An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #104 KJ# to # 64KJ# over #t in [0,5s]#. What is the average speed of the object?

1 Answer
Apr 5, 2017

Answer:

The average speed is #=204.4ms^-1#

Explanation:

The kinetic energy is

#KE=1/2mv^2#

mass is #=4kg#

The initial velocity is #=u_1#

#1/2m u_1^2=104000J#

The final velocity is #=u_2#

#1/2m u_2^2=64000J#

Therefore,

#u_1^2=2/4*104000=52000m^2s^-2#

and,

#u_2^2=2/4*64000=32000m^2s^-2#

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

The points are #(0,52000)# and #(5,32000)#

The equation of the line is

#v^2-52000=(32000-52000)/5t#

#v^2=-4000t+52000#

So,

#v=sqrt((-4000t+52000)#

We need to calculate the average value of #v# over #t in [0,5]#

#(5-0)bar v=int_0^5sqrt((-4000t+52000))dt#

#5 barv=[((-4000t+52000)^(3/2)/(-3/2*4000)]_0^5#

#=((-4000*5+52000)^(3/2)/(-6000))-((-4000*0+52000)^(3/2)/(-6000))#

#=-32000^(3/2)/6000+52000^(3/2)/6000#

#=1022.2#

So,

#barv=1022.2/5=204.4ms^-1#