Question

The displacement $\left(x\right)$ of a particle depends on time $\left(t\right)$ as $x=\alpha t^2-2\beta t^3$.

• A. The particle will return to its starting point after time $\frac{\alpha }{2\beta }$.
• B. The particle will come to rest after time $\frac{\alpha }{3\beta }$.
• C. The initial velocity of the particle was zero but its initial acceleration was $2\alpha$.
• D. No net force will act on the particle at $t=\frac{\alpha }{6\beta }$.

Answer 1

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

The correct options are
A The particle will return to its starting point after time $\frac{\alpha }{2\beta }$.
B The initial velocity of the particle was zero but its initial acceleration was $2\alpha$.
C The particle will come to rest after time $\frac{\alpha }{3\beta }$.
D No net force will act on the particle at $t=\frac{\alpha }{6\beta }$.

It is given that
$x=\alpha t^2-2\beta t^3\Rightarrow v=\frac{dx}{dt}=2\alpha t-6\beta t^2\Rightarrow a=\frac{dv}{dt}=2\alpha -12\beta t$
Put $x=0$
$\Rightarrow t^2(\alpha -2\beta t)=0\Rightarrow t=\frac{\alpha }{2\beta }$

Put $v=0$
$v=2\alpha t-6\beta t^2=0\Rightarrow t=\frac{\alpha }{3\beta }$

initial velocity at time $t=0\Rightarrow v_{initial}=0$

Put $a=0$ for force to be zero because $F=ma$

$a=2\alpha -12\beta =0\Rightarrow t=\frac{\alpha }{6\beta }$

initial acceleration at time $t=0\Rightarrow a_{initial}=2\alpha$