Learn How to Find Velocity from Position: A Step-by-Step Guide

How to Find Velocity from Position

Velocity is a fundamental concept in physics and mathematics that represents the rate of change of position with respect to time. In other words, it tells us how fast an object is moving and in what direction. In this blog post, we will explore different methods of finding velocity from position, including calculus, position-time function, position-velocity equations, and position vector. So, let’s dive in!

Understanding the Basic Concepts

Before we jump into the methods, let’s clarify a few basic concepts. Velocity is a vector quantity, which means it has both magnitude (speed) and direction. It is often represented by the symbol “v.” On the other hand, position is a vector that represents the location of an object in space. It is usually denoted by the symbol “r” or “s.”

Importance of Velocity in Physics and Mathematics

velocity from position 2

Velocity plays a crucial role in both physics and mathematics. In physics, velocity helps us analyze the motion of objects, understand their behavior, and predict their future positions. It is an essential component of kinematics, which is the branch of physics that deals with the motion of objects. In mathematics, velocity is a key concept in calculus, where it is used to calculate derivatives and integrals, allowing us to solve various mathematical problems.

Calculating Velocity from Position in Calculus

Calculus is a powerful mathematical tool that enables us to find the velocity of an object from its position function. The position function, denoted as “r(t),” represents the position of an object as a function of time. To derive the velocity from the position function, we need to take the derivative of the position function with respect to time.

Deriving Velocity from Position Function in Calculus

Let’s say we have a position function, r(t), which gives us the position of an object at a given time, t. To find the velocity, we take the derivative of the position function with respect to time:

v(t) = \frac{{dr}}{{dt}}

Here, v(t) represents the velocity at time t. By calculating the derivative, we obtain the instantaneous velocity at any specific time.

Worked out Examples

  • Example 1: Suppose the position function of an object is given by r(t) = 2t^2 + 3t + 1. To find the velocity at a particular time, say t = 2, we can differentiate the position function with respect to time:

v(t) = \frac{{dr}}{{dt}} = \frac{{d}}{{dt}}(2t^2 + 3t + 1) = 4t + 3

Substituting t = 2 into the derivative, we find:

v(2) = 4(2) + 3 = 11

Therefore, the velocity at t = 2 is 11 units per second.

  • Example 2: Let’s consider a different position function, r(t) = \sin(t). To calculate the velocity, we differentiate the position function with respect to time:

v(t) = \frac{{dr}}{{dt}} = \frac{{d}}{{dt}}(\sin(t)) = \cos(t)

Thus, the velocity at any time t is given by v(t) = \cos(t).

Finding Velocity from Position-Time Function

how to find velocity from position
Image by NSF – Wikimedia Commons, Wikimedia Commons, Licensed under CC BY 4.0.
how to find velocity from position
Image by NSF – Wikimedia Commons, Wikimedia Commons, Licensed under CC BY 4.0.
velocity from position 3

Another method to determine velocity from position is by using the position-time function. The position-time function, denoted as “s(t),” expresses the position of an object as a function of time. To calculate velocity from the position-time function, we need to find the derivative of the position-time function.

Steps to Calculate Velocity from Position-Time Function

To find the velocity from the position-time function, follow these steps:

  1. Differentiate the position-time function, s(t), with respect to time, t.
  2. The derivative of the position-time function represents the velocity as a function of time, v(t).
  3. Substitute the desired value of time into the velocity function to find the velocity at that specific time.

Worked out Examples

  • Example 1: Consider the position-time function s(t) = 3t^2 + 2t + 5. To find the velocity at a particular time, say t = 2, we differentiate the position-time function:

v(t) = \frac{{ds}}{{dt}} = \frac{{d}}{{dt}}(3t^2 + 2t + 5) = 6t + 2

Substituting t = 2 into the derivative, we find:

v(2) = 6(2) + 2 = 14

Hence, the velocity at t = 2 is 14 units per second.

  • Example 2: Let’s consider a different position-time function, s(t) = \cos(t). To determine the velocity, we differentiate the position-time function with respect to time:

v(t) = \frac{{ds}}{{dt}} = \frac{{d}}{{dt}}(\cos(t)) = -\sin(t)

Therefore, the velocity as a function of time is v(t) = -\sin(t).

Using Position-Velocity Equations

Position-velocity equations provide a direct method to determine velocity from position. These equations describe the relationship between position and velocity of an object. By rearranging these equations, we can solve for velocity.

How to Use Position-Velocity Equations to Determine Velocity

There are several position-velocity equations available, but one of the most commonly used is:

v(t) = \frac{{\Delta x}}{{\Delta t}}

Here, \Delta x represents the change in position and \Delta t represents the change in time. By dividing the change in position by the change in time, we obtain the average velocity over that time interval.

Worked out Examples

  • Example 1: Suppose an object moves a distance of 10 meters in a time interval of 2 seconds. To find the average velocity, we use the position-velocity equation:

v(t) = \frac{{\Delta x}}{{\Delta t}} = \frac{{10\, \text{m}}}{{2\, \text{s}}} = 5\, \text{m/s}

Therefore, the average velocity over the given time interval is 5 meters per second.

  • Example 2: Let’s consider a scenario where a car travels 200 meters in 20 seconds. To find the average velocity, we use the position-velocity equation:

v(t) = \frac{{\Delta x}}{{\Delta t}} = \frac{{200\, \text{m}}}{{20\, \text{s}}} = 10\, \text{m/s}

Hence, the average velocity over the given time interval is 10 meters per second.

