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?
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Answered by
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
0
Answer:
845
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