A body covers first one-third of the distance with a velocity 20m/s ,the second one-third with a velocity of 30m/s and the last one third with a velocity of 40m/s. The average velocity is nearly?
Answers
Explanation:
Velocity(v1) upto 1/3rd distance =20m/s
Velocity(v2) upto 2/3rd distance =30m/s
Velocity(v3) of last part=40m/s
Average velocity =3v1v2v3/v1v2+v2v3+v3v1
=3*20*30*40/600+1200+800
=72000/2600=27.6m/s
Let us assume that the total distance covered by the body is 'd'.
Case 1)
A body covers first one-third of the distance with a velocity 20m/s.
Given that, distance = 1/3 × d = d/3 and velocity = 20 m/s.
Time = Distance/Speed (velocity)
t1 = (d/3)/20
t1 = d/60 ..................(1st equation)
Case 2)
The second one-third with a velocity of 30m/s.
Distance = d/3 and velocity = 30 m/s
t2 = (d/3)/30
t2 = d/90 ..................(2nd equation)
Case 3)
The last one third with a velocity of 40m/s.
Distance = d/3 and velocity = 40 m/s
t3 = (d/3)/40
t3 = d/120 ................(3rd equation)
We have to find the average velocity of the body.
Average velocity = (Total distance or displacement covered)/(Total time taken)
Total time taken = t1 + t2 + t3
= d/60 + d/90 + d/120
= (6d + 4d + 3d)/360
= 13d/360
Average velocity = d/(13d/360)
= 360d/13d
= 27.69 m/s