An object has a mass of 3 kg. The object's kinetic energy uniformly changes from 12 KJ to 72KJ over t in [0,5s]. What is the average speed of the object?

1 Answer
Narad T.
May 5, 2017

The average speed is =163.3ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

mass is =3kg

The initial velocity is =u_1

1/2m u_1^2=12000J

The final velocity is =u_2

1/2m u_2^2=72000J

Therefore,

u_1^2=2/3*12000=8000m^2s^-2

and,

u_2^2=2/3*72000=48000m^2s^-2

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

The points are (0,8000) and (5,48000)

The equation of the line is

v^2-8000=(48000-8000)/5t

v^2=8000t+8000

So,

v=sqrt((8000t+8000)

We need to calculate the average value of v over t in [0,5]

(5-0)bar v=int_0^4sqrt((8000t+8000))dt

5 barv=[((8000t+8000)^(3/2)/(3/2*8000)]_0^5

=((8000*5+8000)^(3/2)/(12000))-((8000*0+8000)^(3/2)/(12000))

=48000^(3/2)/12000-8000^(3/2)/12000

=816.7

So,

barv=816.7/5=163.3ms^-1

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