An object has a mass of 2 kg. The object's kinetic energy uniformly changes from 64 KJ to 72 KJ over t in [0, 3 s]. What is the average speed of the object?

1 Answer
Narad T.
Sep 20, 2017

The average speed is =260.7ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is =2kg

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=64000J

The final kinetic energy is 1/2m u_2^2=72000J

Therefore,

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

and,

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

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

The points are (0,64000) and (3,72000)

The equation of the line is

v^2-64000=(72000-64000)/3t

v^2=2666.7t+64000

So,

v=sqrt((2666.7t+64000)

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

(3-0)bar v=int_0^3(sqrt(2666.7t+64000))dt

3 barv=[((2666.7t+64000)^(3/2)/(3/2*2666.7))]_0^3

=((2666.7*3+64000)^(3/2)/(4000))-((2666.7*0+64000)^(3/2)/(4000))

=72000^(3/2)/4000-64000^(3/2)/4000

=782.2

So,

barv=782.2/3=260.7ms^-1

The average speed is =260.7ms^-1

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