The kinetic energy is
KE=1/2mv^2KE=12mv2
The mass is m=1kgm=1kg
The initial velocity is =u_1ms^-1=u1ms−1
The final velocity is =u_2 ms^-1=u2ms−1
The initial kinetic energy is 1/2m u_1^2=48000J12mu21=48000J
The final kinetic energy is 1/2m u_2^2=36000J12mu22=36000J
Therefore,
u_1^2=2/1*48000=96000m^2s^-2u21=21⋅48000=96000m2s−2
and,
u_2^2=2/1*36000=72000m^2s^-2u22=21⋅36000=72000m2s−2
The graph of v^2=f(t)v2=f(t) is a straight line
The points are (0,48000)(0,48000) and (3,36000)(3,36000)
The equation of the line is
v^2-48000=(36000-48000)/3tv2−48000=36000−480003t
v^2=-4000t+48000v2=−4000t+48000
So,
v=sqrt(-4000t+48000)v=√−4000t+48000
We need to calculate the average value of vv over t in [0,3]t∈[0,3]
(3-0)bar v=int_0^3(sqrt(-4000t+48000))dt(3−0)¯v=∫30(√−4000t+48000)dt
3 barv= [(-4000t+48000)^(3/2)/(3/2*-4000)] _( 0) ^ (3)3¯v=⎡⎣(−4000t+48000)3232⋅−4000⎤⎦30
=((-4000*3+48000)^(3/2)/(-6000))-((-4000*0+48000)^(3/2)/(-6000))=⎛⎝(−4000⋅3+48000)32−6000⎞⎠−⎛⎝(−4000⋅0+48000)32−6000⎞⎠
=48000^(3/2)/6000-36000^(3/2)/6000=48000326000−36000326000
=614.3=614.3
So,
barv=614.3/3=204.8ms^-1¯v=614.33=204.8ms−1
The average speed is =204.8ms^-1=204.8ms−1
