Question

A particle moves along a straight line with equation of motion s=t^4-1t^3. Find the value of t (other than 0) at which the acceleration is equal to zero

Answer 1

I get up to 4t^3-3t^2 => 12t^2-6t. then what?

Answer 2

acceleration = second derivative of s. next =0

Answer 3

v=s'=4t^3-3t^2
a=12t^2-6t=0
2t^2-t=0
t(2t-1)=0
t=0 or t=1/2

Answer 4

a=s''(t)

Answer 5

ok?

Answer 6

yes! thanks :]

Answer 7

welcome!