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(B) Let the total distance be $2S$. The first half distance $S$ is covered with speed $v_1 = 2 \,ms^{-1}$. The time taken is $t_1 = \frac{S}{2}$.For t

Vedclass · Accepted Answer

$\frac{8}{3} \,ms^{-1}$

Vedclass · Answer

(B) Let the total distance be $2S$. The first half distance $S$ is covered with speed $v_1 = 2 \,ms^{-1}$. The time taken is $t_1 = \frac{S}{2}$.For the remaining half distance $S$, it is covered in two equal time intervals $t_2$ each, with speeds $v_2 = 3 \,ms^{-1}$ and $v_3 = 5 \,ms^{-1}$.Thus, $S = v_2 t_2 + v_3 t_2 = (3 + 5) t_2 = 8 t_2$. Therefore, $t_2 = \frac{S}{8}$.The total time taken for the second half is $2 t_2 = 2 	imes \frac{S}{8} = \frac{S}{4}$.Total time $T = t_1 + 2 t_2 = \frac{S}{2} + \frac{S}{4} = \frac{3S}{4}$.Average speed = $\frac{	ext{Total distance}}{	ext{Total time}} = \frac{2S}{3S/4} = \frac{8}{3} \,ms^{-1}$.

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