Click here for a version of this paper done in SI units (meters instead of feet)

Think Like a Physicist

or, Why do Physicists Waste So Much Time Talking About Math?

Copyright (c) 1996 by Kenny Felder
Animated GIF by Chris Quarnstrom

After talking to a lot of bright-but-frustrated students, I've come to the conclusion that the "hard part" of Physics is not the Physics per se—the laws of nature and so on. And it isn't the math either! The problem is somewhere in between: it's going back and forth between the Physics and the math that loses people. The "I hate word problems" aspect, if you will.

So, the following experiment. I'm going to do an entire Physics lecture about math, but using no math at all. This may sound odd, or even useless, but I will make a very strong claim: the concepts in this small paper are exactly the concepts that prevent otherwise intelligent people from really enjoying Physics. Give this paper an hour of your life, and the payback—in terms of better grades, or just in terms of opening up a rich new mental world—could be absolutely enormous.

This paper is based on the following scenario: you throw a rock. It goes straight up into the air, and then comes down. That's all.

Now, let's put some numbers to it. First, let's assume that the rock starts at the ground (even though it would be difficult to throw a rock that way in real life). You throw the rock with an initial speed of 32 feet per second. It goes up for exactly one second, reaches a height of 16 feet, and then takes exactly one second to fall back down.

Finally, let's coin some letters. We will call the time t, measured in seconds. t starts at zero when the rock begins its journey, and reaches 2 at the end (because the whole trip is 2 seconds). We will call the height of the rock off the ground h, measured in feet. h starts at zero, when the rock is on the ground, goes all the way up to 16, and ends back at zero when the rock is back on the ground.
Rock thrown 16 ft in the air (Wait a minute, I thought he said no math.)

(It's okay, those are just some numbers and letters. I promise not to try to add them or anything fancy like that.)
Okay, here comes the leap. At any given time t, the rock is at a certain height h. There is a mathematical equation that relates the two variables: that is, it tells you what h is, for any given time t. Let's call this the height equation. I'm not going to tell you what the height equation is (it's printed at the end, and you may already know it, but indulge me and don't use it). The key thing to know about this equation, is simply that it exists. And this whole paper is going to be a series of questions about what you would get, if you were to use this equation.

For each question, please read the question (in boldface) without looking at the answer (anything typed below the boldface). Then write down the answer you think is right, before you look at the real answer. Remember, do all of this without ever looking at, or using the equation!

Question 1: If you plug t=0 into the height equation, what h will you get?

Once again (and for the last time, I promise), I want to stress: please write down your answer before you read any farther. The answer should be in the form of a number. Without using any equation, figure out what you would get, for h, if you set t=0 and went through the math.

In order to answer this question, you have to translate it into words. The word translation is: "what height is the rock at, after no time at all has elapsed?" I hope the answer is obvious by now: it is zero. At time t=0—at the beginning, in other words—the rock is on the ground, which means h=0. If you plugged t=0 and got any other value for h, that wouldn't mean your common sense was wrong. It would mean that the equation was wrong, or else you did your math wrong. That answer has to be right.

Question 2: If you plug t=1 into the height equation, what h will you get?

Answer: you will get 16. At time t=1, one second after launch, the rock is at 16 feet high.

Question 3: If you plug t=2 into the height equation, what h will you get?

Answer: you will get 0. At time t=2, two seconds after launch, the rock is back on the ground, at height 0.

Question 4: If you plug t=3 into the height equation, what h will you get?

Your first impulse might be to say, that you won't get anything at all. But that's wrong. It's an equation, it has to do something!

Your second guess might be that you will get 0. That's a pretty good guess! Because after three seconds, the rock will still be on the ground, at height zero.

But in fact, what you will discover if you do the math (*don't do it now, take my word for it!), is that you get a negative number. What does that mean?

What it means is that the equation is only valid during the time of the rock's flight: it doesn't know about the ground at all. The equation doesn't understand that the rock stops suddenly when it hits the ground. It blindly computes as if the rock was going to keep going, straight down, and hence finds a negative h. So in this case, the equation is just wrong! You know something about the problem that the equation doesn't know. So if you were asked "where will the rock be after three seconds?" you might use the equation: but when it gave you a negative number, you would have to be smart, and realize that it was lying to you. The right answer to that question is, "on the ground."

