# Screw falling in an elevator [closed]

I am trying to solve a problem but I keep getting the wrong signs. Here is the problem:

While standing in an elevator, you see a screw fall from the ceiling. The ceiling is 3m above the floor. How long does it take for the screw to hit the floor if the elevator is moving upward and gaining speed at a constant rate of $a_f=4m/s^2$?

My idea was to combine the two accelerations of gravity and the elevator to get an equivalent acceleration of 13.8 for the screw. Then taking the upward direction as positive, we have that the screw displacement when it hits the floor will be -3, its final velocity will be 0, and its acceleration is -13.8. Thus using s=vt-0.5*at^2 we get -3=-0.5(-13.8)t^2 or -3=0.5*13.8t^2.

Why did I get a wrong equation? It appears that the minus should be a plus and this gives the right answer. However, I am not seeing where I got the signs wrong. If upward is +ve then the downward acceleration is -13.8 and the downward displacement is -3.

## closed as off-topic by sammy gerbil, John Rennie, stafusa, ZeroTheHero, Jon CusterJan 2 '18 at 8:35

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When you write $s=vt-\frac{1}{2}at^2$, you have already set the direction of the acceleration to be negative.

Perhaps a clearer approach is to write the equation of motion more generally, in a way that does not assume anything about the signs of any variable: $s=vt+at^2$.

Note that the initial velocity is zero, not the final velocity. I think that's just a typo, because you do indeed take the initial velocity to be zero.

• It wasn't a typo, I was using the vt-0.5at^2 equation not ut-0.5at^2. (In my book it says that there are two equations: the first is vt where v is final velocity, the second is ut where u is initial velocity. I'm not sure how common this is but I would appreciate if you could comment on why the equation is wrong.) – Raghib Dec 31 '17 at 14:35
• Hi, I am just letting you know that I have solved the problem. My problem was that I took the frame of reference as the elevator without following the consequences of this through properly. In particular, I put the final velocity of the screw as 0, when in fact only the initial velocity is 0. This is because if the elevator is the frame of reference, it can be assumed to be at rest. The problem then simply becomes: Find the time taken for a screw dropped from rest to fall 3m if its acceleration is 9.8+4=13.8. – Raghib Dec 31 '17 at 15:54
• I'm glad you figured it out, and I'm sorry that I didn't understand your notation. BTW, the distinction that your book makes between $u$ and $v$, and the presentation of two versions of that equation, is not at all standard. I've been a professional physicist for 41 years, and I never saw it before today! I suppose someone else has seen it, and down voted me on account of my ignorance. :-) – garyp Dec 31 '17 at 22:54
• @garyp : Try Googling "savtu equations" or "suvat equations" or davtu equations". Five equations seems fairly common... – DJohnM Jan 3 '18 at 5:18

First of all, the equation of motion is $s = ut + \frac{1}{2} at^2.$

As the elevator is moving upwards and $g$ acts downward, resultant acceleration = $g-(-a)$ = $g+a$ which is in the downward direction.

Now, given parameters are:

$u = 0$ (as it is free fall)

$s = 3m$

$a_{res} = 9.8 + 4 = 13.8 m/s^2$

Substituting in the equation of motion, you get

$t = 0.65$ $sec$

The reason why all signs are taken as positive is because $\text{u, } a_{res} \text{ and s}$ are all in the same direction that is downward.