Example: Building equations from the physics

Understanding the situation

Here's an example of building equations from the physics that uses physics principles from the section on kinematics. In this case, we have a physical situation where we know some things about the situation and have something critical we want to infer from that information. We rely on our basic physics principles and map them into this particular situation.

Presenting a sample problem

In The Fellowship of the Ring, the hobbit Peregrine Took (Pippin for short) drops a rock into a well while the travelers are in the caves of Moria. This wakes a Balrog (a bad thing) and causes all kinds of trouble. Pippin heard the rock hit the water 7.5 s after he dropped it. Assuming that the rock fell with a constant acceleration of 10 m/s², how deep is the well?

Solving this problem

Tell the story

Let's begin by telling the story: Pippin drops the rock and it falls into the well, speeding up at a constant acceleration until it hits the bottom 7.5 s after he dropped it.

Create a model

Next we have to create a model: What are we going to pay attention to and what are we going to ignore? Let's treat the rock as a small dense object (ignoring its size). Since it's a rock not going very fast, we can assume air resistance is not very relevant so we'll ignore its effect. When the rock hits the water, it creates a sound wave that takes some time to get up to Pippin. For this example, let's ignore that and assume he hears the sound right away, as soon as it hits the water.

Build the equations

Now let's build our equations from the physics of what's happening. We are given the acceleration and the time it takes the rock to fall. Let's call the acceleration $a_0$ (the "0" subscript reminds us that this is a constant - a parameter - and not a variable) and the time interval $\Delta t$. We want the depth of the well. Let's call that $d$ (for "depth").

What else do we need? Well, let's think about what principles we might use to solve this. Since this is a problem with a constant acceleration, the velocity is changing. We probably need the definitions of (average) velocity and of (average) acceleration:

$$\langle v \rangle = \frac{\Delta y}{\Delta t}$$

$$\langle a \rangle = \frac{\Delta v}{\Delta t}$$

We know that acceleration is a constant and we have a name for the change in height, $\Delta y$, so

$$\langle a \rangle = a_0$$

$$\Delta y = d$$

This simplifies our pair of equations (that is, it gets rid of stuff we don't know or have not named with stuff that we either know or have a name for). Our equations are now

$$\langle v \rangle = \frac{d}{\Delta t}$$

$$a_0 = \frac{\Delta v}{\Delta t}$$

We know $\Delta t$ and $a_0$, and $d$ is our unknown. All we have to do to match our equations to our problem is to deal with the velocity terms. We have two: the average velocity ($\langle v \rangle$), and the change in velocity ($\Delta v$). For our special case of constant acceleration, the average velocity is the initial + final divided by 2 (a typical average). The general definition of the change is the final value minus the initial.

$$\langle v \rangle = \frac{1}{2} (v_f + v_i)$$

$$\Delta v = v_f - v_i$$

These are general expressions. Let's put in values for our problem. Our initial velocity is 0 (he just drops it, doesn't throw it down) and let's call our final velocity $v_0$, to remind ourselves that it's a particular value of the velocity, not the variable $v(t)$. Putting these in we get

$$\langle v \rangle = \frac{v_0}{2}$$

$$\Delta v = v_0$$

Putting these into the equations we got from our basic principles, they become:

$$\frac{v_0}{2} = \frac{d}{\Delta t}$$

$$a_0 = \frac{v_0}{\Delta t}$$

Solving our equations

Now we have two equations expressed in terms of the unknowns and knowns of our problem. What do we know and what do we not know? We are given $\Delta t$ and $a_0$. We don't know $v_0$ or $d$. But we now have two equation in two unknowns, so we expect to be able to solve them. To make it look nice, let's write the bottom one as an equation for the unknown $v_0$ and the top as an equation for the unknown $d$.

$$d = \frac{1}{2}v_0 \Delta t$$

$$v_0 = a_0\Delta t$$

Putting the second into the first we get

$$d = \frac{1}{2}a_0 (\Delta t)^2$$

(which some of you may recognize as a formula from high school physics). Putting in values we get $d$ = (10 m/s²)(7.5 s)²/2 = 281.25 m. Since "10" is only good to a couple of % and we are ignoring air resistance and time of sound travel, a good answer is d = 280 m. We can also calculate the speed at the bottom using $v_0 = a_0 \Delta t$ = (10 m/s²)(7.5 s) = 75 m/s.

Evaluating our answer

Now we can evaluate our model assumptions. How long does it take for a sound wave to travel 280 m? Is that time negligible compared to 7.5 s? How big is the drag force of air resistance on the rock at 75 m/s? Is it negligible compared to the force of gravity, $mg$, that is responsible for accelerating the rock downward? We won't go into these now, but we invite you to consider how you might answer these questions on your own!

Note that we had to introduce a new unknown into the problem — one that was not asked for. This will often be the case. The trick is to give it a name and keep moving forward. Once you are accustomed to working with equations with many symbols, this will become natural and comfortable.

Note also that this particular problem can be easily solved with memorized kinematics equations. We prefer that you don't do that, since the method above is far more general and doesn't clutter up your brain with equations that have limited applicability.

When you memorize any equation, you also need to memorize the situations in which those equations apply and many students forget that crucial step! In this case, the only equations we needed are our general principles — the definition of average velocity and acceleration — which translate the conceptual idea of velocity and acceleration into math, and the special case that when the acceleration is constant, the average velocity is the average of the initial and final velocities.

Joe Redish 7/17/17