Question

The position $x$ of a particle varies with time $t$ as $x=\alpha t^2-\beta t^3$. Which of the following is (are) correct?

• A. The particle will return to its starting point after time $\frac{\alpha }{\beta }$.
• B. The particle will come to rest after time $\frac{2\alpha }{3\beta }$.
• C. The initial velocity of the particle was zero but its initial acceleration was not zero.
• D. The acceleration of the particle is zero at $t=\frac{\alpha }{3\beta }$.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A The particle will return to its starting point after time $\frac{\alpha }{\beta }$.
B The particle will come to rest after time $\frac{2\alpha }{3\beta }$.
C The initial velocity of the particle was zero but its initial acceleration was not zero.
D The acceleration of the particle is zero at $t=\frac{\alpha }{3\beta }$.

Differentiate $x=\alpha t^2-\beta t^3$ w.r.t time to get velocity
$v=\frac{dx}{dt}=2\alpha t-3\beta t^2.....\left(1\right)$
Differentiate $v$ w.r.t time to get acceleration
$a=\frac{dv}{dt}=2\alpha -6\beta t....\left(2\right)$
The particle will return to its initial position when $x=0$
i.e., $0=\alpha t^2-\beta t^3$
which gives $t=\frac{\alpha }{\beta }$
The particle will come to rest when $v=0$
i.e., $0=2\alpha t-3\beta t^2$
which gives $t=\frac{2\alpha }{3\beta }$
Using equation $\left(1\right)$ and $\left(2\right)$, we say that initial velocity of the particle is zero but initial acceleration is $a=2\alpha$
$a=0\Rightarrow 2\alpha -6\beta t=0,$ which gives $t=\frac{\alpha }{3\beta }$
Thus, acceleration is zero at $t=\frac{\alpha }{3\beta }$