An object has a mass of #8 kg#. The object's kinetic energy uniformly changes from #240 KJ# to # 640KJ# over #t in [0, 3 s]#. What is the average speed of the object?

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Answer 1 · 2017-06-02T18:18:38

The average speed is #=328.7ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=8kg#

The initial velocity is #=u_1#

#1/2m u_1^2=240000J#

The final velocity is #=u_2#

#1/2m u_2^2=640000J#

Therefore,

#u_1^2=2/8*240000=60000m^2s^-2#

and,

#u_2^2=2/8*640000=160000m^2s^-2#

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

The points are #(0,60000)# and #(3,160000)#

The equation of the line is

#v^2-60000=(160000-60000)/3t#

#v^2=33333.3t+60000#

So,

#v=sqrt((33333.3t+60000)#

We need to calculate the average value of #v# over #t in [0,3]#

#(3-0)bar v=int_0^3sqrt((33333.3t+60000))dt#

#3 barv=[((33333.3t+60000)^(3/2)/(3/2*33333.3)]_0^3#

#=((33333.3*3+60000)^(3/2)/(50000))-((33333.3*0+60000)^(3/2)/(50000))#

#=160000^(3/2)/50000-60000^(3/2)/50000#

#=986.1#

So,

#barv=986.1/3=328.7ms^-1#