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(D) Let the total distance be $2D$. The rider covers distance $D$ with speed $v_1 = 5\,m/s$. The time taken is $t_1 = \frac{D}{5}$.For the remaining d

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(D) Let the total distance be $2D$. The rider covers distance $D$ with speed $v_1 = 5\,m/s$. The time taken is $t_1 = \frac{D}{5}$.For the remaining distance $D$,the rider travels for time $t_2$ such that the first half of $t_2$ is at $v_2 = 10\,m/s$ and the second half of $t_2$ is at $v_3 = 15\,m/s$.Distance $D = (v_2 \cdot \frac{t_2}{2}) + (v_3 \cdot \frac{t_2}{2}) = (10 \cdot \frac{t_2}{2}) + (15 \cdot \frac{t_2}{2}) = 5t_2 + 7.5t_2 = 12.5t_2$.So,$t_2 = \frac{D}{12.5} = \frac{D}{25/2} = \frac{2D}{25}$.The average speed $\langle v angle = \frac{	ext{Total Distance}}{	ext{Total Time}} = \frac{2D}{t_1 + t_2} = \frac{2D}{\frac{D}{5} + \frac{2D}{25}} = \frac{2D}{\frac{5D + 2D}{25}} = \frac{2D \cdot 25}{7D} = \frac{50}{7}\,m/s$.Comparing this with $\frac{x}{7}\,m/s$,we get $x = 50$.

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