# Linear Motion Problems

The steps that you use to solve most physics problems are very similar to what we did in dimensional analysis. The only difference is that, when using "real" physics, we'd like to know the equation ahead of time, rather than figuring it out on the fly using the units. So, here is the procedure:
• Figure out what you are trying to find, and write down the variable that represents it.
• Figure out what things you have. Write down their values - both variable name, number and units, in the format "vi = 4 m / s". How to identify what variables you have in a problem is discussed in more depth below.
• Convert all the things you have into units that involve kg, m, and s. I should see the conversion process written out on your paper. These units are the standard building blocks for all units in physics; the equations we use assume that you have units made up of these. Thus, you need to get rid of things like cm, g, or h. This is why we studied factor-label conversion first. As a review, here are the conversion equations you need to know: 1000 m = 1 km      60 s = 1 min      1000 g = 1 kg1000 mm = 1 m      60 min = 1 h100 cm = 1 m10 mm = 1 cm
• Now you are ready to approach the problem; you've laid out all the pieces, and you need to figure out how to put them together. Being prepared like this before I start looking for an equation gives me a confident feeling about the problem, sort of like when you're doing a jigsaw puzzle and you go through the box first and pull out all the edge pieces, or all the pieces with blue on them. You have a manageable number of variables to worry about, and you have them all together in one place.

Now we want to find an equation that will give us the thing we want, using what we have. You won't always be lucky enough to get an equation with your wanted variable on one side and all the others clustered on the other side. However, any time that you know all the variables in an equation except the one you want, you can solve for that variable using algebra.

So, you want to look through your list of equations and find one that includes the variable you want, and for which all the other variables are known. The table below gives a list of equations you might need, and what variables are involved in them (I find that this list makes it easier to quickly see which ones I can use). The uses of these equations are described in more detail in section 2 below.
EquationVariables

vavg = (Δx)/(Δt)
vavg, Δx, Δt
vavg = (xf - xi)/(tf - ti)vavg, xf, xi, tf, ti
a = (Δv)/(Δt)a, Δv, Δt
vf = vi + aΔta, vf, vi, Δt
Δx = .5(vf+ vi)ΔtΔx, vf, vi, Δt
Δx = viΔt + .5a(Δt)2Δx, Δt, a, vi

• Plug your numbers into that equation. I find that it's easier just to work with bare numbers in solving the equation, since plugging the units in as well gets me confused; however, it's more professional to put the units in
• Finally, you want to solve your equation using algebra.

## 1. What do all these variables mean?

### 1.1. Position and Displacement: units of (m)

All the variables with "x" in their names have to do with your position or displacement. They all have units that are some sort of distance measurement, like m, km, or cm. Anything dealing with your location, with where you are, is an x. xi is your starting position, the loction or height that you started from. xi is the ending position, what you went to. The change in position, often described as how far you moved or fell, is the Δx.

In free fall problems, we often talk about the height of something above the ground. This is also a position. Here, x increases as you go up and decreases as you go down.

You should be aware that all these x's can be positive or negative. If the height or position of some object decreases during the time the problem is concerned with, then the Δx is negative. So, it's not simply an answer to "how far;" it also tells you how far in what direction.

Examples:
 A ball was thrown from a height of 2.8 m to a height of 4.5 m. These measurements are in m, so they are x's of some sort. The problem uses from and to, and definite starting and ending positions are given relative to something else (the ground). So, xi = 2.8 m and xf = 4.5 m. How far does the car go? Here, the question is a "how far." We don't care where it was at the start and end, just the distance traveled in between. So, it is a Δx.

### 1.2. Time: units of (s)

When dealing with time, most often what you have to deal with is how long it took for something to happen, whether that something is a change in velocity, or an object moving some distance. Any measurement of how much time passed while something happened is a Δt. If time is given by a clock that keeps running, or by the horizontal axis of a graph, you might have to find your Δt by subtracting the start time from the end time for some event.

Time is always measured in something like seconds, hours, or minutes. The standard units for time are seconds.

### 1.3. Velocity: units of (m / s)

Displacement and time are enough to tell us where an object is, and when it was there. But this chapter is about linear motion, not jsut objects standing around on a number line. As soon as an object starts moving, it has a velocity. Velocity is defined as how much your position changes every second. Therefore, it has units of (m / s).

