Question

A man walks on a straight road from his home to a market 2.5 \(\text {km}\) away with a speed of 5 \(\text {km}\) \(\text h^{-1}\). Finding the market closed, he instantly turns and walks back home with a speed of 7.5 \(\text {km}\) \(\text h^{-1}\) What is the 

1. magnitude of average velocity, and 
2. 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 !].

Answer 1

Given:

Distance to market = 2.5 km
Speed while going = 5 km h⁻¹
Speed while returning = 7.5 km h⁻¹

Time taken to reach market:

t = 2.5 / 5 = 0.5 h = 30 min

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(i) Time interval: 0 to 30 min

Distance travelled = 2.5 km
Displacement = 2.5 km
Time = 0.5 h

Magnitude of average velocity:

= Displacement / Time = 2.5 / 0.5 = 5 km h⁻¹

Average speed:

= Total distance / Time = 2.5 / 0.5 = 5 km h⁻¹

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(ii) Time interval: 0 to 50 min

Time = 50 min = 5/6 h

Distance covered while returning in 20 min (1/3 h):

= 7.5 × 1/3 = 2.5 km

Total distance travelled = 2.5 + 2.5 = 5.0 km
Net displacement = 0 (returns home)

Magnitude of average velocity:

= 0 / (5/6) = 0 km h⁻¹

Average speed:

= 5.0 / (5/6) = 6.0 km h⁻¹

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(iii) Time interval: 0 to 40 min

Time = 40 min = 2/3 h

Distance covered while returning in 10 min (1/6 h):

= 7.5 × 1/6 = 1.25 km

Total distance travelled = 2.5 + 1.25 = 3.75 km
Net displacement = 2.5 − 1.25 = 1.25 km

Magnitude of average velocity:

= 1.25 / (2/3) = 1.88 km h⁻¹

Average speed:

= 3.75 / (2/3) = 5.63 km h⁻¹

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Final Answers:

(i) 0–30 min:
Average velocity = 5 km h⁻¹, Average speed = 5 km h⁻¹

(ii) 0–50 min:
Average velocity = 0 km h⁻¹, Average speed = 6 km h⁻¹

(iii) 0–40 min:
Average velocity = 1.88 km h⁻¹, Average speed = 5.63 km h⁻¹