An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 96 KJ to 270 KJ over t in [0, 12 s]. What is the average speed of the object?

1 Answer
Narad T.
May 5, 2017

The average speed is =299.5ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

mass is =4kg

The initial velocity is =u_1

1/2m u_1^2=96000J

The final velocity is =u_2

1/2m u_2^2=270000J

Therefore,

u_1^2=2/4*96000=48000m^2s^-2

and,

u_2^2=2/4*270000=135000m^2s^-2

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

The points are (0,48000) and (12,135000)

The equation of the line is

v^2-48000=(135000-48000)/12t

v^2=7250t+48000

So,

v=sqrt((7250t+48000)

We need to calculate the average value of v over t in [0,12]

(12-0)bar v=int_0^12sqrt((7250t+48000))dt

12 barv=[((7250t+48000)^(3/2)/(3/2*7250)]_0^12

=((7250*12+48000)^(3/2)/(10875))-((7250*0+48000)^(3/2)/(10875))

=135000^(3/2)/10875-48000^(3/2)/10875

=3594.1

So,

barv=3594.1/12=299.5ms^-1

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