An object has a mass of #3 kg#. The object's kinetic energy uniformly changes from #36 KJ# to #135 KJ# over #t in [0, 9 s]#. What is the average speed of the object?

1 Answer
May 1, 2018

Answer:

The average speed is #=235.2ms^-1#

Explanation:

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=3kg#

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=36000J#

The final kinetic energy is #1/2m u_2^2=135000J#

Therefore,

#u_1^2=2/3*36000=24000m^2s^-2#

and,

#u_2^2=2/3*135000=90000m^2s^-2#

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

The points are #(0,24000)# and #(9, 90000)#

The equation of the line is

#v^2-24000=(90000-24000)/9t#

#v^2=7333.3t+24000#

So,

#v=sqrt(7333.3t+24000)#

We need to calculate the average value of #v# over #t in [0,9]#

#(9-0)bar v=int_0^9(sqrt(7333.3t+24000))dt#

#9 barv= [(7333.3t+24000)^(3/2)/(3/2*7333.3)] _( 0) ^ (9)#

#=((7333.3*9+24000)^(3/2)/(11000))-((7333.3*0+24000)^(3/2)/(11000))#

#=90000^(3/2)/11000-24000^(3/2)/11000#

#=2116.5#

So,

#barv=2116.5/9=235.2ms^-1#

The average speed is #=235.2ms^-1#