Question

A man walks on a straight road from his home to a market 2.5 km. away with a speed of 5 kmh^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 kmh^-1. what is the

(a) Magnitude of average velocity and

(b) Average speed of the man over the interval of time (t) 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.)

Answer 1

As, v = s/t ⇒ t = s/v

Time taken by the man to reach market,

t = s/v = 2.5/5 = 0.5 h

Time taken by the man to come back.

t1 = s/v₁ = 2.5/7.5 = 0.33 h.

(i) Average velocity (0 - 30 min)

= Δx/Δt = 2.5/0.5 = 5 kmh²¹

(In half i.e., 1/2 = 0.5 h, distance covered by man = 2.5 km).

(ii) Average-velocity (0 - 50 min)

(iii) Average velocity (0 - 40 min)

[Durimng first 30 min, distance covered = 2.5 km in next 10 min, distance covered = 2.5/2 km in return journey]

Average speed (0 - 40 min)