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) O to
50 min, (iii) 0 to 40 roin ? (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 !]
onood
Answers
Answer:
hope this helps you
Explanation:
Distance to market s=2.5km=2.5×10
3
=2500m
Speed with which he goes to market =5km/h=5
3600
10
3
=
18
25
m/s
Speed with which he comes back =7.5km/h=7.5×
3600
10
3
=
36
75
m/s
(a)Average velocity is zero since his displacement is zero.
(b) (i)Since the initial speed is 5km/s and the market is 2.5 km away,time taken to reach market:
5
2.5
=1/2h=30 minutes.
Average speed over this interval =5km/h
(ii)After 30 minutes,the man is travelling wuth 7.5 km/h speed for 50-30=20 minutes.The distance he covers in 20 minutes :7.5×
3
1
=2.5km
His average speed in 0 to 50 minutes: V
avg
=
time
distancetraveled
=
(50/60)
2.5+2.5
=6km/h
(iii)In 40-30=10 minutes he travels a distance of :7.5×
6
1
=1.25km
V
avg
=
(40/60)
2.5+1.25
=5.625km/h