given by: $h(t) = 20t - 5t^2$ , where $h$ is the height and $t$ is the time. Find the velocity of the ball when it reaches its highest point.

Solution:
To find the velocity of the ball when it reaches its highest point, we need to determine the time at which the height is maximum. The maximum height is reached when the velocity becomes zero.

The equation for velocity is the derivative of the height equation:

$\begin{align}v(t) &= \frac{dh}{dt} \&= \frac{d}{dt}(20t - 5t^2) \&= 20 - 10t\end{align}$

We set the velocity equation equal to zero and solve for $t$ :

$\begin{align}20 - 10t &= 0 \10t &= 20 \t &= 2 \, \text{s}\end{align}$

So, the ball reaches its highest point at $t = 2$ seconds.

Now, to find the velocity at $t = 2$ seconds, we substitute $t = 2$ into the velocity equation:

$\begin{align}v(2) &= 20 - 10(2) \&= 20 - 20 \&= 0 \, \text{m/s}\end{align}$

Therefore, the velocity of the ball when it reaches its highest point is $quicklatex.com 58b01ef4d5f29f603c93f7926a610a01 l3$ m/s.

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