An object has a mass of #2 kg#. The object's kinetic energy uniformly changes from #48 KJ# to # 54KJ# over #t in [0,12s]#. What is the average speed of the object?

1 Answer
May 6, 2017

Answer:

The average speed is #=159.7ms^-1#

Explanation:

The kinetic energy is

#KE=1/2mv^2#

mass is #=2kg#

The initial velocity is #=u_1#

#1/2m u_1^2=48000J#

The final velocity is #=u_2#

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

Therefore,

#u_1^2=2/4*48000=24000m^2s^-2#

and,

#u_2^2=2/4*54000=27000m^2s^-2#

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

The points are #(0,24000)# and #(12,27000)#

The equation of the line is

#v^2-24000=(27000-24000)/12t#

#v^2=250t+24000#

So,

#v=sqrt((250t+24000)#

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

#(12-0)bar v=int_0^12sqrt((250t+24000))dt#

#12 barv=[((250t+24000)^(3/2)/(3/2*250)]_0^12#

#=((250*12+24000)^(3/2)/(375))-((250*0+24000)^(3/2)/(375))#

#=27000^(3/2)/375-24000^(3/2)/375#

#=1916#

So,

#barv=1916/12=159.7ms^-1#