3.14 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 !]
3.14 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 !]
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1 Answer
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3.14
The distance from home to market = 2.5 km = 2500 m
The speed of the walking while going = 5 kmph = 1.388 m/s
The speed of walking back from market to home = 7.5 kmph = 2.083 m/s
(a) Magnitude of the average velocity = Displacement / time = 0, since the total displacement is zero.
(b)
i. Time taken to reach the market = distance / onward speed = 2500/1.388 = 1801 s = 30 minutes. So the average speed over 0 – 30 min is 5 kmph
ii. Time taken to reach back home = 2500 / 2.083 = 20 minutes. So the average speed = total distance covered / total time taken = (2500 +2500)/ (30 +20) m/min = 100 m/min = 6 kmph
iii. Average speed ove
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[P] = MLT-1
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