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The displacement(x) of a particle depends on time (t) as x = $α t^{2} - β t^{3}$ .
Column-I Column-II
i. The particle will be at its starting point after time p. $\frac{2 α}{3 β}$
ii. The particle will be at rest at time q. $\frac{α}{β}$
iii. The average velocity of particle from t = 0 to t = $t_{o}$ is equal to its instantaneous velocity at time t = $t_{o}$ . The value of $t_{o}$ is equal to r. $\frac{α}{3 β}$
iv. No net force will act on the particle at time t is equal to s. $\frac{α}{2 β}$
Now match the given columns and select the correct option from the codes given below.

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i-q, ii-s, iii-r, iv-p

i-q, ii-p, iii-s, iv-r

i-s, ii-r, iii-p, iv-q

i-r, ii-q, iii-s, iv-p

answer is B.

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Detailed Solution

x = $α t^{2} - β t^{2}$

v = $\frac{dx}{dt} = 2 αt - 3 {βt}^{2}$

Putting x = 0, we get t = $\frac{α}{β}$

Putting v = 0, we get t = $\frac{2 α}{3 β}$

Given $v_{av} (t = 0 to t = t_{0}) = v (t = t_{0})$

$\Rightarrow \frac{x_{t = t_{0}}}{t_{0}} = v_{t = t_{0}}$

$\frac{{αt}_{0}^{2} - {βt}_{0}^{3}}{t_{o}} = 2 {αt}_{o} - 3 {βt}_{0}^{2} \Rightarrow t_{0} = \frac{α}{2 β}$

a = $\frac{d v}{d t} = 2 α - 6 β t$

If no net force acts; a = 0

$\Rightarrow 2 α - 6 βt = 0 \Rightarrow t = \frac{α}{3 β}$

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