An object has a mass of 4 kg4kg. The object's kinetic energy uniformly changes from 64 KJ64KJ to 180 KJ180KJ over t in [0, 12 s]t[0,12s]. What is the average speed of the object?

1 Answer
Jun 15, 2017

The average speed is =244.5ms^-1=244.5ms1

Explanation:

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=2464000=32000m2s2

and,

u_2^2=2/4*180000=90000m^2s^-2u22=24180000=90000m2s2

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)/12tv232000=900003200012t

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(120)¯v=120(4833.3t+32000)dt

12 barv=[((4833.3t+32000)^(3/2)/(3/2*4833.3)]_0^1212¯v=(4833.3t+32000)32324833.3120

=((4833.3*12+32000)^(3/2)/(7250))-((4833.3*0+32000)^(3/2)/(7250))=(4833.312+32000)327250(4833.30+32000)327250

=90000^(3/2)/7250-32000^(3/2)/7250=9000032725032000327250

=2934.6=2934.6

So,

barv=2934.6/12=244.5ms^-1¯v=2934.612=244.5ms1

The average speed is =244.5ms^-1=244.5ms1