An object has a mass of 3 kg. The object's kinetic energy uniformly changes from 48 KJ to 32KJ over t in [0,6s]. What is the average speed of the object?

1 Answer
Narad T.
May 7, 2017

The average speed is =163ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

mass is =3kg

The initial velocity is =u_1

1/2m u_1^2=48000J

The final velocity is =u_2

1/2m u_2^2=32000J

Therefore,

u_1^2=2/3*48000=32000m^2s^-2

and,

u_2^2=2/3*32000=21333.3m^2s^-2

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

The points are (0,32000) and (6,21333.3)

The equation of the line is

v^2-32000=(21333.3-32000)/6t

v^2=-1777.8t+32000

So,

v=sqrt((-1777.8t+32000)

We need to calculate the average value of v over t in [0,6]

(6-0)bar v=int_0^6sqrt((-1777.8t+32000))dt

6 barv=[((-1777.8t+32000)^(3/2)/(-3/2*1777.8)]_0^6

=((-1777.8*6+32000)^(3/2)/(-2666.7))-((-1777.8*0+32000)^(3/2)/(-2666.7))

=32000^(3/2)/2666.7-21333.3^(3/2)/2666.7

=978.1

So,

barv=978.1/6=163ms^-1

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