Question

The position $x$ of a particle varies with time $\left(t\right)$ as $x=a{t}^2-b{t}^3$. The acceleration at time $t$ of the particle will be equal to zero, when $t$ is equal to

• A. $\frac{2a}{3b}$
• B. $\frac{a}{b}$
• C. $\frac{a}{3b}$
• D. Zero

Answer 1

The correct option is C $\frac{a}{3b}$

We know,
$x=a{t}^2-b{t}^3$
Velocity of the particle can be given by,
$v=\frac{dx}{dt}=2at-3b{t}^2$
And acceleration of the particle is given by,
$acc=\frac{dv}{dt}=2a-6bt$

We need to find out time when acceleration is zero, so
$acc=2a-6bt=0$
$⟹t=\frac{a}{3b}$