An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 128 KJ to 720 KJ over t in [0, 3 s]. What is the average speed of the object?

1 Answer
Narad T.
May 21, 2017

The average speed is =367.4ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

mass is =6kg

The initial velocity is =u_1

1/2m u_1^2=128000J

The final velocity is =u_2

1/2m u_2^2=720000J

Therefore,

u_1^2=2/6*128000=42666.7m^2s^-2

and,

u_2^2=2/6*720000=240000m^2s^-2

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

The points are (0,42666.7) and (3,240000)

The equation of the line is

v^2-42666.7=(240000-42666.7)/3t

v^2=65777.8t+42666.7

So,

v=sqrt((65777.8t+42666.7)

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

(3-0)bar v=int_0^8sqrt((65777.8t+42666.7))dt

3 barv=[((65777.8t+42666.7)^(3/2)/(3/2*65666.7)]_0^3

=((65777.8*3+42666.7)^(3/2)/(98666.7))-((65777.8*0+42666.7)^(3/2)/(98666.7))

=240000^(3/2)/98666.7-42666.7^(3/2)/98666.7

=1102.3

So,

barv=1103.2/3=367.4ms^-1

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