An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #18 KJ# to # 64KJ# over #t in [0,12s]#. What is the average speed of the object?

1 Answer
Apr 17, 2018

Answer:

The average speed is #=115.3ms^-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=18000J#

The final kinetic energy is #1/2m u_2^2=64000J#

Therefore,

#u_1^2=2/6*18000=6000m^2s^-2#

and,

#u_2^2=2/6*64000=21333.3m^2s^-2#

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

The points are #(0,6000)# and #(12,21333.3)#

The equation of the line is

#v^2-6000=(21333.3-6000)/12t#

#v^2=1277.8t+6000#

So,

#v=sqrt(1277.8t+6000)#

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

#(12-0)bar v=int_0^12(sqrt(1277.8t+6000))dt#

#12 barv= [(1277.8t+6000)^(3/2)/(3/2*1277.8)] _( 0) ^ (12)#

#=((1277.8*12+6000)^(3/2)/(1916.7))-((1277.8*0+6000)^(3/2)/(1916.7))#

#=21333.3^(3/2)/1916.7-6000^(3/2)/1916.7#

#=1383.2#

So,

#barv=1383.2/12=115.3ms^-1#

The average speed is #=115.3ms^-1#