An object has a mass of #3 kg#. The object's kinetic energy uniformly changes from #48 KJ# to # 38KJ# over #t in [0,12s]#. What is the average speed of the object?

1 Answer
Mar 13, 2018

Answer:

The average speed is #=169.2ms^-1#

Explanation:

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=3kg#

The initial velocity is #=u_1ms^-1#

The final velocity is #=u_2 ms^-1#

The initial kinetic energy is #1/2m u_1^2=48000J#

The final kinetic energy is #1/2m u_2^2=38000J#

Therefore,

#u_1^2=2/3*48000=32000m^2s^-2#

and,

#u_2^2=2/3*38000=25333.3m^2s^-2#

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

The points are #(0,32000)# and #(12,25333.3)#

The equation of the line is

#v^2-32000=(25333.3-32000)/12t#

#v^2=-555.6t+32000#

So,

#v=sqrt(-555.6t+32000)#

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

#(12-0)bar v=int_0^12(sqrt(-555.6t+32000))dt#

#12 barv= [(-555.6t+32000)^(3/2)/(-3/2*555.6)] _( 0) ^ (12)#

#=((-555.6*8+32000)^(3/2)/(-833.3))-((-555.6*0+32000)^(3/2)/(-833.3))#

#=32000^(3/2)/833.3-25333.3^(3/2)/833.3#

#=2030.7#

So,

#barv=2030.7/12=169.2ms^-1#

The average speed is #=169.2ms^-1#