Physics, asked by anamikameena1202, 5 months ago

Example 4 A man walks on a straight road from his home to a market 3 km away with a
speed of 6 km/h finding the market closed, he instantly turns and walks back with a speed of 9 km/h.
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?​


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Answers

Answered by syatul1981
4

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

Answered by Anonymous
2

Answer:

Average speed from home to market: 6 kmph

time taken from home to market: 3 km/ 6 kmph = 1/2 hr = 30 min

So this is the average speed for 0 to 30 min.: 6 kmph

Average speed from market to home 9 kmph Time taken from market to home on the way back: 3 km/9kmph = 1/3 hour

Average speed for the total journey: from to market and market to home :

= total distance / total time = (3 km +3 km )/[ 1/2 +1/3 Jhr

= 6/(5/6) = 36/5 = 7.2 kmph

This is average speed for 0 to 50 min

distance traveled from 30 to 40 minutes, that

is 10 min: 1/6 hr

= 9 kmph * 1/6 hour = 1.5 km

Average speed for time =0 to 40 min is = distance traveled / time taken = (3 +

1.5) / (40/60)

= 4.5 * 60 / 40 = 6.75 kmph

Velocity = change in displacement vector = vector joining initial position of man and

final position of man

Velocity is dependent on the current position of man and not the distance traveled by man.

Velocity initially at t=0 is 0, as the man is at

home.

Velocity for t = 0 to 30 minutes is : 6 kmph towards the market.

Velocity at t = 50 minutes is zero as the man is at home. THat is, displacement of the man from initial position is zero. So displacement/time = 0.

Distance of the man from home at t = 40 min is : 3 - 1.5 km = 1.5 km Velocity at time 40 minutes is: 1.5 km/(40/60)

hr = 2.25 kmph

The direction of velocity is in the direction towards market from home.


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