An elevator car whose floor-to-ceiling distan | vedclass

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(D) To find the time taken by the bolt to reach the floor,we analyze the motion relative to the elevator.$1$. The acceleration of the elevator is $a_e

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$\sqrt{\frac{5.4}{11}}\,s$

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(D) To find the time taken by the bolt to reach the floor,we analyze the motion relative to the elevator.$1$. The acceleration of the elevator is $a_e = 1.2\,m/s^2$ (upwards).$2$. The acceleration of the bolt due to gravity is $g = 9.8\,m/s^2$ (downwards).$3$. The relative acceleration of the bolt with respect to the elevator is $a_r = a_{bolt} - a_e = (-9.8) - (1.2) = -11\,m/s^2$. The magnitude is $11\,m/s^2$ downwards.$4$. The initial velocity of the bolt relative to the elevator is $u_r = 0$ because the bolt was at rest relative to the ceiling.$5$. The distance to be covered by the bolt relative to the elevator is $s_r = 2.7\,m$.$6$. Using the equation of motion $s_r = u_r t + \frac{1}{2} a_r t^2$:$2.7 = 0 	imes t + \frac{1}{2} 	imes 11 	imes t^2$$2.7 = 5.5 	imes t^2$$t^2 = \frac{2.7}{5.5} = \frac{5.4}{11}$$t = \sqrt{\frac{5.4}{11}}\,s$.

An elevator car whose floor-to-ceiling distance is $2.7\,m$ starts ascending with a constant acceleration of $1.2\,m/s^2$. After $2\,s$ of the start,a bolt falls from the ceiling of the car. The free fall time of the bolt is $(g=9.8\,m/s^2)$.

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