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(B) Let the total distance be $2d$. The first half distance is $d$ and the second half distance is $d$.For the first half distance $(d)$: Velocity $v_

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$10$

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(B) Let the total distance be $2d$. The first half distance is $d$ and the second half distance is $d$.For the first half distance $(d)$: Velocity $v_1 = 7 \,m/s$. Time taken $t_1 = \frac{d}{v_1} = \frac{d}{7}$.For the second half distance $(d)$: Let the total time taken be $t_2$. The first half of this time $(t_2/2)$ is travelled with $v_2 = 14 \,m/s$ and the second half $(t_2/2)$ with $v_3 = 21 \,m/s$.Distance $d = (v_2 	imes \frac{t_2}{2}) + (v_3 	imes \frac{t_2}{2}) = (14 	imes \frac{t_2}{2}) + (21 	imes \frac{t_2}{2}) = 7t_2 + 10.5t_2 = 17.5t_2$.So, $t_2 = \frac{d}{17.5} = \frac{d}{35/2} = \frac{2d}{35}$.Total distance = $2d$.Total time $T = t_1 + t_2 = \frac{d}{7} + \frac{2d}{35} = \frac{5d + 2d}{35} = \frac{7d}{35} = \frac{d}{5}$.Average velocity $v_{avg} = \frac{	ext{Total distance}}{	ext{Total time}} = \frac{2d}{d/5} = 10 \,m/s$.

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