An object has a mass of 5 kg. The object's kinetic energy uniformly changes from 15 KJ to 42KJ over t in [0,8s]. What is the average speed of the object?

1 Answer
Narad T.
May 14, 2017

The average speed is =105.73ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

mass is =5kg

The initial velocity is =u_1

1/2m u_1^2=15000J

The final velocity is =u_2

1/2m u_2^2=42000J

Therefore,

u_1^2=2/5*15000=6000m^2s^-2

and,

u_2^2=2/5*42000=16800m^2s^-2

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

The points are (0,6000) and (8,16800)

The equation of the line is

v^2-6000=(16800-6000)/8t

v^2=1350t+6000

So,

v=sqrt((1350t+6000)

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

(8-0)bar v=int_0^8sqrt((3200t+10666.67))dt

8 barv=[((1350t+6000)^(3/2)/(3/2*1350)]_0^8

=((1350*8+6000)^(3/2)/(2025))-((1350*0+6000)^(3/2)/(2025))

=16800^(3/2)/2025-6000^(3/2)/2025

=845.81

So,

barv=845.81/8=105.73ms^-1

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