Question

please help:
A particle moves along a straight line and its position at time t is given by s(t)= 2t^3 - 27 t^2 + 84 t where s is measured in feet and t in seconds.
what is the TOTAL distance the particle travels between time 0 and time 18

Answer 1

To solve this problem we need to take notice of when the particle changes direction. This happens when s'(t)=v=0, so
s'(t)=v=6t^2-54t+84=0=t^2-9t+14
(t-7)(t-2)=0
so the particle changes direction at t={2,7}
Now we must integrate within these intervals separately and take absolute values to avoid negative answers. The intervals are
\[0 \le t \le 2\]\[2 \le t \le 7\] and \[7 \le t \le 18\]
So integrate\[\int\limits\limits v(t)dt=\int\limits\limits s'(t)dt\] for each interval and take the absolute value of any negative answers. This will give you your solution.

Answer 2

PS: You must add the absolute value of the results of each of the three integrals to find the TOTAL distance, in case that wasn't obvious.

Answer 3

thank you so much for your time, would there by another way to solve without using integration?

Answer 4

hmmm....
I guess if you just integrate\[\int\limits_{a}^{b} s'(t)dt=s(t)\]evaluated from a to b.
So it looks like you could just find the intervals and do
\[\left| s(2)-s(0) \right|+\left| s(2)-s(7) \right|+\left| s(18)-s(7) \right|=\]and that answer should be the same.
Never thought about it that way though.

Answer 5

sorry, the expression should be
|s(2)−s(0)|+|s(7)−s(2)|+|s(18)−s(7)|=
but because it's absolute value it doesn't even matter!!!
Ia-bI=Ib-aI

Answer 6

thank you so much, you helped me a bunch and made it really clear =)

Answer 7

Anytime