A man walks on a straight road from his home to a market 2.5 km away with a speed of `5km 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 (n) average speed of the man over the time interval 0 to 50 minutes?

Explanation

Time taken by a man to go from home to market=
2.5//5=0.5hr=30min.
Time taken by a man to go from market to home=
2.5/7.5=1/3 hr=20min
Total time `=30+20=50` min
case I from 0 to 30 min
distance covered =2.5km
Net displacement=2.5km
(a) Magnitude of average velocity
=net displacement/total time taken
`=(2.5km)/(1//2h)=5km h^(-1)`
(b) Average speed
`=("Total distance travelled")/("Total time taken")`
`=(2.5km)/(1//2h)=5km h^(-1)`
(ii) From 0 to 50 minutes
Distance travelled in 30min `((1)/(2)h)` while going to the market=2.5km
Distance travelled in 20min `(1//3h)` while coming back to home
`=7.5xx(1)/(3)=.2.5km`
Total distance travelled in 50 min
`=(2.5+2.5)=5.0km`
(b) Average speed
`=("Total distance travelled")/("Total time taken")`
`=(5xx60)/(50)=6km h^(-1)`
(a) Magnitude of average velocity
=displacement time taken
`=((2.5-2.5))/(50//60)=0`
(iii) From 0 to 40 min
Distance travelled in 30min. while going to market =2.5km
Distance travelled in 10min. while coming back from market.
`=7.5xx(1)/(6)=(5)/(4)km=1.25km`
Total distance`=2.5+1.25=3.75km`
Displacement`=2.5-1.25=1.25`
(a) Average speed `=("Total distance travelled")/("Total time taken")`
`=(3.75)/(40//60)=5.725km h^(-1)`
Magnitude of average velocity
`=("Net displacement")/("Total time taken")`
`=(1.25)/(50//60)km h^(-1)`
`=1.875km h^(-1)`