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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 of the man over the interval of time $0$ to $40\; min$ ?
$1.875$
$6$
$5.625$
$2.5$
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 $v=\frac{2.5}{5}=5 km / h$
Distance traveled in first 30 min is $=2.5 km$
distance traveled (while returning) in 10 min is $d=7.5 \times \frac{1}{6}=1.25 km$
Total time is 40 min or $\frac{2}{3}$
So displacement in $0-40$ min is $2.5-1.25=1.25 km$
Average velocity for $0-40$ in is $v=\frac{1.25}{\frac{2}{2}}=1.875 km / h$
Average speed for $0-40$ in is $v=\frac{3.75}{\frac{2}{3}}=5.625 km / h$
