Question

The displacement of a particle varies with time $t$ as: $s=\alpha t^2-\beta t^3$. Determine the time $\left(t\right)$ at which acceleration of the particle becomes $0$.

• A. $\frac{\alpha }{3\beta }$
• B. $\frac{\alpha }{\beta }$
• C. $\frac{3\beta }{\alpha }$
• D. $\frac{2\alpha }{3\beta }$

Answer 1

The correct option is A $\frac{\alpha }{3\beta }$

Given,
$s=\alpha t^2-\beta t^3$
Differentiating $s$ w.r.t $t$ we get,
Velocity, $v=\frac{ds}{dt}=2\alpha t-3\beta t^2$
Differentiating $v$ w.r.t $t$ we get,
Acceleration, $a=\frac{dv}{dt}=2\alpha -6\beta t$
As $a=0$, we get
$2\alpha -6\beta t=0\Rightarrow t=\frac{\alpha }{3\beta }$