An object has a mass of #12 kg#. The object's kinetic energy uniformly changes from #254 KJ# to # 24 KJ# over #t in [0, 5 s]#. What is the average speed of the object?

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Answer 1 · 2018-01-09T11:01:57

The average speed is #=147.1ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=12kg#

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=254000J#

The final kinetic energy is #1/2m u_2^2=24000J#

Therefore,

#u_1^2=2/12*254000=42333.3m^2s^-2#

and,

#u_2^2=2/12*24000=4000m^2s^-2#

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

The points are #(0,42333.3)# and #(5,4000)#

The equation of the line is

#v^2-42333.3=(4000-42333.3)/5t#

#v^2=-7666.7t+42333.3#

So,

#v=sqrt(-7666.7t+42333.3)#

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

#(5-0)bar v=int_0^5(sqrt(-7666.7t+42333.3))dt#

#5 barv= (-7666.7t+42333.3)^(3/2)/(3/2*-7666.7)| _( 0) ^ (5) #

#=((-7666.7*5+42333.3)^(3/2)/(-11500))-((-7666.7*0+42333.3)^(3/2)/(-11500))#

#=42333.3^(3/2)/11500-4000^(3/2)/11500#

#=735.4#

So,

#barv=735.4/5=147.1ms^-1#

The average speed is #=147.1ms^-1#