SOLUTION: A particle moves in a straight line with velocity v(t)=root(3t-1) metres per second where t is time in seconds.At t=2, the particle's distance from the starding point was 8 meters
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Question 1189659: A particle moves in a straight line with velocity v(t)=root(3t-1) metres per second where t is time in seconds.At t=2, the particle's distance from the starding point was 8 meters in the positive direction. What is the particle's position at t=7seconds?
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
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
 = \frac{d}{dt}s(t))
so the integral will reverse this process
 = \int v(t)dt)
 = \int \sqrt{3t-1}dt)
 = \frac{1}{3}\int \sqrt{u}du)
Let u = 3t-1, so du = 3dt which means dt = (1/3)du
 = \frac{1}{3}\int u^{1/2}du)
 = \frac{1}{3}*\frac{1}{1+1/2}u^{1+1/2}+C)
 = \frac{1}{3}*\frac{1}{3/2}u^{3/2}+C)
 = \frac{1}{3}*\frac{2}{3}u^{3/2}+C)
 = \frac{2}{9}u^{3/2}+C)
 = \frac{2}{9}(3t-1)^{3/2}+C)
Plug in u = 3t-1
To verify we have the correct antiderivative, differentiate ^{3/2}+C)
with respect to t, and you should get  = \sqrt{3t-1})
again.
---------------------------
Plug in the condition that s(2) = 8 and solve for C.
In other words, plug in t = 2 and s(t) = 8.
 = \frac{2}{9}(3*2-1)^{3/2}+C)
This particular position function is approximately
 \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
 \approx \frac{2}{9}(3*7-1)^{3/2}+5.51548003)
 \approx 25.39163983)
The particle is roughly 25.39163983 meters from the starting point at t = 7 seconds. Round this value however you need to.
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