The kinetic energy is
KE=1/2mv^2KE=12mv2
The mass is =4kg=4kg
The initial velocity is =u_1=u1
The initial kinetic energy is 1/2m u_1^2=64000J12mu21=64000J
The final velocity is =u_2=u2
The final kinetic energy is 1/2m u_2^2=180000J12mu22=180000J
Therefore,
u_1^2=2/4*64000=32000m^2s^-2u21=24⋅64000=32000m2s−2
and,
u_2^2=2/4*180000=90000m^2s^-2u22=24⋅180000=90000m2s−2
The graph of v^2=f(t)v2=f(t) is a straight line
The points are (0,32000)(0,32000) and (12,90000)(12,90000)
The equation of the line is
v^2-32000=(90000-32000)/12tv2−32000=90000−3200012t
v^2=4833.3t+32000v2=4833.3t+32000
So,
v=sqrt((4833.3t+32000)v=√(4833.3t+32000)
We need to calculate the average value of vv over t in [0,12]t∈[0,12]
(12-0)bar v=int_0^12sqrt((4833.3t+32000))dt(12−0)¯v=∫120√(4833.3t+32000)dt
12 barv=[((4833.3t+32000)^(3/2)/(3/2*4833.3)]_0^1212¯v=⎡⎢⎣⎛⎝(4833.3t+32000)3232⋅4833.3⎤⎦120
=((4833.3*12+32000)^(3/2)/(7250))-((4833.3*0+32000)^(3/2)/(7250))=⎛⎝(4833.3⋅12+32000)327250⎞⎠−⎛⎝(4833.3⋅0+32000)327250⎞⎠
=90000^(3/2)/7250-32000^(3/2)/7250=90000327250−32000327250
=2934.6=2934.6
So,
barv=2934.6/12=244.5ms^-1¯v=2934.612=244.5ms−1
The average speed is =244.5ms^-1=244.5ms−1