An object has a mass of 8 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.
Sep 18, 2017

The average speed is =307.4ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is =8kg

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=144000J

The final kinetic energy is 1/2m u_2^2=640000J

Therefore,

u_1^2=2/8*144000=36000m^2s^-2

and,

u_2^2=2/8*640000=160000m^2s^-2

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

The points are (0,36000) and (3,160000)

The equation of the line is

v^2-36000=(160000-36000)/3t

v^2=41333.3t+36000

So,

v=sqrt((41333.3t+36000)

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

(3-0)bar v=int_0^3(sqrt(41333.3t+36000))dt

3 barv=[((41333.3t+36000)^(3/2)/(3/2*41333.3))]_0^3

=((41333.3*3+36000)^(3/2)/(62000))-((41333.3*0+36000)^(3/2)/(62000))

=160000^(3/2)/62000-36000^(3/2)/62000

=922.1

So,

barv=922.1/3=307.4ms^-1

The average speed is =307.4ms^-1

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