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I had this question, I don't seem to get what I am doing wrong so here it goes:An elevator car whose floor to ceiling distance is equal to $2.7 m $ starts ascending with a constant acceleration of $1.2~m/s^2$. Two seconds after it starts, a bolt begins to fall from the ceiling of the elevator. Find the bolt's free fall time. What I did was let $s$ distance traveled by the bolt in time $t$ be $x$ and the distance traveled by the elevator in the direction of bolt be $y$ Then:$x=-(0.5)9.8t^2$ and $y=2.4t+0.5*1.2*t^2$ and $y-x=2.7$ Solving these we get $t=0.515sec$ however the answer is 0.7 sec

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closed as off-topic by AccidentalFourierTransform, John Rennie, CuriousOne, user10851, Gert Apr 30 '16 at 1:06

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    $\begingroup$ Also, I assume from your equation that the acceleration of the elevator is supposed to be $1.2 \, \text{m/s}^2$, and that the $2.7 \, \text{m/s}^2$ in your first sentence is a typo. If not, that's another error. $\endgroup$ – Michael Seifert Apr 29 '16 at 13:37
  • $\begingroup$ The elevator and bolt have accelerated for 2 sec when the bolt begins to fall. What is the initial velocity of the bolt? $\endgroup$ – mmesser314 Apr 29 '16 at 13:38
  • $\begingroup$ Yes thanks you're correct I automatically assumed that the initial velocity of the bolt is from"a bolt begins to fall" but the answer matches that of textbook If I assumed it to be 2.4m/s $\endgroup$ – bulbasaur Apr 29 '16 at 14:01
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Think about this from the perspective of a person in the elevator. No windows, they can't look outside. As far as they are concerned, they live on a small box-like planet where the acceleration due to gravity is 9.8 + 1.2 = 11 m/s$^2$.

In a system where the acceleration due to gravity appears to be 11 m/s$^2$, a bolt drops 2.7 m. How long does it take to drop?

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  • $\begingroup$ perfect your answer is much shorter than mine as you take elevator as the reference frame $\endgroup$ – bulbasaur Apr 29 '16 at 14:53
  • $\begingroup$ Glad you like it. Yes, the real lesson of solving this question lies in gaining an understanding that an accelerating frame of reference is a valid frame, as long as you are careful. $\endgroup$ – Floris Apr 29 '16 at 15:05

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