Question

# An elevator car whose floor-to-ceiling distance is equal to 2.7 m starts ascending with constant acceleration 1.2 m/s2. 2s after the start, a bolt begins falling from the ceiling of the car. Find: (a) the bolt's free fall time; (b) the displacement and the distance covered by the bolt during the free fall in the reference frame fixed to the ground. t = 0.7 sec, s = 0.7 m, x = 1.3 m t = 0.7 sec, s = 2.7 m, x = 2.7 m  t = 0.7 sec, s = 0.7 m, x = 0.7 m   t = 0.7 sec, s = 0.7 m, x = 2.7 m

Solution

## The correct option is A t = 0.7 sec, s = 0.7 m, x = 1.3 m Lets analyze the Motion: At t=0 lift started, u=0 at=1.2; at t=2 sec, bolt falls Velocity of lift at t=2 ⇒v=u+at v1=+2.4m/s Initial velocity of bolt, Ub=same as lift as velocity gets transferred but not acceleration. So Ub=+2.4m/s Acceleration of bolt ab=−9.8m/s2 as immediately it comes under the influence of gravity Lets be the lift for a while. So what would be the velocity and acceleration of the bolt as seen by the lift? Ub/l=Ub−Ul=2.4−2.4=0 m/s ab/l=ab−al=−9.8−1.2=−11m/s2 Now for the lift the bolt fell from ceilling to floor so displacement relative = sb/l=−2.7m ∴ 2.7 is the height of the lift sb/l=Ub/l(t)+12ab/lt2 −2.7=0+12(−11)t2 t=0.7 sec So it took 0.7 sec for the bolt to fall. i.e. the bolt was in motion for 0.7 sec. Now for second part lets come out of lift and be the ground. The moment the bolt leaves the lift imagine the lift has disappeared and only bolt is there doing motion under gravity. It will look the pic. As the moment the screw is loose it has an upward velocity of 2.4 m/s as that of the lift and a downward acceleration of 9.8 m/s2 lets find the displacement. s=ut+12at2 s=2.4(0.7)+12(−9.8)(0.7)2=−0.7mm So bolt is displaced downwards by 0.7m Now the total distance Total distance will be tx+0.7=2x+0.7 when bolt reaches the highest point its velocity will be 0 ⇒v2−u2=2as 0-(2.4)2=2(-9.8)x x=0.3m So distance = 1.3 m

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