An object has a mass of #9 kg#. The object's kinetic energy uniformly changes from #99 KJ# to # 45 KJ# over #t in [0, 4 s]#. What is the average speed of the object?

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Answer 1 · 2017-05-03T05:57:16

The average speed is #=125.7ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=9kg#

The initial velocity is #=u_1#

#1/2m u_1^2=99000J#

The final velocity is #=u_2#

#1/2m u_2^2=45000J#

Therefore,

#u_1^2=2/9*99000=22000m^2s^-2#

and,

#u_2^2=2/9*45000=10000m^2s^-2#

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

The points are #(0,22000)# and #(4,10000)#

The equation of the line is

#v^2-22000=(10000-22000)/4t#

#v^2=-3000t+22000#

So,

#v=sqrt((-3000t+22000)#

We need to calculate the average value of #v# over #t in [0,4]#

#(4-0)bar v=int_0^4sqrt((-3000t+22000))dt#

#4 barv=[((-3000t+22000)^(3/2)/(-3/2*3000)]_0^4#

#=((-3000*4+22000)^(3/2)/(-4500))-((-3000*0+22000)^(3/2)/(-4500))#

#=22000^(3/2)/4500-10000^(3/2)/4500#

#=502.9#

So,

#barv=502.9/4=125.7ms^-1#