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Chapter 4: Problem 90

Given the following acceleration functions of an object moving along a line, find the position function with the given initial velocity and position. $$a(t)=4 ; v(0)=-3, s(0)=2$$

Short Answer

Expert verified
Answer: The position function, s(t), of the object is s(t) = 2t^2 - 3t + 2.

Step by step solution

01

Integrate the acceleration function to find the velocity function

Integrate the given function a(t) = 4 with respect to time (t). $$v(t) = \int a(t) dt = \int 4 dt = 4t + C_1$$ Here, C_1 is the constant of integration, which we determine using the initial condition v(0) = -3.
02

Apply the initial condition to find C_1

Substitute t=0 and v(0) = -3 into the velocity function, and then solve for C_1: $$v(0) = -3 = 4(0) + C_1 \Longrightarrow C_1 = -3$$ So, the velocity function is v(t) = 4t - 3.
03

Integrate the velocity function to find the position function.

Integrate the velocity function, v(t) = 4t - 3, with respect to time (t) to find the position function. $$s(t) = \int v(t) dt = \int (4t - 3) dt = 2t^2 - 3t + C_2$$ Here, C_2 is the constant of integration, which we determine using the initial condition s(0) = 2.
04

Apply the initial condition to find C_2

Substitute t=0 and s(0) = 2 into the position function, and then solve for C_2: $$s(0) = 2 = 2(0)^2 - 3(0) + C_2 \Longrightarrow C_2 = 2$$ Therefore, the position function is s(t) = 2t^2 - 3t + 2.

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