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

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> 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       Log On


   

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) About Me  (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



so the integral will reverse this process





Let u = 3t-1, so du = 3dt which means dt = (1/3)du











Plug in u = 3t-1

To verify we have the correct antiderivative, differentiate with respect to t, and you should get 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.













This particular position function is approximately


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







The particle is roughly 25.39163983 meters from the starting point at t = 7 seconds. Round this value however you need to.