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 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=1/3h`
`therefore` Total time taken = `t_1+t_2=1/2+1/3=5/6h`
=50 min.
In time interval 0 to 50 min
Total distance travelled =2.5+2.5 =5 km.
Total displacement = zero.
a) Average velocity= `(displacement)/(time)=0`
b) Average speed =`(distance)/(time)=5/(5//6)`
`=6km//h`