An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 640 KJ to 360 KJ over t in [0, 12 s]. What is the average speed of the object?

1 Answer
Narad T.
Dec 20, 2017

The average speed is =406.9ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is m=6kg

The initial velocity is =u_1ms^-1

The final velocity is =u_2 ms^-1

The initial kinetic energy is 1/2m u_1^2=640000J

The final kinetic energy is 1/2m u_2^2=360000J

Therefore,

u_1^2=2/6*640000=213333.3m^2s^-2

and,

u_2^2=2/6*360000=120000m^2s^-2

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

The points are (0,213333.3) and (12,120000)

The equation of the line is

v^2-213333.3=(120000-213333.3)/12t

v^2=-7777.8t+213333.3

So,

v=sqrt(-7777.8t+213333.3)

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

(12-0)bar v=int_0^12(sqrt(-7777.8t+213333.3))dt

12 barv= (-7777.8t+213333.3)^(3/2)/(3/2*-7777.8)| _( 0) ^ (12)

=((-7777.8*12+213333.3)^(3/2)/(-11666.7))-((-7777.8*0+213333.3)^(3/2)/(-11666.7))

=213333.3^(3/2)/11666.7-120000^(3/2)/11666.7

=4882.7

So,

barv=4882.7/12=406.9ms^-1

The average speed is =406.9ms^-1

Related questions
See all questions in Displacement and Velocity
Impact of this question
1352 views around the world
You can reuse this answer
Creative Commons License