"Initial KE"=(E_i)=32kJ=32xx10^3J
After 4s
"Final KE"=(E_f)=72kJ=72xx10^3J
KE uniformly increases, so it will have a constant positive slope when plotted against time and this slope will be (Delta E)/(Deltat)=(E_f- E_i)/(Delta t)=((72-32)xx10^3)/4=10^4J/s
,"where "Delta t=4s
If at t th instant ( t in [0,4]) the velocity of the mass be v then the KE at t th instant
E_t=1/2xxmxxv^2=1/2xx2xxv^2=v^2
",where mass "m= 2kg
Now (E_t-E_i)/(t-0)=(v^2-32xx10^3)/(t-0)=10^4
=>v^2=10^4xxt+32xx10^3
=>v=sqrt(10^4xxt+32xx10^3)
=sqrt(10^4(t+3.2))
=10^2sqrt(t+3.2)
Now distance traversed during 4 s
S=int_0 ^4vdt=int_0 ^4 10^2(sqrt(t+3.2))dt
Let z^2=t+3.2
2zdz=dt
when t=0; then z =sqrt3.2
when t=4; then z =sqrt7.2
S=int_0 ^4vdt=int_0 ^4 10^2(sqrt(t+3.2))dt
S =int_(sqrt(3.2))^(sqrt7.2) 10^2xx2z^2dz
=int_(sqrt(3.2)) ^(sqrt7.2) 200xxz^2dz
=200/3xx[(sqrt7.2)^3-(sqrt3.2)^3]m
Hence the average speed ="distance" /"time"=S/4
=200/3xx[(sqrt7.2)^3-(sqrt3.2)^3]/4=226.6m/s
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