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Question
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 magnitude of average velocity, and
- average speed of the man over the interval of time
- 0 to 30 min,
- 0 to 50 min,
- 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!]
Solution
Time taken by the man to reach the market from home, `t_1 = 2.5/5 = 0.5 = 30 min`
Time taken by the man to reach home from the market, `t_2 = 2.5/7.5 = 0.333 = 20 min`
Total time taken in the whole journey = 30 + 20 = 50 min
(i) Over the interval of time 0 to 30 min:
`"Average velocity" = "Displacement"/"Time" = 2.5/0.5 = 5 "km/h"`
`"Average speed" = "Distance"/"Time" = 2.5/0.5 = 5 "km/h"`
(ii) Over the interval of time 0 to 50 min:
Time = 50 min = `5/6 h`
Net displacement = 0
Total distance = 2.5 + 2.5 = 5 km
`"Average velocity" = "Displacement"/"Time" = 0/(5/6) = 0`
`"Average speed" = "Distance"/"Time" = 5/(5/6) = 6 "km/h"`
(iii) Over the interval of time 0 to 40 min:
Speed of the man = 7.5 km
Distance travelled in first 30 min = 2.5 km
Distance travelled by the man (from market to home) in the next 10 min `= 7.5xx 10/60 = 1.25`
Net displacement = 2.5 - 1.25 = 1.25 km
Total distance travelled = 2.5 + 1.25 = 3.75 km
`"Average velocity" = 1.25/(40/60) = (1.25xx3)/2 = 1.875 "km/h"`
Average speed =`3.75/(40/60) = 5.625 "km/h"`
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