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in Physics by (33.0k points)

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

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 !] 

1 Answer

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Best answer

v1 = 5 kmh-1, v2 = 7.5 kmh-1, d = 2.5 km

Time for the man to reach the market

= t1 = d/v1= 2.5/5 = 0.5 h = 30 min

Time for the man to return from the market

= t= d/v2 = 2.5/7.5 = 1/3 h= 20 min 

1. To find the average velocity,

Average velocity = Total path length/Total time taken

(a) During the interval between 0 to 30 min,

Displacement = 2.5 km

time = 30 minutes

∴ Average velocity = 2.5 km/30 min = 5 kmh-1 

(b) interval between 0 and 50 minutes,

Displacement = 0

∴ Average velocity = 0 

(c) During between 0 to 40 minutes,

Displacement = 2.5 km – 7.5 kmh-1 × 10 min = 1.25 km

time = 40 minutes

∴ Average velocity = 1.25/40 min = 1.875 kmh-1

2. To find the average speed:

Average speed = Total path length/Total time taken

(a) During the interval between 0 to 30 min

Path length = 2.5 km

time = 30 minutes

∴ Average speed = 2.5 km/30 min = 5 kmh-1 

(b) During the interval between 0 and 50 minutes,

Path length = 2.5 km + 2.5 km

Time taken = 50 minutes

∴ Average speed = (2.5 km +  2.5 km)/50 min = 6 kmh-1 

(c) During 0 to 40 minutes

Path length = 5 kmh-1 × 30 min + 7.5 kmh-1 × 10 min

2.5 km + 1.25 km = 3.75 km

time = 40 minutes

∴ Average velocity = 3.75 km/40 min = 5.625 kmh-1

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