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a particle starts moving from rest under a constant acceleration, if it travels a distance x in the sec. what distance will it travel in next t sec |
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Answer» Here INITIAL velocity ,u=0 Distance, S=xAcceleration, a=aTime, t=tAccording to second equation of motionS=ut+12at2⇒ X=0×t+12at2Or x=12at2 −−−−(1)Now when time is doubledi.e. t=2tThen \ S=12a(2t)2Or S=4×12at2Or S=4x [using value of x from equation (1)]Thus distance covered in last t SECONDS= distance covered in 2t seconds - distance covered in first t secondsOr S′=S−x=4x−x=3xAnother approach to this problemHere initial velocity ,u=0Distance, S=xAcceleration, a=aTime, t=tAccording to second equation of motionS=ut+12at2⇒ x=0×t+12at2Or x=12at2Or at2=2x −−−−(1)According to 1st equation of motionFinal velocity, v=u+atOrv=0+at=atLet Distance covered in next t seconds = S'And initial velocity in this CASE is vi.e. u=atAccording to 2nd equation of motionS′=ut+12at2Or S′=at.t+12at2Or S′=at2+12at2Or S′=(1+12).at2Or S′=32at2Using value of at^2 from equation (1)S′=32×2x=3x |
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