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

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Question

(D) Time taken to reach the market: $t_1 = \frac{d}{v_1} = \frac{2.5 \,km}{5 \,km/h} = 0.5 \,h = 30 \,min$.Since the total time interval is $40 \,min$

Vedclass · Accepted Answer

$\frac{45}{8} \,km/h$

Vedclass · Answer

(D) Time taken to reach the market: $t_1 = \frac{d}{v_1} = \frac{2.5 \,km}{5 \,km/h} = 0.5 \,h = 30 \,min$.Since the total time interval is $40 \,min$,the time spent walking back is $t_2 = 40 \,min - 30 \,min = 10 \,min = \frac{10}{60} \,h = \frac{1}{6} \,h$.Distance covered while returning: $d_2 = v_2 	imes t_2 = 7.5 \,km/h 	imes \frac{1}{6} \,h = 1.25 \,km$.Total distance covered: $D = 2.5 \,km + 1.25 \,km = 3.75 \,km$.Total time taken: $T = 40 \,min = \frac{40}{60} \,h = \frac{2}{3} \,h$.Average speed = $\frac{	ext{Total distance}}{	ext{Total time}} = \frac{3.75 \,km}{(2/3) \,h} = 3.75 	imes 1.5 = 5.625 \,km/h = \frac{45}{8} \,km/h$.

$A$ man walks on a straight road from his home to a market $2.5 \,km$ away with a speed of $5 \,km/h$. Finding the market closed,he instantly turns and walks back home with a speed of $7.5 \,km/h$. The average speed of the man over the interval of time $0$ to $40 \,min$ is equal to

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