Expert: Abe Mantell - 1/9/2006

Question
Hi,
I hope I'm finding you in the best of health.
A particle starts at time t=0 and moves along the x-axis so that its position at any time t>=0 is given by x(t)=[(t-1)^3](2t-3)
a. find the velocity of the particle at any time t>=0
b. find all times when the acceleration is zero.
c. find v(2)
d. If x(t)=t^2-4t+3, find the total distance traveled from t=0 to t=4.  

Answer
Hello again Jeff,

I just realized I answered 'd' for the original x(t), not
the one you gave in 'd'!  So, with x(t)=t^2-4t+3...
Displacement=x(4)-x(0)=3-3=0 (ends where it started)
Distance=the integral of the absolute value of v(t)
from t=0 to t=4...which gives: 8

Abe