Question

A man walks on a straight road from his home to a market 2.5 km 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. what is the( a) magnitude of average velocity and (b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min and ( iii) 0 to 40 min ?

Answer 1

Answer:

Average speed =totaltimetotaldistance

Time taken to go to the market =speeddistance=52.5hour=21hours=30 minutes

Time taken to return back =speeddistance=7.52.5hour=1/3hours=20 minutes 

Distance travelled in 10 minutes while returning back=6010×7.5=1.25km ( Speed is constant at 7.5 km/hr )

Hence Distance traveled in 40 minutes(30 mins towards the market and 10 mins towards home) =2.5+1.25=3.75km 

So average speed in 0 to 40 minutes =6040hr3.75km

 =845 km/hr

Answer 2

Please refer to the above picture