An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #640 KJ# to # 360 KJ# over #t in [0, 12 s]#. What is the average speed of the object?

1 Answer
Dec 20, 2017

Answer:

The average speed is #=406.9ms^-1#

Explanation:

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=6kg#

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=640000J#

The final kinetic energy is #1/2m u_2^2=360000J#

Therefore,

#u_1^2=2/6*640000=213333.3m^2s^-2#

and,

#u_2^2=2/6*360000=120000m^2s^-2#

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

The points are #(0,213333.3)# and #(12,120000)#

The equation of the line is

#v^2-213333.3=(120000-213333.3)/12t#

#v^2=-7777.8t+213333.3#

So,

#v=sqrt(-7777.8t+213333.3)#

We need to calculate the average value of #v# over #t in [0,12]#

#(12-0)bar v=int_0^12(sqrt(-7777.8t+213333.3))dt#

#12 barv= (-7777.8t+213333.3)^(3/2)/(3/2*-7777.8)| _( 0) ^ (12) #

#=((-7777.8*12+213333.3)^(3/2)/(-11666.7))-((-7777.8*0+213333.3)^(3/2)/(-11666.7))#

#=213333.3^(3/2)/11666.7-120000^(3/2)/11666.7#

#=4882.7#

So,

#barv=4882.7/12=406.9ms^-1#

The average speed is #=406.9ms^-1#