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$\newcommand{\accunit}{\mathrm{m}/\mathrm{s}^2}$

A coin is dropped 60.0m above the top of an elevator car. The coin is dropped just as the car begins to move upwards. The elevator first accelerates from rest at a constant rate equal to one tenth the acceleration due to gravity for 3 s, then continues upward at a constant speed.

How long does it take the coin to hit the top of the car? If the speed of sound is 340 m/s,

how long does it take the repairman to hear the coin hit the car.

I know the initial velocity of the coin is $0$ since it wasn't thrown. The acceleration of the coin would be $-9.8\accunit$ due to gravity. The elevator has an initial velocity of $0$ since it is at rest. The acceleration of the elevator is $0.98\accunit$ which is one-tenth of the force due to gravity for the first three seconds.

Can I solve this problem with two equations? One for the position of the coin and the other for the position of the elevator car. I would set those two equations equal to each other and solve for $t$.

$X_c = \dfrac{1}{2}(-9.8\accunit)t^2$ and $X_e = (-60.0+2.94\accunit)(t)$

I solved for $t$ using the quadratic equation and got $3.31\mathrm{s}$ However when I plug the time back into the two equations I get different $x$ positions for the car and the coin. Can someone explain what I am doing wrong?

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1 Answer

Whoever set the question made it awkward by specifying that the lift only accelerates for 3 seconds, and thereafter continues at constant speed. In those 3 seconds the lift moves about 4.4m and the coin falls about 44m. That means the coin doesn't hit the lift until after the lift has stopped accelerating.

The distance moved by the lift during the first three seconds is the usual $s = 0.5 a t^2 = 4.5a$ and the velocity after 3 seconds is $v = at = 3a$, so after a time $t > 3$ the distance moved by the lift is given by:

$$ s(t) = 4.5 a + 3a (t - 3) $$

Use this in your expression for the lift position and you should get the correct answer.

Response to comment:

The two equations we'll be using are:

$$\begin{align} s &= ut + \frac{1}{2} a t^2 \\ v &= u + at \end{align}$$

We take lift to be at $h = 0$ at $t = 0$ and the coin to be at $h = 60$.

The height of the coin is simply:

$$ h_{coin} = 60 - \frac{1}{2} g t^2 $$

The lift accelerates at $g$/10 for 3 seconds, so during this time it travels a distance:

$$ h_{lift}(3) = \frac{1}{2} \frac{g}{10} 3^2 = 0.45g $$

and after 3 seconds the lift velocity is:

$$ v_{lift}(3) = \frac{g}{10} t = 0.3g $$

At some time $t > 3$ the distance travelled is the $0.45g$ travelled in the first three seconds plus the distance travelled at constant velocity $0.3g$ for $t - 3$ seconds so:

$$ h_{lift} = 0.45g + 0.3g(t - 3) $$

The coin and lift meet when $h_{coin} = h_{lift}$ so:

$$ 60 - \frac{1}{2} g t^2 = 0.45g + 0.3g(t - 3) $$

This is a quadratic equation for $t$. Solve it and you'll get the time the coin and lift meet. The answer isn't 3.08s.

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I got $3.08s$ for the coin to hit the elevator car at position $-46.5m$. I was not sure how you got $s(t)=4.5a+3a(t-3)$. –  Aaron Wong Sep 22 '13 at 21:08
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