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Chapter 4: Problem 89

Given the following acceleration functions of an object moving along a line, find the position function with the given initial velocity and position. $$a(t)=-32 ; v(0)=20, s(0)=0$$

Short Answer

Expert verified
Answer: The position function of the object moving along a line is $$s(t) = -16t^2 + 20t$$.

Step by step solution

01

Write down the given information

We are given: - Acceleration function: $$a(t) = -32$$ - Initial velocity: $$v(0) = 20$$ - Initial position: $$s(0) = 0$$
02

Integrate the acceleration function to find the velocity function

We need to integrate the acceleration function with respect to time to find the velocity function: $$v(t) = \int a(t) dt = \int(-32) dt = -32t + C_1$$ where $$C_1$$ is the constant of integration.
03

Use the initial velocity to find the constant of integration C_1

We know the initial velocity is $$v(0) = 20$$. Substituting this into the velocity function we found: $$20 = -32(0) + C_1$$ Solving for $$C_1$$: $$C_1 = 20$$ Now we have the velocity function: $$v(t) = -32t + 20$$
04

Integrate the velocity function to find the position function

Now we need to integrate the velocity function with respect to time to find the position function: $$s(t) = \int v(t) dt = \int (-32t + 20) dt = -16t^2 + 20t + C_2$$ where $$C_2$$ is another constant of integration.
05

Use the initial position to find the constant of integration C_2

We know that the initial position is $$s(0) = 0$$. Substituting this into the position function we found: $$0 = -16(0)^2 + 20(0) + C_2$$ Solving for $$C_2$$: $$C_2 = 0$$ Now we have the position function: $$s(t) = -16t^2 + 20t$$
06

Write the final position function

The final position function of the object moving along a line is: $$s(t) = -16t^2 + 20t$$

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