The kinetic energy is
KE=1/2mv^2KE=12mv2
The mass is =4kg=4kg
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=144000J12mu21=144000J
The final kinetic energy is 1/2m u_2^2=120000J12mu22=120000J
Therefore,
u_1^2=2/4*144000=72000m^2s^-2u21=24⋅144000=72000m2s−2
and,
u_2^2=2/4*120000=60000m^2s^-2u22=24⋅120000=60000m2s−2
The graph of v^2=f(t)v2=f(t) is a straight line
The points are (0,72000)(0,72000) and (2,60000)(2,60000)
The equation of the line is
v^2-72000=(60000-72000)/2tv2−72000=60000−720002t
v^2=-6000t+72000v2=−6000t+72000
So,
v=sqrt((-6000t+72000)v=√(−6000t+72000)
We need to calculate the average value of vv over t in [0,2]t∈[0,2]
(2-0)bar v=int_0^2(sqrt(-6000t+72000))dt(2−0)¯v=∫20(√−6000t+72000)dt
2 barv=[((-6000t+72000)^(3/2)/(3/2*-6000))]_0^22¯v=⎡⎣⎛⎝(−6000t+72000)3232⋅−6000⎞⎠⎤⎦20
=((-6000*2+72000)^(3/2)/(-9000))-((-6000*0+72000)^(3/2)/(-9000))=⎛⎝(−6000⋅2+72000)32−9000⎞⎠−⎛⎝(−6000⋅0+72000)32−9000⎞⎠
=72000^(3/2)/9000-60000^(3/2)/9000=72000329000−60000329000
=513.6=513.6
So,
barv=513.6/2=256.8ms^-1¯v=513.62=256.8ms−1
The average speed is =256.8ms^-1=256.8ms−1