An object has a mass of #8 kg#. The object's kinetic energy uniformly changes from #24 KJ# to # 42KJ# over #t in [0, 9 s]#. What is the average speed of the object?

1 Answer
Apr 21, 2017

Answer:

The average speed is #90.5ms^-1#

Explanation:

The kinetic energy is

#KE=1/2mv^2#

mass is #=8kg#

The initial velocity is #=u_1#

#1/2m u_1^2=24000J#

The final velocity is #=u_2#

#1/2m u_2^2=42000J#

Therefore,

#u_1^2=2/8*24000=6000m^2s^-2#

and,

#u_2^2=2/8*42000=10500m^2s^-2#

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

The points are #(0,6000)# and #(9,10500)#

The equation of the line is

#v^2-6000=(10500-6000)/9t#

#v^2=500t+6000#

So,

#v=sqrt((500t+6000)#

We need to calculate the average value of #v# over #t in [0,9]#

#(9-0)bar v=int_0^9sqrt((500t+6000))dt#

#9 barv=[((500t+6000)^(3/2)/(3/2*500)]_0^9#

#=((500*9+6000)^(3/2)/(750))-((500*0+6000)^(3/2)/(750))#

#=10500^(3/2)/750-6000^(3/2)/750#

#=814.9#

So,

#barv=814.9/9=90.5ms^-1#