Physics Elevator Question (EK1001 #293) Question posted here

[SSerenity]

Heres what I've come up with. We just want to know how high to throw the box to reach the 6th floor in the time it takes to ride up there.

6 Floors corresponds to 25 meters
It takes 5 seconds to get halfway up, therefore its a 10 second ride.

I use:

x=v*t - 1/2(g)(t)^2

25 = 10v - 500
v = 52.5

The answer is allegedly D. I used almost no rounding, why is my answer still off? Is my process just incorrect? EK shows no actual work for this question.

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[dcamkl1]

Well, my answer was further off than yours. The good thing is that the MCAT is a multiple-choice answer, and there will be problems where you have to round and approximate and choose the best answer. There's no time to be nitpicky about precise numbers.

Here's how I solved it, if you're interested:

First, I assumed the elevator was at rest. So, solving for the time for the elevator to get to the 6th floor: d = 1/2(a)(t)^2, t = 5 sec, as you figured in your post.

So, the stickman wants to catch the box right when he gets to the 6th floor. Throwing an object up will, of course, fall back down. So the object has to peak right when he gets to the 6th floor (5 sec). The time it takes for any object to reach the apex is t = voy/a. So, voy=(a)(t) = 50 m/s. Close enough...

 

[discowisco]

Also why is acceleration 10 when using d=vit + 1/2at^2 instead of the given acc of the elevator?

 

[SSerenity]

Thanks Dcam.

Disco,
Im pretty sure the elevator's acceleration will only affect whats touching the elevator floor. He threw the box at the same moment the elevator started to move. Really, this whole question is crap.

If I threw a box up at ~50m/s in an elevator, it would definitely hit the ceiling.

 

[dcamkl1]

Well, my answer was further off than yours. The good thing is that the MCAT is a multiple-choice answer, and there will be problems where you have to round and approximate and choose the best answer. There's no time to be nitpicky about precise numbers.

Here's how I solved it, if you're interested:

First, I assumed the elevator was at rest. So, solving for the time for the elevator to get to the 6th floor: d = 1/2(a)(t)^2, t = 5 sec, as you figured in your post.

So, the stickman wants to catch the box right when he gets to the 6th floor. Throwing an object up will, of course, fall back down. So the object has to peak right when he gets to the 6th floor (5 sec). The time it takes for any object to reach the apex is t = voy/a. So, voy=(a)(t) = 50 m/s. Close enough...

So, I re-checked my math and the bolded part above is wrong; the time would have been the sqrt (50) = 7 sec, and even that is wrong for this problem. And also, I shouldn't have assumed that the elevator started at rest.

You have to use v = d/t. So, t = d/v = 25/5 = 5 sec. Then plug that into Voy = (a)(t) = (10)(5) = 50 m/s

I think the acceleration of the elevator has nothing to do with this problem, since it is seems that they're saying that it accelerates and decelerates when approaching a floor. Idk.. I wouldn't get stressed out about the 1001 problems. Many of the Q's aren't written well.

 

[FuzzyMemory]

I got 55m/s exactly:

Distance between 1st & 6th floor: 30m

Distance the elevator takes to reach maximum velocity (5m/s):

Vf^2 = Vo^2 + 2ad
5m/s^2 = 0 + 2(1m/s2)(d)
d = 12.5m

The distance the elevator travels while accelerating and decelerating: 12.5 x 2 = 25m, which leaves a (30m-25m) 5m distance in which the elevator is traveling at 5m/s. and so the elevator covers this distance in 1s.

The time required to travel 12.5m is:

d = Vot + 1/2at^2
12.5 = 0 + 1/2 * 1m/s2 * t^2
25 = t^2
t = 5s to reach 5m/s, and another 5s to decelerate back down to 0m/s = 10s

Total time: 10s + 1s = 11s total elevator time.

Then, it's just a simple projectile motion question for the initial velocity of the box:

Time required for the box to reach terminal velocity in 11s/2 = 5.5s

Vf = Vo + at
0 = Vo + (-10m/s2) * 5.5s
Vo = 55m/s

This question seems awfully more involved than most of the physics questions in the MCAT, imo.