An object has a mass of 2 kg. The object's kinetic energy uniformly changes from 32 KJ to 80 KJ over t in [0, 4 s]. What is the average speed of the object?

1 Answer
Narad T.
Aug 16, 2017

The average speed is =234.8ms^-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=32000J

The final kinetic energy is 1/2m u_2^2=80000J

Therefore,

u_1^2=2/2*32000=32000m^2s^-2

and,

u_2^2=2/2*80000=80000m^2s^-2

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

The points are (0,32000) and (4,80000)

The equation of the line is

v^2-32000=(80000-32000)/4t

v^2=12000t+32000

So,

v=sqrt((12000t+32000)

We need to calculate the average value of v over t in [0,4]

(4-0)bar v=int_0^4(sqrt(12000t+32000))dt

4 barv=[((12000t+32000)^(3/2)/(3/2*12000))]_0^4

=((12000*4+32000)^(3/2)/(18000))-((12000*0+32000)^(3/2)/(18000))

=80000^(3/2)/18000-32000^(3/2)/18000

=939.1

So,

barv=939.1/4=234.8ms^-1

The average speed is =234.8ms^-1

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