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Chapter 4: Problem 92

Given the following acceleration functions of an object moving along a line, find the position function with the given initial velocity and position. $$a(t)=2 \cos t ; v(0)=1, s(0)=0$$

Short Answer

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

Step by step solution

01

Integrate the acceleration function to find the velocity function

To find the velocity function, v(t), we need to integrate the acceleration function, \(a(t) = 2 \cos t\). $$v(t) = \int a(t) dt = \int 2 \cos t dt$$ The integral of \(\cos t\) is \(\sin t\), so we have: $$v(t) = 2 \sin t + C_1$$ where \(C_1\) is the constant of integration.
02

Use the initial condition for velocity to find the constant of integration \(C_1\)

The initial condition for velocity is \(v(0) = 1\). We'll use this to find the value of \(C_1\). $$1 = 2 \sin(0) + C_1$$ Since \(\sin(0) = 0\), we have: $$1 = C_1$$ Thus, the velocity function is: $$v(t) = 2 \sin t + 1$$
03

Integrate the velocity function to find the position function

To find the position function, s(t), we need to integrate the velocity function, \(v(t) = 2 \sin t + 1\). $$s(t) = \int v(t) dt = \int (2 \sin t + 1) dt$$ To find the integral, we can integrate each term separately: $$s(t) = \int 2 \sin t dt + \int 1 dt$$ The integral of \(2 \sin t\) is \(-2 \cos t\), and the integral of 1 is t: $$s(t) = -2 \cos t + t + C_2$$ where \(C_2\) is the constant of integration.
04

Use the initial condition for position to find the constant of integration \(C_2\)

The initial condition for position is \(s(0) = 0\). We'll use this to find the value of \(C_2\). $$0 = -2 \cos(0) + 0 + C_2$$ Since \(\cos(0) = 1\), we have: $$0 = -2(1) + C_2$$ Thus, $$C_2 = 2$$
05

Write down the position function

Now that we have the constant of integration, \(C_2\), we can write down the position function, s(t): $$s(t) = -2 \cos t + t + 2$$ So, the position function for the given object is: $$s(t) = -2 \cos t + t + 2$$

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