An object has a mass of 1 kg. The object's kinetic energy uniformly changes from 64 KJ to 36 KJ over t in [0, 3 s]. What is the average speed of the object?

1 Answer
Narad T.
May 8, 2017

The average speed is =315.2ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

mass is =1kg

The initial velocity is =u_1

1/2m u_1^2=64000J

The final velocity is =u_2

1/2m u_2^2=36000J

Therefore,

u_1^2=2/1*64000=128000m^2s^-2

and,

u_2^2=2/1*36000=72000m^2s^-2

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

The points are (0,128000) and (3,72000)

The equation of the line is

v^2-128000=(72000-128000)/3t

v^2=-18666.7t+128000

So,

v=sqrt((-18666.7t+128000)

We need to calculate the average value of v over t in [0,3]

(3-0)bar v=int_0^3sqrt((-18666.7t+128000))dt

3 barv=[((-18666.7t+128000)^(3/2)/(-3/2*18666.7)]_0^3

=((-18666.7*3+128000)^(3/2)/(-28000))-((-18666.7*0+128000)^(3/2)/(-28000))

=128000^(3/2)/28000-72000^(3/2)/28000

=945.5

So,

barv=945.5/3=315.2ms^-1

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