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.