Physics, asked by arjungupta0335, 10 months ago

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. 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 and (iii) 0 to 40 min?

Answers

Answered by maniyachawla12
3

Answer:This may help you

Explanation:

Average speed =

totaltime

totaldistance

Time taken to go to the market =

speed

distance

=

5

2.5

hour=

2

1

hours=30 minutes

Time taken to return back =

speed

distance

=

7.5

2.5

hour=1/3hours=20 minutes

Distance travelled in 10 minutes while returning back=

60

10

×7.5=1.25km ( Speed is constant at 7.5 km/hr )

Hence Distance traveled in 40 minutes(30 mins towards the market and 10 mins towards home) =2.5+1.25=3.75km

So average speed in 0 to 40 minutes =

60

40

hr

3.75km

=

8

45

km/hr

Answered by Anonymous
0

Answer:

845

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