Table of Contents

1. Velocity as an Integral
   1. Definition
   2. Total Change in Velocity
   3. Velocity at a Particular Point
   4. Velocity Expression over Time
   5. Indefinite Integral Approach
   6. Exam Tip
   7. Worked Example
2. Position as an Integral
   1. Definition
   2. Total Change in Displacement
   3. Displacement at a Particular Point
   4. Displacement Expression over Time
   5. Indefinite Integral Approach
   6. Exam Tip
   7. Worked Example
3. Glossary
4. Summary and Key Takeaways
5. Practice Questions

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Velocity as an Integral

Definition

$$v(t) = \int a(t) , dt$$

This follows because acceleration is the derivative of velocity:

$$\frac{d}{dt} v(t) = a(t)$$

and differentiation and integration are inverse operations.

Total Change in Velocity

The definite integral

$$\int_{t_1}^{t_2} a(t) , dt$$

represents the total change in velocity between $t = t_1$ and $t = t_2$.

• Acceleration is the rate of change of velocity.
• Thus, $a(t) \cdot \Delta t$ is the change in velocity over a small time interval $\Delta t$.
• $a(t) , dt$ is the limit of this change as $\Delta t \to 0$.
• The integral $\int_{t_1}^{t_2} a(t) , dt$ represents the total change in velocity.

You can also think of this as the integral calculating the area under an acceleration-time graph. The change in velocity is equal to this area.

Velocity at a Particular Point

To find the velocity at a particular point in time, you need to:

1. Determine the change in velocity between times $t = t_1$ and $t = t_2$.
2. Add this change to the velocity at time $t = t_1$.

$$v(t_2) = v(t_1) + \int_{t_1}^{t_2} a(t) , dt$$

Velocity Expression over Time

To find an expression for the velocity at any point in time, you need to:

1. Determine the change in velocity between a time $t = t_0$ and any other time $t$.
2. Add this change to the velocity at $t = t_0$.

$$v(t) = v(t_0) + \int_{t_0}^{t} a(w) , dw$$

$w$ here is simply a dummy variable used for the integration.

Indefinite Integral Approach

Alternatively, find the indefinite integral

$$\int a(t) , dt$$

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.

Worked Example

Example 1

The acceleration of a particle for 0 \le t \le 60 seconds is given by the function $a(t) = \frac{1}{4}\sqrt{t}$, 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 the change in velocity, we can use the definite integral of the acceleration:

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

Evaluate the integral; factoring out the constant can help:

$$\frac{1}{4} \int_{10}^{20} t^{1/2} , dt$$

Using the power rule for integration:

$$\frac{1}{4} \left[ \frac{2}{3} t^{3/2} \right]_{10}^{20}$$

Calculate the definite integral:

$$\frac{1}{4} \left( \frac{2}{3} (20)^{3/2} - \frac{2}{3} (10)^{3/2} \right) = 9.637$$

(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 $t = 45$:

$$\int_{3}^{45} \frac{1}{4}\sqrt{t} , dt$$

Using the power rule for integration:

$$\frac{1}{4} \left[ \frac{2}{3} t^{3/2} \right]_{3}^{45}$$

Calculate the definite integral:

$$\frac{1}{4} \left( \frac{2}{3} (45)^{3/2} - \frac{2}{3} (3)^{3/2} \right) = 49.446$$

Add this change to the initial velocity at $t = 3$ seconds:

$$2 + 49.446 = 51.446$$

Position as an Integral

Definition

$$s(t) = \int v(t) , dt$$

This follows because velocity is the derivative of displacement:

$$\frac{d}{dt} s(t) = v(t)$$

and differentiation and integration are inverse operations.

Total Change in Displacement

The definite integral

$$\int_{t_1}^{t_2} v(t) , dt$$

represents the total change in displacement between $t = t_1$ and $t = t_2$.

• Velocity is the rate of change of displacement.
• Thus, $v(t) \cdot \Delta t$ is the change in displacement over a small time interval $\Delta t$.
• $v(t) , dt$ is the limit of this change as $\Delta t \to 0$.
• The integral $\int_{t_1}^{t_2} v(t) , dt$ represents the total change in displacement.

