Question 1189659
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To find the position function s(t), we integrate the velocity function v(t)


Recall that the velocity is the derivative of the position function


*[Tex \Large v(t) = \frac{d}{dt}s(t)]


so the integral will reverse this process


*[Tex \Large s(t) = \int v(t)dt]


*[Tex \Large s(t) = \int \sqrt{3t-1}dt]


*[Tex \Large s(u) = \frac{1}{3}\int \sqrt{u}du] Let u = 3t-1, so du = 3dt which means dt = (1/3)du


*[Tex \Large s(u) = \frac{1}{3}\int u^{1/2}du]


*[Tex \Large s(u) = \frac{1}{3}*\frac{1}{1+1/2}u^{1+1/2}+C]


*[Tex \Large s(u) = \frac{1}{3}*\frac{1}{3/2}u^{3/2}+C]


*[Tex \Large s(u) = \frac{1}{3}*\frac{2}{3}u^{3/2}+C]


*[Tex \Large s(u) = \frac{2}{9}u^{3/2}+C]


*[Tex \Large s(t) = \frac{2}{9}(3t-1)^{3/2}+C] Plug in u = 3t-1


To verify we have the correct antiderivative, differentiate *[Tex \Large \frac{2}{9}(3t-1)^{3/2}+C] with respect to t, and you should get *[Tex \Large v(t) = \sqrt{3t-1}] again.


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Plug in the condition that s(2) = 8 and solve for C.
In other words, plug in t = 2 and s(t) = 8.


*[Tex \Large s(t) = \frac{2}{9}(3t-1)^{3/2}+C]


*[Tex \Large s(2) = \frac{2}{9}(3*2-1)^{3/2}+C]


*[Tex \Large 8 \approx 2.48451997 +C]


*[Tex \Large 8-2.48451997 \approx C]


*[Tex \Large 5.51548003 \approx C]


*[Tex \Large C \approx 5.51548003]


This particular position function is approximately
*[Tex \Large s(t) \approx \frac{2}{9}(3t-1)^{3/2}+5.51548003]


Now plug in t = 7 to find the position of the particle at the time of 7 seconds


*[Tex \Large s(t) \approx \frac{2}{9}(3t-1)^{3/2}+5.51548003]


*[Tex \Large s(7) \approx \frac{2}{9}(3*7-1)^{3/2}+5.51548003]


*[Tex \Large s(7) \approx 25.39163983]


The particle is <font color=red>roughly 25.39163983</font> meters from the starting point at t = 7 seconds. Round this value however you need to.
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