An elevator moves floor to ceiling distance is 2 to 5 m. It starts ascending with constant accelaration 1.25 ms^2. One second after start a bolt starts falling from ceiling elevator. Find the time in which bolt strikes the floor of elevator. Also, find displacement of bolt during this time.

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Solution

$T h e d i s \tan c e m o v e d u p b y e l e v a t o r i n o n e s e c o n d h = \frac{1}{2} \times 1.25 \times 1 \times 1 = \frac{1.25}{2} m a n d t h e v e l o c i t y a c q u i r e d v = 1.25 \times 1 = 1.25 m / s W h e n a b o l t f a l l s, l e t t h e t i m e t a k e n b y i t t o r e a c h t h e f l o o r b e t . S o (5 - 2) = - 1.25 t + \frac{1}{2} \times 10 \times t^{2} o r 3 = - 1.25 + 5 t^{2} o r 5 t^{2} - 1.25 t - 3 = 0 o r 20 t^{2} - 5 t - 12 = 0 o r t = \frac{5 \pm \sqrt{25 + 960}}{40} = 0.9 s S o t h e t i m e t a k e n t o r e a c h t h e f l o o r t = 0.9 s T o t a l t i m e t a k e n b y e l e v a t o r t_{0} = 1 + 0.9 = 1.9 s T h e t o t a l d i s \tan c e c o v e r e d i n 1.9 s i s h_{0} = \frac{1}{2} \times 1.25 \times 1.9 \times 1.9 = 2.26 m T h e r e f o r e t o t a l d i s p l a c e m e n t o f t h e b o l t = 2.26 - \frac{1.25}{2} = \frac{4.52 - 1.25}{2} o r (h_{0} - h) = \frac{3.27}{2} = 1.635 m$

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