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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 (iii)

Q 3.14     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

           (b)     average speed of the man over the interval of time 

                     (iii) 0 to 40 min  

Answers (1)

Time taken by the man to reach the market from home, $t_{1}=2.5 / 5=1 / 2 \mathrm{h}=30 \mathrm{min}$
Time taken by the man to reach home from the market,$t_{2}=2.5 / 7.5=1 / 3 \mathrm{h}=20 \mathrm{min}$
Total time taken in the whole journey =30+20=50 min

(iii) 0 to 40 min
Speed of the man $=7.5 \mathrm{km} / \mathrm{h}$ 
Distance travelled in first $30 \mathrm{min}=2.5 \mathrm{km}$ 
Distance travelled by the man (from market to home) in the next 10 min  $=7.5 \times 10 / 60=1.25 \mathrm{km}$
Net displacement $=2.5-1.25=1.25 \mathrm{km}$ 
Total distance travelled $=2.5+1.25=3.75 \mathrm{km}$ 
Average velocity = Displacement / Time $=1.25 /(40 / 60)=1.875 \mathrm{km} / \mathrm{h}$ 
Average speed = Distance / Time $=3.75 /(40 / 60)=5.625 \mathrm{km} / \mathrm{h}$ 

Posted by

Satyajeet Kumar

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