a man walks on a straight road from his home to market 2.5 away with a speed of 5km/h finding the market closed , he intantly turns and walks back to home with a speed of 7.5km h. what is the a. magnitude of average velocity and b. average speed of the man over the time interval 0- 50 mins.
Answers
Answer:
Solution
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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