An object has a mass of 2 kg. The object's kinetic energy uniformly changes from 18 KJ to 64KJ over t in [0,12s]. What is the average speed of the object?

1 Answer
Narad T.
Mar 15, 2017

The average speed of the object is =199.6ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The initial velocity is =u_1

1/2m u_1^2=18000J

The final velocity is =u_2

1/2m u_2^2=64000J

Therefore,

u_1^2=2/2*18000=18000m^2s^-2

and,

u_2^2=2/2*64000=64000m^2s^-2

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

The points are (0,18000) and (12,64000)

The equation of the line is

v^2-18000=(64000-18000)/12t

v^2=3833.3t+18000

So,

v=sqrt((3833.3t+18000)

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

(12-0)bar v=int_0^12sqrt((3833.3t+18000)dt

12 barv=[((3833.3t+18000)^(3/2)/(3/2*3833.3)]_0^12

=((3833.3*12+18000)^(3/2)/(5750))-((3833.3*0+18000)^(3/2)/(5750))

=63999.6^(3/2)/5750-18000^(3/2)/5750

=2395.8

So,

barv=2395.8/12=199.6ms^-1

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