Finding Velocity from Position Vector

In some cases, position is represented by a position vector, which is a mathematical object that specifies both magnitude and direction. To find velocity from a position vector, we can take the derivative of the position vector with respect to time.

Calculating Velocity from Position Vector

Let’s consider a position vector \mathbf{r}(t). To calculate velocity, we take the derivative of the position vector with respect to time:

\mathbf{v}(t) = \frac{{d\mathbf{r}}}{{dt}}

Here, \mathbf{v}(t) represents the velocity vector at time t. By differentiating the position vector, we obtain the instantaneous velocity vector.

Worked out Examples

  • Example 1: Suppose a particle’s position vector is given by \mathbf{r}(t) = 2t\mathbf{i} + 3t^2\mathbf{j}, where \mathbf{i} and \mathbf{j} are unit vectors in the x and y directions, respectively. To find the velocity vector at a specific time, say t = 2, we differentiate the position vector with respect to time:

\mathbf{v}(t) = \frac{{d\mathbf{r}}}{{dt}} = \frac{{d}}{{dt}}(2t\mathbf{i} + 3t^2\mathbf{j}) = 2\mathbf{i} + 6t\mathbf{j}

Substituting t = 2 into the derivative, we find:

\mathbf{v}(2) = 2\mathbf{i} + 6(2)\mathbf{j} = 2\mathbf{i} + 12\mathbf{j}

Hence, the velocity vector at t = 2 is 2\mathbf{i} + 12\mathbf{j}.

  • Example 2: Let’s consider a different position vector, \mathbf{r}(t) = \cos(t)\mathbf{i} + \sin(t)\mathbf{j}. To determine the velocity vector, we differentiate the position vector with respect to time:

\mathbf{v}(t) = \frac{{d\mathbf{r}}}{{dt}} = \frac{{d}}{{dt}}(\cos(t)\mathbf{i} + \sin(t)\mathbf{j}) = -\sin(t)\mathbf{i} + \cos(t)\mathbf{j}

Therefore, the velocity vector at any time t is given by -\sin(t)\mathbf{i} + \cos(t)\mathbf{j}.

And there you have it! We have explored different methods of finding velocity from position, including calculus, position-time function, position-velocity equations, and position vector. Understanding how to find velocity from position is crucial for analyzing motion and solving various physics and mathematical problems. So go ahead, apply these methods, and unlock the fascinating world of velocity!

Numerical Problems on how to find velocity from position

Problem 1:

A particle moves in a straight line with a position function given by (s(t) = 2t^2 – 3t + 5) where (s) represents the position of the particle at time (t).

Find the velocity of the particle at time (t = 2) seconds.

Solution:

To find the velocity of the particle, we need to differentiate the position function with respect to time. The derivative of the position function gives us the velocity function.

Given position function: (s(t) = 2t^2 – 3t + 5)

Differentiating with respect to time:

(\frac{ds}{dt} = \frac{d}{dt}(2t^2 – 3t + 5))

Using the power rule of differentiation:

(\frac{ds}{dt} = 4t – 3)

Now, substituting (t = 2) into the velocity function:

(v = \frac{ds}{dt}\bigg|_{t=2})

(v = 4(2) – 3)

(v = 8 – 3)

(v = 5) m/s

Therefore, the velocity of the particle at (t = 2) seconds is (5) m/s.

Problem 2:

A car accelerates uniformly from rest and reaches a position of (s = 100) m in (t = 5) seconds. Find the velocity of the car at (t = 5) seconds.

Solution:

Given that the car accelerates uniformly, we can use the equation of motion:

(s = ut + \frac{1}{2}at^2)

Where:
– (s) is the position of the car,
– (u) is the initial velocity of the car (which is zero in this case since the car starts from rest),
– (a) is the acceleration of the car, and
– (t) is the time in seconds.

Since the car starts from rest, (u = 0). The equation simplifies to:

(s = \frac{1}{2}at^2)

Substituting the given values:

(100 = \frac{1}{2}a(5^2))

Simplifying:

(100 = \frac{25}{2}a)

(a = \frac{100 \times 2}{25})

(a = 8) m/s²

Now, to find the velocity at (t = 5) seconds, we can use the equation:

(v = u + at)

Since (u = 0) (starting from rest), the equation becomes:

(v = at)

Substituting the values:

(v = 8 \times 5)

(v = 40) m/s

Therefore, the velocity of the car at (t = 5) seconds is (40) m/s.

Problem 3:

velocity from position 1

A ball is thrown vertically upwards with an initial velocity of (20) m/s. The equation of its position function is given by (s(t) = 20t – 5t^2). Find the velocity of the ball at its maximum height.

Solution:

To find the velocity at the maximum height, we need to determine the time when the ball reaches its peak. At the maximum height, the velocity of the ball is zero.

Given position function: (s(t) = 20t – 5t^2)

To find the time at the maximum height, we differentiate the position function and set it equal to zero:

(\frac{ds}{dt} = 20 – 10t = 0)

Solving for (t):

(10t = 20)

(t = 2) seconds

At (t = 2) seconds, the ball reaches its maximum height.

To find the velocity at the maximum height, we differentiate the position function with respect to time:

(\frac{ds}{dt} = \frac{d}{dt}(20t – 5t^2))

Using the power rule of differentiation:

(\frac{ds}{dt} = 20 – 10t)

Substituting (t = 2) into the velocity function:

(v = \frac{ds}{dt}\bigg|_{t=2})

(v = 20 – 10(2))

(v = 20 – 20)

(v = 0) m/s

Therefore, the velocity of the ball at its maximum height is (0) m/s.

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