An object has a mass of #5 kg#. The object's kinetic energy uniformly changes from #36 KJ# to #135 KJ# over #t in [0, 9 s]#. What is the average speed of the object?

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Answer 1 · 2017-05-25T10:11:18

The average speed is #=182.2ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=5kg#

The initial velocity is #=u_1#

#1/2m u_1^2=36000J#

The final velocity is #=u_2#

#1/2m u_2^2=135000J#

Therefore,

#u_1^2=2/5*36000=14400m^2s^-2#

and,

#u_2^2=2/5*135000=54000m^2s^-2#

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

The points are #(0,14400)# and #(9,54000)#

The equation of the line is

#v^2-14400=(54000-14400)/9t#

#v^2=4400t+14400#

So,

#v=sqrt((4400t+14400)#

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

#(9-0)bar v=int_0^9sqrt((4400t+14400))dt#

#9 barv=[((4400t+14400)^(3/2)/(3/2*4400)]_0^9#

#=((4400*9+14400)^(3/2)/(6600))-((4400*0+14400)^(3/2)/(6600))#

#=54000^(3/2)/6600-14400^(3/2)/6600#

#=1639.5#

So,

#barv=1639.5/9=182.2ms^-1#