An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #640 KJ# to # 320 KJ# over #t in [0, 12 s]#. What is the average speed of the object?

Question

1294 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-04T10:09:51

The average velocity is #=487.6ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=4kg#

The initial velocity is #=u_1#

#1/2m u_1^2=640000J#

The final velocity is #=u_2#

#1/2m u_2^2=320000J#

Therefore,

#u_1^2=2/4*640000=320000m^2s^-2#

and,

#u_2^2=2/4*320000=160000m^2s^-2#

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

The points are #(0,320000)# and #(12,160000)#

The equation of the line is

#v^2-320000=(160000-320000)/12t#

#v^2=-13333.3t+320000#

So,

#v=sqrt((-13333.3t+320000)#

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

#(12-0)bar v=int_0^12sqrt((-13333.3t+320000))dt#

#12 barv=[((-13333.3t+320000)^(3/2)/(-3/2*13333.3)]_0^12#

#=((-13333.3*12+320000)^(3/2)/(-20000))-((-13333.3*0+320000)^(3/2)/(-20000))#

#=-160000^(3/2)/20000+320000^(3/2)/20000#

#=5851#

So,

#barv=5851/12=487.6ms^-1#