A man walks on a straight road from his home to a market \(2.5~\text {km}\) away with a speed of \(5~\text{km/h}.\) Finding the market closed, he instantly turns and walks back home with a speed of \(7.5~\text{km/h}.\) What will be the magnitude of the average velocity and the average speed of the man over the interval of time from \(0\) to \(30\) min?
1. \(0~\text{km/h},~6~\text{km/h}\)
2. \(0~\text{km/h},~5~\text{km/h}\)
3. \(0~\text{km/h},~3.75~\text{km/h}\)
4. \(2~\text{km/h},~5~\text{km/h}\)

Time taken by the man to reach the market from home,t1=2.55=12h=30min

Time taken by the man to reach home from the market,t2=2.57.5=13h=20min

Total time taken in the whole journey = 30 + 20 = 50 min

i0to30min
Avragevelocity=DisplacementTime=2.51/2=5km/h
Averagespeed=DistanceTime=2.51/2=5km/h
ii0to50min
Time=50min=5060=56h
Netdisplacement=0
Totaldistance=2.5+2.5=5km
Averagevelocity=DisplacementTime=0
Averagespeed=DistanceTime=55/6=6km
iii0to40min
Speedoftheman=7.5km/h
Distancetravelledinfirst30min=2.5km
Distancetravelledbythemanfrommarkettohomeinthenext10min
=7.5×1060=1.25km
Netdisplacement=2.5-1.25=1.25km
Totaldistancetravelled=2.5+1.25=3.75km
Andaveragevelocity=DisplacementTime=1.2540/60=1.875km/h
Averagespeed=DistanceTime=3.7540/60=5.625km/h