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
$5\, km/h$
$\frac{{25}}{4} \,km/h$
$\frac{{30}}{4} \,km/h$
$\frac{{45}}{8} \,km/h$
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
A man walks from his home to market with a speed of $5\;km/h$.
Distance $ = 2.5\;km$ and time$ = \frac{d}{v} = \frac{{2.5}}{5} = \frac{1}{2}hr$.
and he returns back with speed of $7.5\;km/h$ in rest of time of 10 minutes.
Distance $ = 7.5 \times \frac{{10}}{{60}} = 1.25\;km$
So, Average speed = ${Total\ distance}\over{Total\ time}$
$ = \frac{{(2.5 + 1.25)km}}{{(40/60)hr}} = \frac{{45}}{8}km/hr$.
Distance $ = 2.5\;km$ and time$ = \frac{d}{v} = \frac{{2.5}}{5} = \frac{1}{2}hr$.
and he returns back with speed of $7.5\;km/h$ in rest of time of 10 minutes.
Distance $ = 7.5 \times \frac{{10}}{{60}} = 1.25\;km$
So, Average speed = ${Total\ distance}\over{Total\ time}$
$ = \frac{{(2.5 + 1.25)km}}{{(40/60)hr}} = \frac{{45}}{8}km/hr$.
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