Expert: Ahmed Salami - 1/9/2006

Question
Hi,
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
Hi Jeff,
Please forgive the delay.
First you simplify [(t-1)^3](2t-3) to get
x(t) = (t^3 - 3t^2 + 3t -1)(2t - 3)
    = 2t^4 - 9t^3 + 15t^2 - 11t + 3
v = dx/dt
we therefore differentiate to get
v(t) = 8t^3 - 27t^2 + 30t - 11
a = dv/dt
 = 24t^2 - 54t + 30
we find the solution of this quadratic equation to get all the times when the acceleration is zero.
Therefore,
t = 1s , 1.25s
v(2) = 8(2)^3 - 27(2)^2 + 30(2) - 11
    = 64 - 108 + 60 - 11
    = 5
I hope it helps.
Regards.