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In section 1 of the Physics guide, topics such as basic one dimensional movement will be discussed.
Some General Terms
In physics, there are a lot of terms that need understanding before actual application becomes possible. These are a list of terms essential to the understanding of this section.
 Magnitude
 Direction
 Scalar
 Vector
 Distance
 Displacement
 Speed
 Velocity
 Acceleration
Magnitude is a measurement with a unit. Pretty basic, right? Example: 20 grams
Direction is the degree of variance by which an object is moving to the East axis. In better words, it is where the object is going compared to a positive axis. An example of this is 20^{o} east. This means the subject is going slightly north but mostly east.
A Scalar is a quantity represented by magnitude. It is unaffected by direction.
A Vector, on the other hand, is represented by both magnitude and direction.
Distance, a scalar quantity, is how much space the object covered in total. This represents the whole path the object has taken.
Displacement, a vector quantity, is the change in position of an object. This is pretty much the measurement in a straight line from point A to point B. Remember, displacement requires a direction.

Let's say the left point is Point A and the right point is Point B. If someone ran from Point A to Point B on the red path, then the red path is the distance, and the black path is displacement. 
Speed is a scalar quantity that represents how fast an object was going in meters per second. The speed of an object is determined by its change in distance divided by the amount of time it took traveling.
Velocity is a vector quantity that represents an object's speed and direction. It is represented by an object's displacement divided by time.
Note the difference between Speed and Velocity!
Acceleration is an object's change in velocity divided by time. Acceleration requires a force as described by Newton's Second Law (F=ma). Acceleration is a vector quantity that must go the direction of the force.
Motion in one dimension is like motion on a straight path, or line. There are only two directions: forward (positive) and backward (negative). This makes many of the calculations pretty straightforward. In order to determine the attributes of an object in motion, some formulas relating velocity, displacement, time, and acceleration are needed. Remember, displacement is change of position of an object. In one dimension, the only real difference between displacement and distance is that if an object changes direction at any time, the two measurements will differ. Otherwise, they are congruent in magnitude. Let's look at a few equations.
Finding the Average
These first two equations deal with finding the average of a measurement over a given time. When an object is in motion, it's movement is not always constant. It can stop, speed up, slow down, or change direction. These two equations find the average, or what the acceleration/velocity would have been if the object was constant.
The first equation is how to find the average velocity of an object between any two times.
What?! Triangle x? The triangle is Delta (a Greek letter) that means "the change in". So, the equation translates to velocity (v) is the change in displacement (x) divided by the change in time (t). So, to find the average velocity of an object, you would find the difference in the final position of an object and its initial position and divide that by the objects difference in final time and initial time. Here is what is happening in the equation: the velocity of an object is how many meters (or other unit) its traveling for every second with a direction, so to find that average velocity, you would divide the amount it moved by the amount of time that passed by.
Here's an example. Let's say there is a player on a football field. He is at the 10 yard line. He runs to the 40 yard line in 6 seconds. What is his average velocity? Using our equation (1), we can plug in our values. $\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} v = {40 yd10 yd \over 6 s0 s}$ When we solve, we get 30 yards divided by 6 seconds which when simplified means the football player was running 5 ^{yd}/_{s}.
The equation can be used to find the time or displacement of an object if the velocity is known. Let's take that football player again and say that from the 40 yard line, he ran into the end zone at the same average velocity. How many seconds did it take him? Let's observe our equation again. (1) Substitute what you know: velocity v, position 1 x_{0}, and position 2 x_{f}. Here is what happens. $\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} 5^{yd}/_{s} = {0yd40yd \over t_{f}0s}$ The reason the velocity is negative is because the football player is now going the opposite direction. Remember velocity needs direction, and direction in one dimension is represented by positive and negative numbers! Using algebraic solving, we need to find t_{f}. First, get rid of the 0's. Multiply both sides by t_{f}, and then divide by 5^{yd}/_{s}. In the end, we get t_{f} = 8 seconds. It took the football player 8 seconds to run from the 40 yard line to the end zone.
The second equation is how to find the average acceleration between any two times.
This equation translates to acceleration (a) is the change in velocity (v) over the change in time (t). Acceleration is different from velocity. Acceleration is the rate of increase/decrease of velocity rather than the rate of distance. Another way this can be thought of is the rate of the rate of increase/decrease of displacement. So, for a positive acceleration, the velocity of that object is increasing. For a negative acceleration, the force on the object is going the opposite of the positive direction meaning that an object already going the positive direction will slow down and start going the negative direction.
