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
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, (iii) 0 to 40 min ? [Note : You will appreciate from this exercise why it is better to define average speed as total path length divided by time and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]

Explanation

Time taken by man to go from his home to market `t_(1)="Distance"/"Speed"=(2.5)/5=1/2h`
Time taken by man to go from market to his home `t_(2)=(2.5)/(7.5)=1/3 h`
Total time taken `=t_(1)+t_(2)=1/2+1/3=5/6`
h = 50 min
i) 0 to 30 min
a) Average velocity
`" "="Displacement"/"Time"=(2.5)/(1//2)=5 km//h`
b) Average speed
`" "="Distance"/"Time"=(2.5)/(1//2)=5 km//h`
ii) 0 to 50 min
Total distance travelled
`" "=2.5+2.5=5km`
Total displacement = zero
a) Average velocity `= "Displacement"/"Time"=0`
b) Average speed `="Distance"/"Time"=5/(5//6)=6km//h`
iii) 0 to 40 min
`" "`Distance moved in 30 min (from home to market) = 2.5 km
Distance moved in 10 min (from market to home) with speed 7.5 km/h = `7.5 times 10/60`
= 12.5 km
So displacement = 2.5 - 1.25 = 1.25 km
Distance travelled = 2.5 + 1.25 = 3.75 km
a) Avg velocity `=(1.25)/((40//60))=1.875km//h`
b) Avg speed `=(3.75)/((40//60))=5.625km//h`