A man walks on a straight road from his home | vedclass

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(C) Time taken to reach the market: $t_1 = \frac{2.5 \; km}{5 \; km \; h^{-1}} = 0.5 \; h = 30 \; min$.Time taken to return home: $t_2 = \frac{2.5 \;

Vedclass · Accepted Answer

$5.625 \; km \; h^{-1}$

Vedclass · Answer

(C) Time taken to reach the market: $t_1 = \frac{2.5 \; km}{5 \; km \; h^{-1}} = 0.5 \; h = 30 \; min$.Time taken to return home: $t_2 = \frac{2.5 \; km}{7.5 \; km \; h^{-1}} = \frac{1}{3} \; h = 20 \; min$.Total distance traveled in $40 \; min$:In the first $30 \; min$,the man travels $2.5 \; km$ to reach the market.In the remaining $10 \; min$ $(= \frac{1}{6} \; h)$,he walks back at $7.5 \; km \; h^{-1}$.Distance covered while returning: $d_2 = 7.5 \; km \; h^{-1} 	imes \frac{1}{6} \; h = 1.25 \; km$.Total distance = $2.5 \; km + 1.25 \; km = 3.75 \; km$.Average speed = $\frac{	ext{Total distance}}{	ext{Total time}} = \frac{3.75 \; km}{40/60 \; h} = \frac{3.75}{2/3} \; km \; h^{-1} = 5.625 \; km \; h^{-1}$.

$A$ man walks on a straight road from his home to a market $2.5 \; km$ away with a speed of $5 \; km \; h^{-1}$. Finding the market closed,he instantly turns and walks back home with a speed of $7.5 \; km \; h^{-1}$. What is the average speed of the man over the interval of time $0$ to $40 \; min$?

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