Here is an example. A car is at an intersection on a road with a speed limit of 40 mph. When the light turns green, the car accelerates to that speed limit in 6 seconds. What was the car's average acceleration? First, identify the equation for finding average acceleration. (2) What do you know? We know the car's beginning velocity, 0, it's final velocity, 40 mph, and the amount of time it took, 6 seconds. Warning! Notice that there is a problem with the units. The velocity is in miles per hour while the time is in seconds. Problems cannot be solved when different units are used! So, we need to convert miles per hour to miles per second. To do this, we multiply by the number one using separate units. Here is what we get. $\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} {40mi \over hr}*{hr \over 3600s} = 0.011^{mi}/_s$ The hours cancel out as a result of reciprocal multiplication, and by simplification, we get our answer. Now that seconds are being used, we can solve our problem. Plug in our equation (2) what you know. $\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} {0.011^{mi}/_s \over 6s}$ When we simplify the problem, we get 0.00183 ^{mi}/_{s2}. So, the car's average acceleration was 0.00183 ^{mi}/_{s2}. Note that this equation can also solve for velocity or time if the acceleration is known.
Graphical Application
These equations (1)(2) are used for finding the average measurement of an object's movement. What the object actually does is not taken into account in the equations. So how do we represent the object's real motion? Graphs are used to represent what an object does over a given amount of time. The graph gives a simple two dimensional representation of what the object does on its one dimensional path. Each equation has its own type of graph.
Position vs. Time
A position vs. time graph shows the object's total displacement over a given amount of time. The xaxis represents time, and the yaxis represents displacement. Here is the path of an object taken over ten seconds using position vs. time graph.
This is a graph of a professor's walk with position being distance from the school. With this graph, we can answer questions like:
1) How far from the school was the professor at 2 seconds?
2) What was the professor's maximum distance from the school?
3) What was the professor's average velocity from 0 seconds to 4 seconds?
4) What was the professor's average velocity from 4 seconds to 7 seconds?
5) What was the professor's average velocity from 7 seconds to 10 seconds?
6) What was the professor's average velocity for the whole trip?
7) Why was the professor's walk so short?
Let's answer each question while explaining the process.
Question 1
This is a simple question. To find the position of the professor at any time, all you have to do is find the specified time on the xaxis and move up or down until you find the line. Identify the point's y location, and that it the position. So, in this example, we need to find the professor's location at two seconds. So, we find two seconds on the xaxis, and move up till we find the line. It turns out that the professor was three meters from the school at two seconds.
Question 2
Here is another simple question. To find the maximum distance from a starting point, all you have to do is find the highest point on the graph regardless of time. Find the y component of that highest point, and the question is answered. In the professor's walk, his maximum distance is where the line is highest. By looking at the graph, it can be assumed that the professor's farthest distance from the school is 4 meters.
Question 3
Ok, here is where our formulas come in handy. When finding the average velocity on a graph, we are finding the constant velocity of an object between two points even though the velocity was not necessarily constant. It would be like drawing a straight line between the two point's of interest. We need to find the professor's average velocity between 0 seconds and 4 seconds. First, we need to identify the formula we are going to use. Let's use the first one we learned. (1) This formula represents the change in position over the change in time. Substituting into the formula, we get $\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} {4m0m \over 4s0s}$. When simplified, we find that the professor's average velocity between 0 seconds and four seconds was 1 ^{m}/_{s}.
Question 4
Again, we have to find the average velocity. This time, we are doing it between 4 seconds and 7 seconds. Notice that the line between these two points is flat, or parallel to the xaxis. Hmm, what does this mean? Let's see, first, identify your formula. (1) Substitute into the formula what you know. For this example, we know that x_{f} is 4 meters, x_{0} is 4 meters, t_{f} is 7 seconds, and t_{0} is 4 seconds. This yields $\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} {4m4m \over 7s4s}$ which when simplified is 0 ^{m}/_{s}. This means that the professor did not move for three seconds.
