An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 128 KJ to 36 KJ over t in [0, 12 s]. What is the average speed of the object?

1 Answer
Narad T.
Nov 14, 2017

The average speed is =200ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is m=4kg

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=128000J

The final kinetic energy is 1/2m u_2^2=36000J

Therefore,

u_1^2=2/4*128000=64000m^2s^-2

and,

u_2^2=2/4*36000=18000m^2s^-2

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

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

The equation of the line is

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

v^2=-3833.3t+64000

So,

v=sqrt(-3833.3t+64000)

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

(12-0)bar v=int_0^12(sqrt(-3833.3t+64000))dt

12 barv=[((-3833.3t+64000)^(3/2)/(-3/2*3833.3))]_0^12

=((-3833.3*12+64000)^(3/2)/(-5750))-((-3833.3*0+64000)^(3/2)/(-5750))

=64000^(3/2)/5750-18000^(3/2)/5750

=2395.8

So,

barv=2395.8/12=200ms^-1

The average speed is =200ms^-1

Related questions
See all questions in Displacement and Velocity
Impact of this question
1465 views around the world
You can reuse this answer
Creative Commons License