An object has a mass of 12 kg. The object's kinetic energy uniformly changes from 81 KJ to 18 KJ over t in [0, 4 s]. What is the average speed of the object?

1 Answer
Narad T.
Mar 15, 2017

The average speed is =89.16ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The initial velocity is =u_1

1/2m u_1^2=81000J

The final velocity is =u_2

1/2m u_2^2=18000J

Therefore,

u_1^2=2/12*81000=13500m^2s^-2

and,

u_2^2=2/12*18000=3000m^2s^-2

The graph of v^2=f(t) is a straight line

The points are (0,13500) and (4,3000)

The equation of the line is

v^2-13500=(3000-13500)/4t

v^2=-2625t+13500

So,

v=sqrt((-2625t+13500)

We need to calculate the average value of v over t in [0,4]

(4-0)bar v=int_0^5sqrt(-2625t+13500) dt

4 barv=[(-2625t+13500)^(3/2)/(3/2*-2625)]_0^4

=((-2625*4+13500)^(3/2)/(-3937.5))-((-2625*0+13500)^(3/2)/(-3937.5))

=3000^(3/2)/-3937.5-13500^(3/2)/-3937.5

=1/3937.5(13500^(3/2)-3000^(3/2)))

=356.63

So,

barv=356.63/4=89.16ms^-1

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