Question

Question 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!]

Class XI Physics Motion In A Straight Line Page 57

Answer 1

Answer:-

Time taken by the man to reach the market from home,t1 = 2.5/5 = 1/2 h = 30 min
Time taken by the man to reach home from the market, t2 = 2.5/7.5 = 1/3 h = 20 min
Total time taken in the whole journey = 30 + 20 = 50 min

(i) 0 to 30 min
Average velocity = Displacement/Time = 2.5/(1/2) = 5 km/h
Average speed = Distance/Time = 2.5/(1/2) = 5 km/h

(ii) 0 to 50 min
Time = 50 min = 50/60 = 5/6 h
Net displacement = 0
Total distance = 2.5 + 2.5 = 5 km
Average velocity = Displacement / Time = 0
Average speed = Distance / Time = 5/(5/6) = 6 km/h

(iii) 0 to 40 min
Speed of the man = 7.5 km/h
Distance travelled in first 30 min = 2.5 km
Distance travelled by the man (from market to home) in the next 10 min
= 7.5 × 10/60 = 1.25 km
Net displacement = 2.5 – 1.25 = 1.25 km
Total distance travelled = 2.5 + 1.25 = 3.75 km
Average velocity = Displacement / Time = 1.25 / (40/60) = 1.875 km/h
Average speed = Distance / Time = 3.75 / (40/60) = 5.625 km/h

Answer 2

heya !
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Time taken by men to go from home to market = t1 
t1 = distance / time = 2.5/5 = 1/2 h = 30 minute 

Time taken by man to go from market to home 2.5/7.5 =1/3h = 20 minutes

Total time taken = t1+t2 = 30 + 20 = 50 minutes

I) 0 to 30 minutes
 
Average velocity = displacement / time
                              = 2.5/1/2 = 5 km/ hr

Average velocity =  2.5/ 1/2  = 5 km/hr 

ii) 0 to 50 

 total path length covered = 2.5 + 2.5 = 5 km 
Total displacement = 0 

a) Av. Velocity = 0 

b) Av. Speed =  5 / 5/6  = 6 km/h 

iii) 0 to 40 

Distance moved in 30 min = 2.5 km 
Distance moved in 10 minutes = speed x time = 5 x 10 /60 = 0.8 km 
displacement =  2.5 - 0.8 = 1.7 km 
Total path length travelled = 2.5 + 1.7 = 4.2 km

a) Av . Velocity = 1.7 / 40/ 60 = 2.52 km/hr 

B) Av. speed = 4.2 / 40/60    = 6.3 km/hr 
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