Position, Velocity & Acceleration (College Board AP® Calculus AB)

Revision Note

Author

Jamie Wood

Expertise

Maths

Velocity as an integral

How is velocity defined as an integral?

Velocity is the integral of acceleration with respect to time
- $v (t) = \int a (t) d t$
  - This follows because acceleration is the derivative of velocity, $\frac{d}{d t} v (t) = a (t)$
  - and differentiation and integration are inverse operations
The definite integral $\int_{t_{1}}^{t_{2}} a (t) d t$ represents the total change in velocity between $t = t_{1}$ and $t = t_{2}$
- Acceleration is the rate of change of velocity
  - Therefore $a (t) \cdot Δ t$ is the change in velocity over a small time interval $Δ t$
  - $a (t) d t$ is the limit of this change as $∆ t \to 0$
  - The integral $\int_{t_{1}}^{t_{2}} a (t) d t$ sums up all these infinitesimal changes between $t = t_{1}$ and $t = t_{2}$
- You may also think of this as the integral calculating the area under an acceleration-time graph
  - The change in velocity is equal to this area
To find the velocity at a particular point in time, you need to find
- the change in velocity between times $t = t_{1}$ and $t = t_{2}$
- then add this on to the velocity at time $t = t_{1}$
  - $v (t_{2}) = v (t_{1}) + \int_{t_{1}}^{t_{2}} a (t) d t$
To find an expression for the velocity at any point in time, you need to find
- the change in velocity between a time $t = t_{0}$ and any other time $t$
- then add this on to the velocity at $t = t_{0}$
  - $v (t) = v (t_{0}) + \int_{t_{0}}^{t} a (w) d w$
  - $w$ here is simply a dummy variable used for the integration
Alternatively, find the indefinite integral $\int a (t) d t$
- This produces an expression for the velocity, including a constant of integration, $+ C$
- Use information in the question about the velocity at a particular point in time to find the value of $C$
- This also gives you an expression describing the velocity at any point in time

Exam Tip

Remember that $\int_{t_{1}}^{t_{2}} a (t) d t$ represents the total change in velocity over a period of time, not the final velocity - this is a common error!

Worked Example

The acceleration of a particle for $0 \leq t \leq 60$ seconds is given by the function $a$ defined by $a (t) = \frac{1}{4} \sqrt{t}$ , where $a (t)$ is measured in meters per second squared.

(a) Find the total change in velocity of the particle between $t = 10$ and $t = 20$ .

Answer:

To find a change in velocity, we can use a definite integral of the acceleration

$\int_{10}^{20} \frac{1}{4} \sqrt{t} d t$

Evaluate the integral; factoring out the constant can help

For a question like this, it is likely you could use your calculator to find this integral

$\begin{array}{rcl} \int_{10}^{20} \frac{1}{4} \sqrt{t} d t & = & \frac{1}{4} \int_{10}^{20} t^{\frac{1}{2}} d t \\ = & \frac{1}{4} {[\frac{2}{3} t^{\frac{3}{2}}]}_{10}^{20} \\ = & \frac{1}{4} ([\frac{2}{3} {(20)}^{\frac{3}{2}}] - [\frac{2}{3} {(10)}^{\frac{3}{2}}]) \\ = & 9.636657 . . . \end{array}$

Round the answer to 3 decimal places and state appropriate units

The question asks for a change so you should state if it is an increase or decrease

Increase of 9.637 meters per second

(b) Given that the particle has a velocity of 2 meters per second at time $t = 3$ seconds, find the velocity of the particle at $t = 45$ seconds.

Answer:

We need to find the change in velocity between $t = 3$ and $t = 45$ , and add this on to the "starting" velocity of 2 meters per second at $t = 3$

For a question like this, it is likely you could use your calculator to find this integral

Change in velocity:

$\begin{array}{rcl} \int_{3}^{45} \frac{1}{4} \sqrt{t} d t & = & \frac{1}{4} \int_{3}^{45} t^{\frac{1}{2}} d t \\ = & \frac{1}{4} {[\frac{2}{3} t^{\frac{3}{2}}]}_{3}^{45} \\ = & \frac{1}{4} (\frac{2}{3} {(45)}^{\frac{3}{2}} - (\frac{2}{3} {(3)}^{\frac{3}{2}})) \\ = & 49.445504 . . . \end{array}$

This is the change in velocity between $t = 3$ and $t = 45$ , so we need to add on the starting value at $t = 3$ , which is 2 meters per second

$2 + 49.445504 . . . = 51.445507 . . .$

Round the answer to 3 decimal places and state appropriate units

51.446 meters per second

Position as an integral

How is position defined as an integral?

