An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 144 KJ to 640KJ over t in [0, 3 s]. What is the average speed of the object?

1 Answer
Narad T.
May 14, 2017

The average speed is =434.68ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

mass is =4kg

The initial velocity is =u_1

1/2m u_1^2=144000J

The final velocity is =u_2

1/2m u_2^2=640000J

Therefore,

u_1^2=2/4*144000=72000m^2s^-2

and,

u_2^2=2/4*640000=320000m^2s^-2

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

The points are (0,72000) and (3,320000)

The equation of the line is

v^2-72000=(320000-72000)/3t

v^2=82666.67t+72000

So,

v=sqrt((82666.67t+72000)

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

(3-0)bar v=int_0^3sqrt((82666.67t+72000))dt

3 barv=[((82666.67t+72000)^(3/2)/(3/2*82666.67)]_0^3

=((82666.67*3+72000)^(3/2)/(124000))-((82666.67*0+72000)^(3/2)/(124000))

=320000^(3/2)/124000-72000^(3/2)/124000

=1304.03

So,

barv=1304.03/3=434.68ms^-1

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