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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 (in $km/h$) of the man over the interval of time $0$ to $50\; min$ ?
$4$
$6$
$0$
$10$
Solution
Time taken to reach market $t_{1}=\frac{2.5}{5}=0.5$ hour $=30 min$
time taken to get back to home is $t_{2}=\frac {2.5}{7 .5}=.33$hour$=20 min$
Average velocity for $0-30$ in is $v=\frac{2.5}{5}=5 km / h$
Average speed for $0-30$ in is $s=\frac{2.5}{0.5}=5 km / h$
total time he took for travelling $t=30+20=50 min =\frac{5}{6}$ hour
When he reached back then net displacement is zero
so for $0-50$ min
Average velocity for $0-50$ in is $v=\frac{0}{\frac{5}{6}}=0 km / h$
Total distance he traveled when he arrive back is $2.5+2.5=5 km$
Average speed for $0-50$ in is $v=\frac{{5}}{\frac{5}{6}}=6 km / h$
