An object has a mass of #2 kg#. The object's kinetic energy uniformly changes from #32 KJ# to #96 KJ# over #t in [0, 4 s]#. What is the average speed of the object?

1 Answer
Apr 24, 2017

Answer:

The average speed is #=250.2ms^-1#

Explanation:

The kinetic energy is

#KE=1/2mv^2#

mass is #=2kg#

The initial velocity is #=u_1#

#1/2m u_1^2=32000J#

The final velocity is #=u_2#

#1/2m u_2^2=96000J#

Therefore,

#u_1^2=2/2*32000=32000m^2s^-2#

and,

#u_2^2=2/2*96000=96000m^2s^-2#

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

The points are #(0,32000)# and #(4,96000)#

The equation of the line is

#v^2-32000=(96000-32000)/4t#

#v^2=16000t+32000#

So,

#v=sqrt((16000t+32000)#

We need to calculate the average value of #v# over #t in [0,4]#

#(4-0)bar v=int_0^4sqrt((16000t+32000))dt#

#4 barv=[((16000t+32000)^(3/2)/(3/2*16000)]_0^4#

#=((16000*4+32000)^(3/2)/(24000))-((16000*0+32000)^(3/2)/(24000))#

#=96000^(3/2)/24000-32000^(3/2)/24000#

#=1000.8#

So,

#barv=1000.8/4=250.2ms^-1#