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

Average velocity `= ("Total displacement")/("Time interval")`
and average speed `= ("Path length")/("Time interval")`
(i) For 0 to 30 minutes time interval,
Distance covered from home to market , `=2.5 km`
Speed =5 km/hr
`therefore` Time taken in market `= (2.5)/(5) = 0.5hours = 30` minutes
(a) Average velocity `= ("displacement")/("time") = (2.5)/(0.5) =6km//hr`
(b) Average speed `= (2.5)/(0.5) = 5 km//hr`
(ii) For 0 to 50 minutes time interval,
Velcity while returning back `=7.5 km//hr`
distance `=2.5 km`
`therefore ` Time taken to cover this distance `= (2.5)/(7.5) = 1/3` hours = 20 minutes
Time taken to complete the travelling `=30+20 ` (home to market and market to home)
= 50 minutes `=5/6` hours
Displacement `=2.5 -2.5 =0 km`
`therefore` Average velocity `= (0)/(5//6)=0`
Total distance covered `=2.5 + 2.5 =5 km`
Average speed `= (5)/(5//6) =6` km/hr
(iii) For time interval 0 to 40 minutes,
Time taken by man to reach market from home,
=30 minutes
`therefore ` Time to return back,
`= 40 -30= 10` minutes
Distance covered in 10 minuutes,
= velocity while returning `xx` time
`=7.5 xx (10)/(60) = 1.25 km`
Displacement `=2.5 - 1.25 =1.25 km`
`therefore ` Average velocity `= (1.25)/(40//60)=1.875km//hr`
Avereage speed `= ("total distance")/("time")`
`=(2.5 +1.25)/(40//60) =3.75 xx (60)/(40)=5.625` km/hr
In second case, average velocity is zero but not average speed.