wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5kmh1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5kmh1. 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!]

Open in App
Solution

Time taken by the man to reach the market from home,
t1=2.55=12 h=30 min
Time taken by the man to reach home from the market, t2=2.57.5=13 h=20 min
Total time taken in the whole journey = 30 + 20 = 50 min

(i) 0 to 30 min

Average velocity=DisplacementTime=2.512=5 km/hAverage speed=DistanceTime=2.512=5 km/h

(ii) 0 to 50 min = 56 h

Net displacement=0
Average velocity=DisplacementTime=0Average speed=DistanceTime=2.5+2.5(56)=6 km/h

(iii) 0 to 40 min = 23 h

Distance travelled in first 30 min = 2.5 km
Speed of the man while returning= 7.5 km/h
Distance travelled by the man (from market to home) in the next 10 min
=7.5×1060=1.25 km
Net displacement = 2.5 - 1.25 = 1.25 km
Total distance travelled = 2.5 + 1.25 = 3.75 km
Average velocity=1.25(23)=1.875 km/hAverage speed=3.75(23)=5.625 km/h


flag
Suggest Corrections
thumbs-up
83
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon