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 (a) magnitude of average velocity and (b) average speed of the man over the time interval 0 to 50 min.

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Detailed Solution

$(a) Average velocity = \frac{total displacemnt}{total timetaken}$
$= \frac{Zero}{t} = zero$
$(b) Average speed = \frac{total distance}{total time taken}$
$\begin{array}{l} = \frac{2 .5 + 2 .5}{t_{1} + t_{2}} \\ = \frac{5}{\frac{2 .5}{5} + \frac{2 .5}{7.5}} = 6 km / h \end{array}$

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