The Physics Philes, lesson 2: Math for Miles!
In which math is relearned, more books are purchased, and heads are banged.
Last week was my first full week of lessons. It nearly killed me. OK. Not really. But I have been sleeping really well.
As I wrote last week, I’m really not a math person. Which makes my choice of physics as a topic a bit curious. But I was convinced that, provided with an explanation. I was humming along fine until I came across this phrase:
“…on a graph of position as a function of time…”
I also ran across some problems that required me to know and execute derivative equation rules. I had no idea what to do. I thought I could make it work. But I can’t. Not on my own. So I did what any self-respecting student would do. I bought a study book.
It turns out that the warnings issued by many a math teacher are true: Math is, indeed, cumulative knowledge. If you don’t grasp a basic concept, it’s hard to move on to more advanced topics. Since it’s been about 10 years since I’ve had a math class (and I’ve never had calculus) I’ve forgotten most of what I used to know. So I have to start from the beginning. And by “the beginning,” I mean algebra. This project has morphed into my own part-time class schedule: physics and calculus. So in addition to discussing a physics concept every week I’ll discuss some math that I am either relearning or learning for the first time.
Let me apologize if this is incredibly boring for you. I didn’t fully realize when I started how much algebra, geometry, and trigonometry I had forgotten. The examples I use are taken from my study book, because I want to make sure: a. I got the answer write and b. the problem is set up correctly. I’ll get to the physics in a bit.
Graphs seem to be an important part of physics so far, likewise with calculus. So I needed to brush up on my linear equations.
Linear equations always have three components: two variable terms and a constant term. Standard form of a linear equation will look like this:
Ax + By = C
OK. That looks familiar. I seem to remember solving equations like that years and years ago. I wonder if I can still hack it. I’ll put this equation in standard form:
3x – 4y – 1 = 9x + 5y – 12
I’ll first need to subtract 9x from both sides:
(3x – 9x) – 4y – 1 = (9x – 9x) + 5y – 12
-6x – 4y – 1 = 5y – 12
Next, I’ll need to subtract 5y from both sides:
-6x (-4y – 5y) – 1 = (5y – 5y) – 12
-6x – 9y – 1 = -12
-6x – 9y – 1 = -12
Hey Mr. 1! What are you doing on that side of the equation? Get over there with Ms. 12:
-6x – 9y – (1 + 1) = (-12 + 1)
-6x – 9y = -11
Whoops! See that -6x? That’s supposed to be a positive number in standard form. I better multiply each side by -1:
-1 (-6x – 9y) = -1 ( -11)
6x + 9y = 11
Ah yes. It’s coming back to me. Slowly, and I need a lot more practice, but it’s there. A rich vein of algebra-ore waiting to be mined. But there is oh-so-much more. Next I need to relearn the two major ways to create the equation of a line.
The first is the slope-intercept form: y = mx + b, where m represents the slope of the line and b represents the y-intercept of the line. Solve for y if the slope of the line is -3 and the y-intercept is 5:
y = -3x + 5
See what I did there? I just plugged in the values. Easy peasy lemon squeezy. The second major way to create the equation of a line is point-slope form. You can use this form when you only have a point and a slope, but no y-intercept. If we have point (x1, y1) and slopem, the equation will be:
y – y1 = m(x – x1)
The x and y with subscripts represent the point coordinates. Let’s try one. Find the equation of the line through point (0, -2) with slope 2/3. First, let’s plug in the values we have into the above equation:
y – (-2) = 2/3(x – 0)
y + 2 = 2/3(x – 0)
Oh noes! Fractions. It’s possible that I got the mechanics of this next part wrong. If I did, please tell me in the comments. Now, we multiply each side by 3 to get rid of the fraction:
3(y + 2 = 2/3(x – 0)
3y + 6 = 2(x – 0)
Let’s get that right side simplified.
3xy + 6 = 2x
Subtract 3x from both sides:
3y + 6 -3y = 2x – 3y
6 = 2x – 3y
Ta-da! I did it! I think. Right? Someone please check my work.
That’s not quite all the math I retaught myself this week, but you came here for some physics. So that’s what you’re going to get.
This week was dominated by velocity and acceleration. This is how we can talk about the motion of an object. Today I’ll be discussing average velocity and average acceleration (as opposed to instantaneous velocity and instantaneous acceleration. Mostly because the instantaneous stuff completely confounds me.) First, velocity!
Average velocity is the change in position of an object divided by the change in time. To find the average velocity (represented by v), all we have to do is divide the change in position (represented by x) by the change in time (represented by t). Or, in math-speak:
v = Δx/Δt
Whoa there. What’s with those shapes? That’s the Greek letter delta, used in algebra to represent “change in.” Let’s say we have a starting place of x0 and an ending point of xf. To find Δx, we just subtract xf from x0. We do the same with Δt. We subtract the final time (tf) from the initial time (t0). Like this:
v = xf – x0 / tf – t0 = Δx/Δt
Hm. That might not be very clear. Let’s do an example. Let’s say you’re on your bike. You start from rest and ride for 10 s. In that 10 s you go 50 m. What is your velocity? All we have to do is plug the values into equation:
v = 50 m/10 s = 5 m/s
Pretty easy, right? Once you get past all the Greek letters and subscripts, it’s a pretty straight forward concept. We can also use this equation to solve for the position of the object by solving for x.
Now that we have average acceleration down, let’s figure out average acceleration. Acceleration is the change in velocity divided by the time in which that change takes place. We come across our good friend Δ again:
a = vf – v0/tf – t0 = Δv/Δt
OK! Let’s plug in some values! Let’s say that you’re at a stop light. When the light turns green you accelerate. After 10 s the car is moving at 50 m/s. What is the acceleration?
a = (50 m/s – 0 m/s) / (10 s – 0 s)
a = 50 m/s / 10 s = 5 m/s2
Notice the difference between the answers for velocity and acceleration? For velocity, the unit is m/s. For acceleration, the unit is m/s2. That’s how you know when we’re talking about velocity and when we’re talking about acceleration.
I mentioned above that I was having a little trouble with instantaneous acceleration and instantaneous velocity. But my existing math experience does not encompass derivative equations, which is evidently necessary to calculate instantaneous acceleration and velocity. This is cause for great concern; I’m afraid I won’t be able to catch up on the math in time to figure out the physics. I know we have a bunch of smart-pants readers. You will get 100 gold stars if you can explain the concept of limits and derivative equation rules to me in a useful way. The challenge is issued. Will you accept? (Please accept!)
This is, of course, only a fraction of what I actually covered this week. Did I make any mistakes? Should I have gone about something in a different manner? Let me know in the comments!