A man walks on a straight road from his home to a market 2.5 km/h 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
1. 5 km/h
2. 25/4 km/h
3.45/8 km/h

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Solution

Dear student,
As the man is moving initially with speed of 5km/hr it will cover 2.5 km in 30 min i.e he will reach market
$T i m e t a k e n = \frac{D i s \tan c e}{s p e e d} = \frac{5}{2.5} = \frac{1}{2} h r o r 30 m i n s$
$0 t o 40 m i n = D i s \tan c e t r a v e l l e d i n f i r s t 30 m i n = 2.5 k m D i s \tan c e t r a v e l l e d i n n e x t 10 m i n = \frac{7.5 x 10}{60} = 1.25 k m a v e r a g e s p e e d f o r 40 m i n = \frac{T o t a l d i s \tan c e}{t o t a l t i m e} = \frac{3.75 k m}{40 m i n} = \frac{3.75 \times 60}{40} k m / h r = 45 / 8 k m / h r = 5.625 k m / h r$

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