If a car of mass 1000 kg accelerates uniformly from rest to a velocity of 54 km/h in 5 s, then determine its acceleration and kinetic energy gained by it.

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Kunal Kumar

m = 1000kg u(initial velocity) = 0 ; as the body starts from rest . v( final velocity ) = 54 km/h = 15 m/s t(time) = 5 sec using equation of motion , v = u + at , or 15 = 0 + a*5, or a = (15-0) / 5 so , a = 3 m /s^2 now , we know kinetic energy = 1/2 M V^2 so...
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m = 1000kg u(initial velocity) = 0 ; as the body starts from rest . v( final velocity ) = 54 km/h = 15 m/s t(time) = 5 sec using equation of motion , v = u + at , or 15 = 0 + a*5, or a = (15-0) / 5 so , a = 3 m /s^2 now , we know kinetic energy = 1/2 M V^2 so K.E = 1/2 * 1000 *15 *15 = 112500 Joules . read less
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mass=1000kg u=0km/hr=0m/s v=54km/hr=15m/s work= change in kinetic energy = 1/2xmxv^2 - 1/2xmxu^2 = 1/2x1000x15^2 - 0 =112500 J power = work/time = 112500/5 = 22500 J/s=22.5 kw from first eqution of motion v=u+at u=0 therfore v=at 15=5a a=3m/s^2 from second equation of motion s=ut+/2at^2 0+1/2*3*5*% s=37.5...
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mass=1000kg u=0km/hr=0m/s v=54km/hr=15m/s work= change in kinetic energy = 1/2xmxv^2 - 1/2xmxu^2 = 1/2x1000x15^2 - 0 =112500 J power = work/time = 112500/5 = 22500 J/s=22.5 kw from first eqution of motion v=u+at u=0 therfore v=at 15=5a a=3m/s^2 from second equation of motion s=ut+/2at^2 0+1/2*3*5*% s=37.5 m now v avg=dsance/time=37.5/5=7.5 m/s gaain in KE=KE fina=KE initial =1/2mv^2-1/2mv2^2 =1/2*1000*15^=112.5kj read less
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