Physics, asked by leelavenkat0303, 11 months ago

5. A man walks on a straight road from
his home to a market 2.5 km away with a
speed of 5 km/h. Finding the market closed,
he instantly turns and walks back home
with a speed of 7.5 km/h. The average
speed of the man over the interval of time
0 to 40 min, is equal to
1) 5 km/h
km/h
30
km/h​

Answers

Answered by PayelSen
0

Answer:

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

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