Position, or displacement, is the integral of velocity with respect to time
- $s (t) = \int v (t) d t$
  - This follows because velocity is the derivative of displacement, $\frac{d}{d t} s (t) = v (t)$
  - and differentiation and integration are inverse operations
The definite integral $\int_{t_{1}}^{t_{2}} v (t) d t$ represents the total change in displacement between $t = t_{1}$ and $t = t_{2}$
- Velocity is the rate of change of displacement
  - Therefore $v (t) \cdot Δ t$ is the change in velocity over a small time interval $Δ t$
  - $v (t) d t$ is the limit of this change as $∆ t \to 0$
  - The integral $\int_{t_{1}}^{t_{2}} v (t) d t$ sums up all these infinitesimal changes between $t = t_{1}$ and $t = t_{2}$
- You may also think of this as the integral calculating the area under a velocity-time graph
  - The change in displacement is equal to this area
To find the displacement at a particular point in time, you need to find
- the change in displacement between times $t = t_{1}$ and $t = t_{2}$
- then add this on to the displacement at time $t = t_{1}$
  - $s (t_{2}) = s (t_{1}) + \int_{t_{1}}^{t_{2}} v (t) d t$
To find an expression for the displacement at any point in time, you need to find
- the change in displacement between a time $t = t_{0}$ and any other time $t$
- then add this on to the displacement at $t = t_{0}$
  - $s (t) = s (t_{0}) + \int_{t_{0}}^{t} v (w) d w$
  - $w$ here is simply a dummy variable used for the integration
Alternatively, find the indefinite integral $\int v (t) d t$
- This produces an expression for the displacement, including a constant of integration, $+ C$
- Use information in the question about the displacement or position at a particular point in time to find the value of $C$
- This also gives you an expression describing the displacement at any point in time

Exam Tip

Remember that $\int_{t_{1}}^{t_{2}} v (t) d t$ represents the total change in displacement over a period of time

It is not the final displacement or position
It is also not the total distance traveled
Both of these are common errors!

Worked Example

A particle moves along the $x$ -axis with a velocity described by the function

$v (t) = 12 t - \frac{1}{20} t^{3}$ for $0 \leq t \leq 24$

$v$ is measured in feet per second and $t$ is measured in seconds.

(a) Given that at time $t = 3$ the particle is at a displacement of 40 feet from the origin, find the displacement of the particle from the origin at time $t = 10$ .

Answer:

We need to find the total change in displacement from $t = 3$ to $t = 10$ , and add this on to 40 feet, which was the displacement at $t = 3$

The total change in displacement is found by integrating the velocity

For a question like this, it is likely you could use your calculator to find this integral

$\begin{array}{rcl} \int_{3}^{10} 12 t - \frac{1}{20} t^{3} d t & = & {[6 t^{2} - \frac{1}{80} t^{4}]}_{3}^{10} \\ = & 6 {(10)}^{2} - \frac{1}{80} (10^{4}) - (6 {(3)}^{2} - \frac{1}{80} (3^{4})) \\ = & 422.0125 \end{array}$

Add this on to the displacement at $t = 3$ and round to 3 decimal places

$422.0125 + 40 = 462.0125$

At $t = 10$ , the particle will be 462.013 feet from the origin

(b) The particle starts it's motion at $t = 0$ , with a displacement of zero feet.

Find the length of time it takes for the particle to return the same position that it started in.

Answer:

When the particle is back in the same place it started, its displacement will be zero again

You could think of this as being a total change of zero from the starting point

Because $s (0) = 0$ , the definite integral of the velocity starting at $t = 0$ will be equal to the total displacement

Use this fact, and solve for an unknown upper limit, $T$

$s = \int_{0}^{T} 12 t - \frac{1}{20} t^{3} d t = 0$

Integrate and substitute in the limits

$\begin{array}{rcl} {[6 t^{2} - \frac{1}{80} t^{4}]}_{0}^{T} & = & 0 \\ [6 T^{2} - \frac{1}{80} T^{4}] - [0 - 0] & = & 0 \\ 6 T^{2} - \frac{1}{80} T^{4} & = & 0 \end{array}$

Factorize and solve

$T^{2} (6 - \frac{1}{80} T^{2}) = 0$

$T^{2} = 0$ or $6 - \frac{1}{80} T^{2} = 0$

$T = 0$ or $T = \sqrt{480} = 4 \sqrt{30} = 21.908902 . . .$

$T = 0$ corresponds to the start of the motion when the displacement was also 0

Round the answer to 3 decimal places

It takes 21.909 seconds for the particle to return to its starting place

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Position, Velocity & Acceleration (College Board AP® Calculus AB)

Revision Note

Velocity as an integral

How is velocity defined as an integral?

Exam Tip

Worked Example

Position as an integral

How is position defined as an integral?

Exam Tip

Worked Example

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Author: Jamie Wood