Q2. A lift starts from rest. Its acceleration is plotted against time in the following graph. When it comes to rest its height above its starting point is

(1) 20 m              (2) 64 m                (3) 32 m                  (4) 128 m

Dear student
Area under velocity time graph gives the change in velocity of the particle.Area from t=0 to t=4 A1=4×4=16so velocity of the particle at t=4 sec isv=16 m/sFrom t=4 to t=8 s it moves with constant velocity of 16 m/s as its acceleration is zero Area from t=8s to t=12sA2=-4×4=-16m/sSo particle comes to rest at t=12 sec as total area becomes zero at this time .distance moved from t=0 to t=4 secx1=12×a1×t21=12×4×42=32mdistance moved from t=4 to t=8 secx2=v×t=16×4=64mdistance moved from t=8 to t=12 secx3=ut-12×a2×t21x3=16×4-12×4×42=64-32=32mtotal distance s=x1+x2+x3s=32+64+32=128mRegards

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