neededthings said:energy= Pt= 3x10^2 * 4 = 1.2 x10^3 J
Useful energy = 1.2 x10^3 *0.92 = 1.104 x10^3
KE= 1/2m( v)^2
root(1.104 x10^3 *2 /78) = change in v = +/-5.32 m/s (3 S.F.). He is accelerating hence +5.32m/s
21km/h = 21000m/60x60s= 5.83 m/s (3 S.F)
Total speed = 5.32+5.83 = 11.15m/s
Kishlay said:useful energy =1.104x10^3 ... right
then change in kinetic energy will be equal to useful energy given in 4 seconds...
the formula 1/2mv^2, 'V' is the instataneous velocity, not the change in velocity
so we get
change in kinetic energy = useful energy given..
mx(Vfinal2-Vinitial2)/2
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