Question 5
We are finding average velocity again. This time, it's between 7 seconds and 10 seconds. When we look at the line, we notice that the line is going in the general down direction. This is important, and you're about to see why. Identify the formula. You should hopefully know it by now. When we substitute our values, we get something like $\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} {0m4m \over 10s7s}$. The 0 is first because that is x_{f}, the professor's final position. When we simplify all the way, we get ^{4}/_{3} ^{m}/_{s}. The answer is negative meaning the professor was going the opposite direction, towards the school in this case. Always remember that velocity must have a direction, and in the first dimension, direction is symbolized by positive and negative values!
Question 6
Now, we are finding the professor's average velocity for the whole trip, or from 0 seconds to 10 seconds. Identify the formula (1) and plug in the values. We get [[$ \definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} {0m0m \over 10s0s} which is, surprisingly or not, 0 ^{m}/_{s}. Remember that average velocity is finding what the object's constant velocity would have been. Even though the professor did move, his overall change in position was 0 meters, so he technically did not go anywhere within that time frame. His distance, however, is 8 meters, so it can be confirmed that he did go somewhere, but his overall change in position in the given amount of time is 0.
Question 7
It's funny how the professor's trip was so short. His total trip from the school and back lasted 10 seconds, and he only traveled 4 meters from the school. Technically, this question cannot be answered, but whatever happened can be inferred. What type of trip would make the professor take only 10 seconds? We can come up with possibilities such as maybe he needed to get something out of his car which was parked on the curb. Maybe there was a stray student outside. Maybe the professor spotted a penny on the ground. Who really knows? However, we do know what it's not such as he went to eat lunch or he wanted to jog a mile. If you are to find the most logical explanation for any particular model know what it might be, but also know what it can't be.
Velocity vs. Time
A velocity vs. time graph shows the object's velocity change over a given amount of time regardless of displacement. The xaxis is time, and the yaxis is velocity. Here is the velocities of an object over 10 seconds.
This is a graph of a professor's run with his varying velocity. With this graph, we can answer these questions.
1) What was the professor's velocity at 4 seconds?
2) What was the professor's maximum velocity?
3) What was the professor's average acceleration from 0 seconds to 2 seconds?
4) What was the professor's average acceleration from 2 seconds to 3 seconds?
5) What was the professor's average acceleration from 3 seconds to 8 seconds?
6) What was the professor's average acceleration over the whole trip?
Let's answer each question while explaining the process.
Question 1
Just like in the first graph, this question is an easy one to answer. To find the velocity at any given moment of time, all you have to do is find the time on the xaxis and move vertically until you find the point of intersection. Simply, in this example, we would find the time of four seconds and move up. It seems to cross at six meters per second. So, the professor was running at six meters per second at four seconds.
Question 2
Just like in the first graph, the maximum velocity is where the graph reaches its highest. On this graph, the line maxes out at 7 meters per second. The professor's maximum velocity was 7 meters per second.
Question 3
This is where the difference becomes apparent. This time, we must find the acceleration of the object. This would be the median acceleration between two points on the graph. So, like before, we would take the velocity with its time for both points and input those values into the acceleration equation. (2) So, for this example, we need to find the velocity of the professor at each given time. Then, we must input the values into the equation. At 0 seconds, the professor was going 0 ^{m}/_{s}. At 2 seconds, he was going 7 ^{m}/_{s}. When we put the values into the equation, we get $\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} {70} \over {20}$. This is simplified to 3.5 ^{m}/_{s2}. The professor's average acceleration between 0 seconds and 2 seconds was 3.5 ^{m}/_{s2}.
Question 4
Again, we have to find the professor's average acceleration, except this time, the line is flat. Before, the velocity was 0. Logically, since the professor had no change in velocity, his acceleration was 0, but to be sure, let's check our formula. (2) Put in the values to get $\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} {77} \over {32}$. This simplified is 0 ^{m}/_{s2}.
Question 5
More acceleration. Now, we are finding his acceleration from 3 seconds to 8 seconds. Notice that the line goes down. In our previous graph, the velocity was negative where the line went down. In this example, the acceleration exhibits the same behavior. Input the numbers to get $\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} {07} \over {83}$. This simplified is 1.4 ^{m}/_{s2}. The acceleration was negative meaning the professor was slowing down. In this case, he slowed down to a stop. Just like before, the acceleration was negative for a downward trend on the line.
Question 6
Now, we need to find the average acceleration over his whole trip from 0 to 10. Remember, this is like drawing a line between the two points. Input your values for the equation to get $\definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} {10} \over {100}$. When the equation is simplified, we get 0.1 ^{m}/_{s2}. His overall acceleration was very slow.