An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 48 KJ to 225 KJ over t in [0, 8 s]. What is the average speed of the object?

1 Answer
Narad T.
Apr 7, 2017

The average speed is =209.2ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

mass is =6kg

The initial velocity is =u_1

1/2m u_1^2=48000J

The final velocity is =u_2

1/2m u_2^2=225000J

Therefore,

u_1^2=2/6*48000=16000m^2s^-2

and,

u_2^2=2/6*225000=75000m^2s^-2

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

The points are (0,16000) and (8,75000)

The equation of the line is

v^2-16000=(75000-16000)/8t

v^2=7375t+16000

So,

v=sqrt((7575t+16000)

We need to calculate the average value of v over t in [0,8]

(8-0)bar v=int_0^8sqrt((7375t+16000))dt

8 barv=[((7375t+16000)^(3/2)/(3/2*7375)]_0^8

=((7375*8+16000)^(3/2)/(11062.5))-((7375*0+16000)^(3/2)/(11062.5))

=75000^(3/2)/11062.5-16000^(3/2)/11062.5

=1673.7

So,

barv=1673.7/8=209.2ms^-1

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