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One question about Calculus (Rectilinear Motion)?
A particle moves along a straight line and its position at time t is given by s(t)= 2t^3 - 27 t^2 + 108 t where s is measured in feet and t in seconds.
Find the velocity (in ft/sec) of the particle at time t=0: I figured out that this is 108.
The particle stops moving (i.e. is in a rest) twice, once when t=A and again when t=B where A < B. I also figured out that A is 3 and B is 6.
What is the position of the particle at time 18? I'm pretty sure that this is 4860.
Finally, what is the TOTAL distance the particle travels between time 0 and time 18? This is the part that I'm stuck on because I'm not sure how to calculate total distance.
-Thanks for the help.
1 Answer
- simplicitusLv 71 decade agoFavorite Answer
It is a pleasure helping someone who has done as much as possible on the problem.
The difference between the total difference traveled and the net difference in positions is any extra distnce to the particle going "backwards".
Since the particle is moving in a straight line with differentiable velocity, in order for it to switch directions, it first has to stop.
You know that it stopped at t = A and t = B. This defines the following regions:
t in [0, A)
t in (A,B)
t in (B, 18]
for the first and last intervals, it was moving forward. Was it moving forward or backward for the middle interval?
If forward, then you are all set - the total distance is just ending point - starting point.
If it went backward, then you have to add in the distance it went back times 2 (it went over that distance three times)