Question 5: If you plug h=16 into the height equation, what t will you get?

The big point I want to make with this question is that equations go both ways. We have an equation that relates t to h. You can plug in a t and get an h. But you can also plug in an h to get a t! In this case, the question you are asking the equation is, "When will the rock be at height 16?" The answer, of course, is after one second. So if you do the math right, you will wind up with t=1.

Question 6: If you plug h=0 into the height equation, what t will you get?

This might look like a rerun of question 1, but don't answer too quickly! The answer is t=0, of course, because at time zero, the height is zero. But wait a minute: isn't the height also zero after two seconds have elapsed? So t=2 is also a valid answer! What will the math say?

The answer is, it will say both. You will find both t=0 and t=2 come out of this one question.

Here I want to make a brief mathematical digression, which you can feel free to ignore if you like. h is a function of t. That means that for any given t there is exactly one h. But the reverse is not necessarily true. In principle, one h could have one t, or two, or none at all, or twenty. In this particular case, we will see 0, 1, or 2 answers, but never more. This might tip you off, if you remember your math, that a quadratic equation is at work. If it doesn't, that's okay too.

Question 7: If you plug h=10 into the height equation, what kind of values for t will you get? (ie how many answers; positive, negative, or zero?)

In this case, there is no way to actually figure out the number without going through the math. But you can figure out what kind of answers you would expect to get. This is very important, because if you go through the math and get answers that don't make sense, you know you did the math wrong!

In this case, what you are asking the equation is: "when will the rock be at height 10?" The answer is, of course, twice: once on the way up, and once on the way down. So you would expect to get two answers back, both positive. One answer should be between 0 and 1 seconds, and the other should be between 1 and 2 seconds; since those represent the times that the rock was moving up and moving down, respectively. Going further, you might even recognize that the number 1 should be exactly in between the two numbers: so you might get 0.8 and 1.2, or 0.7 and 1.3, or something like that. But that is as far as you can go without crunching the math.

Question 8: If you plug h=20 into the height equation, what kind of values for t will you get? (ie how many answers; positive, negative, or zero?)

This is a very different question from the last one, since the rock never actually reaches that height! You might expect to get a bogus answer, like we did in number 4. If you wrote that down, give yourself credit, you're doing fine. But in fact, although the equation doesn't know about the ground, it does know about the rock reaching the top. Because our equation is valid for the entire period of the toss, even if it isn't valid before and after. So as it turns out, you will get…no answer at all.

Question 9: If you plug h=-4 into the height equation, what kind of values for t will you get? (ie how many answers; positive, negative, or zero?)

By now, I hope you've become pretty familiar with our little equation. You know that it understands what the rock does between the throw, and hitting the ground; but it doesn't know about the ground. So you can probably guess that you will get some answer greater than 2; the equation believes that after two seconds, the rock keeps going down, and eventually reaches height -4.

But in fact, if you do the math, you will get two answers. Because the equation also doesn't know about the start of the rock's journey. So you will get one answer greater than 2, but you will also get a negative answer…signifying that the rock was below the ground, some time before we started counting!

Does that mean that the rock really started below the ground? No: once again, we know better than the equation. The rock was probably sitting on the ground, or maybe you were just putting it on the ground, or who knows what? before time t=0. But we know that we can't expect the equation to know that either.

For questions 10-13, we are going to introduce one more variable: v for the velocity, or speed, of the stone, measured in feet per second. v is positive when the stone is going up, and negative when the stone is going down. Like h, v is a function of time: that is, there is an equation that tells you v for any given value of t. I'll call this the velocity equation.

Question 10: If you plug t=0 into the velocity equation, what kind of values for v will you get? (ie how many answers; positive, negative, or zero?)

Remember, a positive v means you are going up: a negative v means you are going down. I hope it is clear that you will get one answer, and it will be positive. In fact, it will be 32. How do I know? Because we already said that the rock starts off at 32 feet per second. It pays to remember details.

Question 11: If you plug t=1½ into the velocity equation, what kind of values for v will you get? (ie how many answers; positive, negative, or zero?)

After 1½ seconds, the stone is going down. So you will get one negative answer. The exact number is impossible to determine without going through the math.