You can also think of this as the integral calculating the area under a velocity-time graph. The change in displacement is equal to this area.

Displacement at a Particular Point

To find the displacement at a particular point in time, you need to:

1. Determine the change in displacement between times $t = t_1$ and $t = t_2$.
2. Add this change to the displacement at time $t = t_1$.

$$s(t_2) = s(t_1) + \int_{t_1}^{t_2} v(t) , dt$$

Displacement Expression over Time

To find an expression for the displacement at any point in time, you need to:

1. Determine the change in displacement between a time $t = t_0$ and any other time $t$.
2. Add this change to the displacement at $t = t_0$.

$$s(t) = s(t_0) + \int_{t_0}^{t} v(w) , dw$$

$w$ here is simply a dummy variable used for the integration.

Indefinite Integral Approach

Alternatively, find the indefinite integral

$$\int v(t) , dt$$

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.

Worked Example

Example 2

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

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

$t$ is measured in seconds.

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

Answer:

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

The total change in displacement is found by integrating the velocity:

$$\int_{3}^{10} (12t - \frac{1}{20} t^3) , dt$$

Using the power rule for integration:

$$\left[ 6t^2 - \frac{1}{80} t^4 \right]_{3}^{10}$$

Calculate the definite integral:

$$6(10)^2 - \frac{1}{80}(10)^4 - \left( 6(3)^2 - \frac{1}{80}(3)^4 \right) = 422.013$$

Add this to the displacement at $t = 3$:

$$422.013 + 40 = 462.013$$

(b) The particle starts its motion at $t = 0$ seconds with a displacement of zero feet. Find the length of time it takes for the particle to return to 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.

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

$$\int_{0}^{T} (12t - \frac{1}{20} t^3) , dt = 0$$

Integrate and substitute in the limits:

$$\left[ 6t^2 - \frac{1}{80} t^4 \right]_{0}^{T} = 0$$

Factorize and solve:

$$6T^2 - \frac{1}{80} T^4 = 0$$

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

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

$$T = 0$$ or $$T = \sqrt{480} = 4\sqrt{30} \approx 21.909$$

Glossary

• Velocity: The rate of change of displacement with respect to time.
• Acceleration: The rate of change of velocity with respect to time.
• Displacement: The overall change in position of an object.
• Definite Integral: An integral evaluated over a specific interval, representing the total accumulation of a quantity.
• Indefinite Integral: An integral without specified limits, representing a family of functions and includes a constant of integration.
• Constant of Integration: The arbitrary constant added to the function after integrating, denoted as $C$.

Summary and Key Takeaways

• Velocity is the integral of acceleration with respect to time.
• The definite integral of acceleration represents the total change in velocity over a time interval.
• To find the velocity at a specific point, add the change in velocity to the initial velocity.
• Displacement is the integral of velocity with respect to time.
• The definite integral of velocity represents the total change in displacement over a time interval.
• To find the displacement at a specific point, add the change in displacement to the initial displacement.
• Use the indefinite integral to find general expressions for velocity and displacement, and apply initial conditions to determine constants of integration.

Practice Questions

Practice Question

1. Given the acceleration function $a(t) = 2t + 1$, find the velocity function $v(t)$ if the initial velocity at $t = 0$ is 3 m/s.

Practice Question

2. A particle has a velocity function $v(t) = 3t^2 - 2t + 1$. Calculate the total change in displacement between $t = 1$ and $t = 4$.

Practice Question

3. If a particle starts from rest and accelerates according to the function $a(t) = 4$, find its velocity function and its displacement function.

Practice Question

4. The velocity of a car is given by $v(t) = 5t^3 - t^2 + 2t$. Determine the displacement of the car after 3 seconds if it starts from rest.

Practice Question

5. A particle's acceleration is given by $a(t) = -6t$. If the initial velocity is 12 m/s, find the time when the particle comes to rest.