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Dear Student , After touching down the aeroplane's velocity was 225 km/hr and it stops after 2 min . Let us take that the aeroplane stops after going through s distance and a be the accelration of it . Now as the aeroplane stops after s distance so , final velocity is zero and initial velocity is 225 km/hr .
(i)Now from the equation of motion we can write that ,v=u+at⇒0=225×518+120a t=2 min=120 sec⇒120a=-62·5⇒a=-0·52 m/s2Now here the negative sign indicates the acceleration is negative ,i.e it is retardation .(ii)Again from the equation of motion we can write that ,v2=u2+2as⇒0=62·52+2×-0·52×s⇒s=62·522×0·52=3750 m
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