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Chapter 4: Problem 94

Given the following acceleration functions of an object moving along a line, find the position function with the given initial velocity and position. $$a(t)=2 e^{-t / 6} ; v(0)=1, s(0)=0$$

Short Answer

Expert verified
Question: Determine the position function, $$s(t),$$ given the acceleration function $$a(t) = 2 e^{-t/6}$$ and the initial conditions $$v(0) = 1$$ and $$s(0) = 0.$$ Answer: The position function is $$s(t) = -72 e^{-t/6} + 13t + 72.$$

Step by step solution

01

Integrate the acceleration function to find the velocity function

To find the velocity function, $$v(t),$$ integrate the acceleration function, $$a(t),$$ with respect to time: $$v(t)=\int a(t) dt = \int 2 e^{-t/6} dt.$$ To integrate this function, make a substitution $$u = -t/6,$$ so $$du = -1/6 dt$$ and $$dt = -6 du.$$ Now, the integral becomes: $$v(t) = \int 2 e^{u} (-6 du) = -12 \int e^{u} du.$$ After integrating, we get: $$v(t) = -12 e^{u} + C = -12 e^{-t/6} + C.$$ Since $$v(0) = 1,$$ we can find the constant $$C$$: $$1 = -12 e^{0} + C \Rightarrow C = 1 + 12.$$ Thus, the velocity function is: $$v(t) = -12 e^{-t/6} + 13.$$
02

Integrate the velocity function to find the position function

Now we need to integrate the velocity function, $$v(t)$$, with respect to time to find the position function, $$s(t):$$ $$s(t) = \int v(t) dt = \int (-12 e^{-t/6} + 13) dt.$$ Split the integral: $$s(t) = \int -12 e^{-t/6} dt + \int 13 dt.$$ Use the same substitution as before: $$u = -t/6, du = -1/6 dt, dt = -6 du.$$ $$s(t) = -72\int e^{u} du + 13\int dt.$$ After integrating, we get: $$s(t) = -72 e^{u} + 13t + D = -72 e^{-t/6} + 13t + D.$$ Since $$s(0) = 0,$$ we can find the constant $$D$$: $$0 = -72 e^{0} + 13(0) + D \Rightarrow D = 72.$$ Thus, the position function is: $$s(t) = -72 e^{-t/6} + 13t + 72.$$

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