Velocity cannot be measured directly. It can only be measured by observing how far an object moves in a known time interval, and dividing.

There are four different types of velocities that you should be aware of. When an object is moving at a more or less constant pace, we say that it is moving at some average velocity vavg. You will know when a problem deals with average velocity becasue it will always say "average." Otherwise, the assumption is that if you are told a velocity, that velocity is accurate only for the exact second that you measured it at.

When objects are accelerating, their velocity is changing. In this situation, we start to have to deal with three more types of velocity: initial velocity, final velocity, and change in velocity. vi is the speed or velocity that an object starts at, the velocity that it accelerates from. vf is the speed or velocity that an object ends up at after some change, the velocity that it accelerates to. The difference between these two is Δv, the amount that the speed of an object changes by in a given time.

The final thing you need to know about velocity is that one particular velocity, 0 m / s, has a lot of other sneaky names. This is the velocity that means that an object's position is not changing as time goes on - it is the velocity of an unmoving object. We might also say that the object is "at rest," or "still." If an object is "released" at the start of its motion, or if it "stops" at the end of its motion, this means that the start or end velocity, respectively, is zero.

Examples:
 A car going 2 m / s stops. The 2 m / s is clearly a velocity, by its units. To figure out what velocity it was, I might try to explain the situation with different words: "A car was going at 2 m / s and then it stopped." Now it's clear that vi = 2 m / s. The problem also tells us that the car stopped. "Stop" means a velocity of 0 m / s, and it is clearly at the end of the motion, so vf = 0 m / s. The car sped up by 4 m / s. This deals with a change in the velocity of the car. The original velocity could be 0 m / s or 120 m / s, for all we know. So, all we have is Δv = +4 m / s. Actually, if the car is going backwards, we might have Δv = -4 m / s; all we know is that it "sped up."

### 1.4. Acceleration: units of ((m / s) / s)

An acceleration is easy to recognize, because it always has units of m / s2, or at least, of some sort of velocity per time. It might be called "speeding up" or "slowing down." It's also not hard to tell when it's what the problem wants; the problem will have to say "at what rate is he accelerating?" or something of the sort.

The one tricky thing here is in free fall problems, where the acceleration is given to you for free. Any accelerating object has an acceleartion of g = -9.8 m / s2. That is to say, its velocity decreases by 9.8 m / s every second that it is in the air.

## 2. What were those equations, again?

Listed below are all the equations that you will need for linear motion problems. Some are just definitions, telling how a variable si found; others give quite useful relationships between the variables.   This is how all the Δ_ variables are defined. Change is some quantity, whether position, time, or velocity, is just the ending value minus the starting value. This equation defines what average velocity is: the distance that you travelled, divided by the time it took. You might want to use this equation in some other form. For example, Δx = vavgΔt. That is, the change in your position if you move at a certain average velocity, a certain distance per second, is just the time you were traveling for, times the displacemnt you gain each second. With this equation, we introduce the idea of acceleration. Acceleration is the amount that your velocity changes by every seond; thus, it is found by dividing the amount that your velocity changed by by the time that it took for your velocity to change. You can use this equation in different forms; for example, vf = vi + aΔt. This equation means that if you start at some velocity vi and add an acceleration of a to your velocity every second for Δt seconds, you will end up with a velocity of vi.  These two equations are the ones that you will find yourself using most in acceleration problems. They tell you how to find the distance that an objects travels, while accelerating, in a given time. One uses the beginning and ending velocities, one uses just the starting velocity and the acceleration.

## 3. Tips for solving algebraic equations

Solving the equation may be the hardest part for some of you. Here are some tips on algebra techniques you may find useful: If the variable you are trying to solve for is on the bottom, multiply both sides by that variable. This will cancel it out from the bottom and bring it up to the top on the other side of the equation. If the variable you want is being multiplied by some coefficient, divide both sides by that coefficient. If at all possible, simplify expressions involving just numbers before you go on to solve for the variable. Don't forget that you can take a square root! Also, remember that it's easier to figure out what the square root does if you get the squared variable all by itself on a side first.

Remember, I won't take many points off for algebra mistakes. The physics part of the problem is the part that takes you from a description of a situation into an algebraic equation. However, you do need to know algebra to do well in this class. If you see one of the situations above, it should leap off the page and tell you that it's time to use one of these tricks.