a motorcycle starts at point A at time t=0 s and stops at point B. The distance ..?

Question

a motorcycle starts at point A at time t=0 s and stops at point B. The distance covered in t seconds is given by x=t^2 [2-(t/3)]m. Show that at ts, its velocity is t(4-t) m/s. Find the time taken by the motorcycle to reach point B and the distance between A and B.

Answer 1

v(t) = dx(t)/dt = d/dt (2t^2-t^3/3) = 4t-t^2 = t*(4-t) {m/s}

When is the motorcycle stopped?
t*(4-t) = 0
t = 0s or t = 4s

At t =4s the motorcycle is at point B
x(4) = 4^2 (2-4/3) ~ 10.67 m