First slide

Average Speed

Question

A man walks on a straight road from his home to a market 2.5 km away with a speed of $5 {kmh}^{- 1}$ . Finding the market closed, he instantly turns and walks back home with a speed of $7.5 {kmh}^{- 1}$ . The average speed of the man over the interval of time 0 to 50 min is equal to

Moderate

A

$5 {kmh}^{- 1}$
B

$\frac{25}{4} {kmh}^{- 1}$
C

$\frac{30}{5} {kmh}^{- 1}$
D

$\frac{45}{8} {kmh}^{- 1}$

Solution

Man walks from his home to market with a speed of $5 {kmh}^{- 1}$ . Distance = 2.5 km and time
$= \frac{d}{v} = \frac{2.5}{5} = \frac{1}{2} = 30 \min$
and he returns back with speed of $7.5 {kmh}^{- 1}$ in rest a time of 20 min.
Distance $= 7.5 \times \frac{20}{60} = 2.5 km$
So, average speed $= \frac{total distance}{total time} = \frac{(2.5 + 2.5)}{(50 / 60) h} = \frac{30}{5} km / h$

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