An object has a mass of #2 kg#. The object's kinetic energy uniformly changes from #18 KJ# to # 54KJ# over #t in [0,12s]#. What is the average speed of the object?

1 Answer
Apr 14, 2017

Answer:

The average speed is #=187.7ms^-1#

Explanation:

The kinetic energy is

#KE=1/2mv^2#

mass is #=2kg#

The initial velocity is #=u_1#

#1/2m u_1^2=18000J#

The final velocity is #=u_2#

#1/2m u_2^2=54000J#

Therefore,

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

and,

#u_2^2=2/2*54000=54000m^2s^-2#

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

The points are #(0,18000)# and #(12,54000)#

The equation of the line is

#v^2-18000=(54000-18000)/12t#

#v^2=3000t+18000#

So,

#v=sqrt((3000t+18000)#

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

#(12-0)bar v=int_0^12sqrt((3000t+18000))dt#

#12 barv=[((3000t+18000)^(3/2)/(3/2*3000)]_0^12#

#=((3000*12+18000)^(3/2)/(4500))-((3000*0+18000)^(3/2)/(4500))#

#=54000^(3/2)/4500-18000^(3/2)/4500#

#=2251.9#

So,

#barv=2251.9/12=187.7ms^-1#