Question on acceleration of gravity

Question

In pre-calculus last week my teacher explained a scenario in which a man throws a rock down a well and takes 4 seconds for you to hear the rock hit the bottom. How far downs the well? Last years I took regular physics and learned that distance equals Velocity x Time. So knowing that gravity pulls with 9.8m/s/s I assumed I would just multiply 4 by 9.8. Sadly I was wrong and my quick method of solving this problem was incorrect, we ended up using a very very large quadratic equation. My question Is why I was wrong in my answer? 

When I think about the problem assume I went wrong by using 9.8 as if it were a regular velocity when in fact it is a constant acceleration, so the speed is constantly increasing. If that is the case then can anyone explain mathematically how quickly gravity accelerates in a 4 second period and also what the easiest and most efficient way of solving this problem would be to your knowledge?

Ps, this isn't homework I'm curious as to how this works

Answer 1

In the question, you clearly stated

it takes 4 seconds for you to hear the rock hit the bottom of the well

In 4 seconds, a rock (ignoring air drag) would drop $\frac12 9.81 (4)^2=78.5 m$ - but it would take sound about 1/4 of a second to reach your ear, so the well must be a little bit less deep (1/4 second is almost 10 meters at the velocity near the bottom).

The correct equation is:

$$t = \sqrt{\frac{2 h}{g}} + \frac hc$$

Namely, the time for the rock to drop, plus the time for the sound to get back to you. In this equation, $c$ is the speed of sound (340 m/s) and $h$ is the depth of the well (to be determined).

Solving for $h$ with $t=4$ we get approximately $71\ m$ - which is what you said the answer was (according to your comment to Jim's answer). I will leave the messy bits of the math to you (or you can go to Wolfram Alpha)

The other answers given (Jim, Alfred Centauri) focus on the basic equation of motion for an object accelerating with constant acceleration, namely

$$ x = x_0 + v_0 t + \frac12 a t^2$$

That equation is needed to find the time for the rock to hit the bottom - if you start with $x_0=0, v_0=0$, set the acceleration equal to $g$, and solve for $t$, you find the first term in my expression above; the second term (which accounts for the velocity of sound) is needed to get to the "right" answer (the difference is about 10% - which I consider significant).

Answer 2

One of the five most common equations of motion in high school physics is $$\Delta d=v_0\Delta t+^1\!\!/_2\,a\Delta t^2$$ $v_0$ is your initial velocity (0 for your case), a is the acceleration (g for your case) and $\Delta t$ is the time. That's the easiest equation you'll find for something like this, from here they get a bit messier.

Answer 3

Last years I took regular physics and learned that distance equals Velocity x Time.

That's not quite correct. In one dimension, the distance travelled equals the average velocity multiplied by the change in time

$$\Delta x = \bar v \cdot \Delta t$$

where $\bar v$ is the average velocity. If the velocity is constant, we can write

$$\Delta x = v \cdot \Delta t$$

Since there is acceleration in this problem, the velocity is not constant so, to use the first formula, you would need to know the average velocity. It turns out that, for constant acceleration $a$, the average velocity is just

$$\bar v = v_0 +\frac{a}{2}\Delta t$$

where $v_0$ is the initial velocity. Thus, the first equation becomes

$$\Delta x = \bar v \cdot \Delta t = \left(v_0 +\frac{a}{2}\Delta t\right)\Delta t = v_0 \Delta t + \frac{a}{2}\left(\Delta t\right)^2$$

Answer 4

The way you described it you said you solved it as 4(9.8). You should have done 1(9.8) + 2(9.8) + 3(9.8) + 4(9.8), or 10(9.8), or 98.