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

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Question

(A) The time taken to reach the market is $t_1 = \frac{	ext{distance}}{	ext{speed}} = \frac{2.5 \; km}{5 \; km \; h^{-1}} = 0.5 \; h = 30 \; min$.Si

Vedclass · Accepted Answer

$5 \; km \; h^{-1}$

Vedclass · Answer

(A) The time taken to reach the market is $t_1 = \frac{	ext{distance}}{	ext{speed}} = \frac{2.5 \; km}{5 \; km \; h^{-1}} = 0.5 \; h = 30 \; min$.Since the time interval requested is $0$ to $30 \; min$,we only consider the motion from home to the market.During this interval,the total distance covered is $2.5 \; km$.The total time taken is $30 \; min = 0.5 \; h$.Average speed is defined as $\frac{	ext{Total distance}}{	ext{Total time}}$.Average speed $= \frac{2.5 \; km}{0.5 \; h} = 5 \; km \; h^{-1}$.

$A$ 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 $30 \; min$?

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