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(C) Let the total distance of the track be $2d$.The first half distance $d$ is covered with velocity $v_0 = 10 \ m \ s^{-1}$. The time taken is $t_1 =

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$10$

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(C) Let the total distance of the track be $2d$.The first half distance $d$ is covered with velocity $v_0 = 10 \ m \ s^{-1}$. The time taken is $t_1 = \frac{d}{v_0} = \frac{d}{10}$.The remaining distance $d$ is covered in time $T$, where for the first half of time $T/2$, velocity is $v_1$ and for the second half of time $T/2$, velocity is $v_2$.The distance covered in the second half is $d = v_1(T/2) + v_2(T/2) = (v_1+v_2) \frac{T}{2}$.Given $v_1+v_2 = 20 \ m \ s^{-1}$, so $d = 20 	imes \frac{T}{2} = 10T$. Thus, $T = \frac{d}{10}$.The total time taken for the journey is $t_{total} = t_1 + T = \frac{d}{10} + \frac{d}{10} = \frac{2d}{10} = \frac{d}{5}$.The average velocity is $v_{avg} = \frac{	ext{Total Distance}}{	ext{Total Time}} = \frac{2d}{d/5} = 10 \ m \ s^{-1}$.

Column $I$	Column $II$
$(A)$ Change in velocity	$(p)$ $-5/3\,SI \text{ unit}$
$(B)$ Average acceleration	$(q)$ $-20\,SI \text{ unit}$
$(C)$ Total displacement	$(r)$ $-10\,SI \text{ unit}$
$(D)$ Acceleration at $t=3\,s$	$(s)$ $-5\,SI \text{ unit}$

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