Question 12: If you plug v=0 into the velocity equation, what of value for t will you get? (Give an actual numerical answer.)

If you're having trouble with this one, let me urge you to turn the question into words. At what time is the speed of the stone zero?

If you think about it, I think you will see that there is only one answer. As the stone is going up, its speed is always positive. On the way down, its speed is always negative. But right at the top—for just a split instant—the stone stops, as it turns around. Its speed is neither positive nor negative. It is zero. So the answer to this question is, you will get 1. Because at time t=1 second, the stone is still.

Question 13: When v=0 what is h? (Give an actual numerical answer.)

This is a sneaky problem, in terms of the math. There is no equation (at least, we haven't introduced one) that relates v to h! So what you would actually have to do is use both of our equations, the height and velocity. First, you would use the velocity equation to find the time when v=0. As we discussed in question 12, you would find t=1. Then you would plug that into the height equation, and we're back to question 2: we wind up with h=16.

So it's getting more roundabout here. But we're also starting to answer some harder questions. Because in fact, this is exactly how a Physicist would go about figuring out how high something will go when you throw it! He would start with the observation that at the top, the velocity is zero—and he would work backward from there, to find the height. This is a great example of a problem where knowing all the equations in the world won't help you solve the problem, if you don't also think about the Physics of the situation.

Question 14: What's the point of all this?

I mean this one very seriously. Indulge me one more time and write down what you think the point of going through all this math, without the math, is.

If you're scoring yourself, give yourself credit for whatever you wrote down. (No credit if you wimped out and didn't write anything.) But here is how I would put it. Mathematical equations represent reality. When we do Physics, we go through a translation process: we translate a physical question into math, we solve the math, and then we translate the math back into a physical answer. This indirect sort of approach turns out to be a very powerful, very rich way of solving all kinds of problems which would be next to impossible otherwise.

But the danger is that we will leave our common sense behind, and trust the math too much! Math can never represent reality fully. And even when it gives the right answer, if you don't understand the Physics behind the problem, you haven't really learned anything useful or interesting. It's when you are using the math as a tool to give answers to problems that you fully understand—and you fully understand what the math is really doing for you—that it becomes useful, and fun. And the key to that fun is never abandoning your common sense.

That way, when you reach Relativity and Quantum Mechanics, and your common sense is suddenly wrong about everything, the real fun begins!

Epilogue or Appendix or Whatever: The Equations

After you have gone through all of the above, you may find it worthwhile to actually work through the numbers and confirm that you get what you expected. In fact, I would highly recommend it. When you see that the numbers do actually turn out the way you predicted, your confidence level goes way up.

So: throughout our period of concern, the rock is under the force of gravity, which happens to exert a constant acceleration. That isn't obvious, it's just a law of Physics. But that's where we start.

The equation for motion under constant acceleration is x=xo+vot+½at2 where x is the position at any given time, xo is the initial position, vo is the initial velocity, and a is the acceleration. This equation is also not obvious; although with a really good understanding of velocity and acceleration, you can derive it slowly using algebra, or very quickly using calculus.

In our case, the position x was actually a height h. The initial position xo was zero: that is, we started at height zero. The initial velocity vo was 32: we started at 32 feet per second. And the acceleration was -32, since the acceleration due to gravity happens to be -32 feet/second/second. (Also not obvious: this is just an experimental result.) So the actual height equation we were working with comes from plugging in all these numbers:

h = 32t - 16t2

The general velocity equation under constant acceleration turns out to be v=vo+at, so our velocity equation is:

v = 32 - 32t

Using those two equations, you should be able to get all the answers. Most of the questions require only very basic algebra: questions 5-9 involve quadratic equations.

As an example, let's take problem 1. We started by saying that t=0. So by the above equation, we see that h=32(0)-16(0)2 which is 0. So sure enough, the math gives us what we thought it would.

Note that the "no-answer" problem, number 8, winds up mathematically resolving to the square root of a negative number. This is one of my favorite examples of math, almost magically, doing what it has to in order to give you the right answer. Similarly, you will get one valid answer for question 5, and two valid answers for question 6. Try it!

Gary and Kenny Felder's Math and Physics Help Home Page
Send comments or questions to the author