Expert: Ahmed Salami - 4/6/2011

Question
Given that the position of a particle is found by x(t)= (t^3)-(6t^2)+1; t>0, find the distance that the particle travels from t=2 to t=5.
I understand that i need to take the derivative of the equation, and then set it to zero to find the critical values, but I'm lost after that.
Thank you for your time.

Answer
Hi Elinor,
Well, distance is a scalar and so we do not consider the direction in which it takes place but only the length of it. Basically, we could just have said the distance we need is the position at t = 5 less that at t = 2 i.e x(5) - x(2). But there might be times in that interval when the particle moves in opposite directions and that distance would be lost since it cancels out. We therefore need to find the(se) point(s) and that is the idea of the critical value(s).
Now,
x = t³ - 6t² + 1
v = dx/dt = 3t² - 12t
Equating to zero,
3t² - 12t = 0
3t(t - 4) = 0
t = 0 or 4
and we can immediately see that at t = 4 (which is the one in the concerned interval) the particle changes direction.
So, the total distance travelled from t = 2 to t = 5 is given by
|x(4) - x(2)| + |x(5) - x(4)|
= |(-31) - (-15)| + |(-24) - (-31)|
= |-16| + |7|
= 16 + 7
= 23

Regards