8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals

In this guide, we will look at how to use integrals to find the position, velocity, and acceleration of a function.

To begin, it is important to understand the definitions of position, velocity, and acceleration. Position, or displacement, is the measure of how far an object has moved from a reference point. Velocity is the rate of change of position with respect to time. Acceleration is the rate of change of velocity with respect to time.

To find the position of a function, we need to use the definite integral. The definite integral of velocity with respect to time gives us the change in position, or displacement, over a given interval. The definite integral of velocity with respect to time is represented by the symbol ∫.

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

To find the acceleration of a function, we need to use the derivative of the velocity function. The derivative of the velocity function with respect to time gives us the acceleration of the function.

Example 1: Consider the function s(t) = 3t^2. To find the position of the function, we need to find the definite integral of the velocity function, which is s(t) = ∫(3t^2) dt = t^3 + C. To find the velocity, we need to find the derivative of the position function, which is s'(t) = 3t^2. To find the acceleration, we need to find the derivative of the velocity function, which is s''(t) = 6t.

Example 2: Consider the function v(t) = -2t^3 + 6t. To find the position of the function, we need to find the definite integral of the velocity function, which is s(t) = ∫(-2t^3 + 6t) dt = -2/4 t^4 + 6/2 t^2 + C. To find the velocity, we need to find the derivative of the position function, which is v(t) = -2t^3 + 6t. To find the acceleration, we need to find the derivative of the velocity function, which is v''(t) = -6t.

Example 3: Consider the function a(t) = -9t^2. To find the position of the function, we need to find the definite integral of the velocity function, which is s(t) = ∫(-9t^2) dt = -3/2 t^3 + C. To find the velocity, we need to find the derivative of the position function, which is a(t) = -9t^2. To find the acceleration, we need to find the derivative of the velocity function, which is a''(t) = -18t.

Example 4: Consider a ball being thrown upwards with an initial velocity of 15 m/s. The position function of the ball is given by:

s(t) = 15t - 4.9t^2

To find the velocity function, we need to take the derivative of the position function:

v(t) = ds/dt = 15 - 9.8t

To find the acceleration function, we need to take the derivative of the velocity function:

a(t) = dv/dt = -9.8

Since the acceleration is constant, the ball is under the influence of constant gravity and the velocity function is linear.

Example 5: Consider a car moving along a straight road. The position function of the car is given by:

s(t) = 3t^3 - 12t^2 + 20t

To find the velocity function, we need to take the derivative of the position function:

v(t) = ds/dt = 9t^2 - 24t + 20

To find the acceleration function, we need to take the derivative of the velocity function:

a(t) = dv/dt = 18t - 24

The acceleration function is a quadratic function and has a turning point at t = 1.33, which is the time at which the car changes direction.

Example 6: Consider a roller coaster ride. The position function of the roller coaster is given by:

s(t) = sin(t) + cos(2t)

To find the velocity function, we need to take the derivative of the position function:

v(t) = ds/dt = cos(t) - 2sin(2t)

To find the acceleration function, we need to take the derivative of the velocity function:

a(t) = dv/dt = -sin(t) - 4cos(2t)

The acceleration function is a combination of sine and cosine functions, which indicates that the roller coaster ride is oscillating and changing direction frequently.

Example 7: Consider a rocket launching into space. The position function of the rocket is given by:

s(t) = 10e^(-0.5t)

To find the velocity function, we need to take the derivative of the position function:

v(t) = ds/dt = -5e^(-0.5t)

To find the acceleration function, we need to take the derivative of the velocity function:

a(t) = dv/dt = 2.5e^(-0.5t)

The acceleration function is an exponential function, which indicates that the rocket is accelerating rapidly at the beginning of the launch and slowing down as it reaches space.

Example 8: Consider a person jumping off a diving board. The position function of the person is given by:

s(t) = -4.9t^2 + 10t + 2

To find the velocity function, we need to take the derivative of the position function:

v(t) = ds/dt = -9.8t + 10

To find the acceleration function, we need to take the derivative of the velocity function:

a(t) = dv/dt = -9.8

The acceleration function is a constant function, which indicates that the person is under the influence of constant gravity and the velocity function is linear.

In conclusion, connecting position, velocity, and acceleration of functions using integrals is an important concept in AP Calculus AB that deals with the relationship between position, velocity, and acceleration of a function. By using integrals and derivatives, we can find the position, velocity, and acceleration of a function and understand how they are related.