Question

The position x of a particle varies with time ,(t) as x=a t square -b t cube .the acceleration will zero at time t equals to ????????????????????

Answer 1

Question 36133

The position of a particle is $x(t) = at^2 - bt^3$ , so velocity is $v = x'(t) = 2at - 3bt^2$ and acceleration is $a(t) = x''(t) = 2a - 6bt$ .

The acceleration is zero when $2a - 6bt = 0 \Rightarrow t = \frac{a}{3b}$ .