5-20-98
We live in a 3-dimensional world, so why bother analyzing 1-dimensional situations? Basically, because any straight-line (as opposed to rotational) motion problem can be separated into one or more 1-dimensional problems. Problems are often analyzed this way in physics; a complex problem can often be reduced to a series of simpler problems.
In analyzing the motion of objects, there are four basic parameters to keep track of. These are time, displacement, velocity, and acceleration. Time is a scalar, while the other three are vectors. In 1 dimension, however, it's difficult to see the difference between a scalar and a vector! The difference will be more obvious in 2 dimensions.
The displacement represents the distance travelled, but it is a vector, so it also gives the direction. If you start in a particular spot and then move north 5 meters from where you started, your displacement is 5 m north. If you then turn around and go back, with a displacement of 5 m south, you would have travelled a total distance of 10 m, but your net displacement is zero, because you're back where you started. Displacement is the difference between your final position (x) and your starting point (xo) :
Imagine that on your way to class one morning, you leave home on time, and you walk at 3 m/s towards campus, and let's say you have to walk east to get to campus. After exactly one minute you realize that you've left your physics assignment at home, so you turn around and run, at 6 m/s, back to get it. You're running twice as fast as you walked, so it takes half as long (30 seconds) to get home again.
There are several ways to analyze those 90 seconds between the time you left home and the time you arrived back again. One number we can calculate is your average speed, which is defined as the total distance covered divided by the time. If you walked for 60 seconds at 3 m/s, you covered 180 m. You covered the same distance on the way back, so you went 360 m in 90 seconds.
Average speed = distance / elapsed time = 360 / 90 = 4 m/s.
The average velocity, on the other hand, is given by:
Average velocity = displacement / elapsed time.
In this case, your average velocity for the round trip is zero, because you're back where you started so the displacement is zero
We usually think about speed and velocity in terms of their instantaneous values, which tell us how fast, and in what direction, an object is travelling at a particular instant. The instantaneous velocity is defined as the rate of change of position with time, for a very small time interval. In a particular time interval delta t, if the displacement changes by 
, the velocity during that time interval is:
The instantaneous speed is simply the magnitude of the instantaneous velocity.
An object accelerates whenever its velocity changes. Going back to the example we used above, let's say instead of instantly breaking into a run the moment you turned around, you steadily increased your velocity from 3m/s west to 6 m/s west in a 10 second period. If your velocity increased at a constant rate, you experienced a constant acceleration of 0.3 m/s per second (or, 0.3 m/s2).
We can figure out the average velocity during this time. If the acceleration is constant, which it is in this case, then the average velocity is simply the average of the initial and final velocities. The average of 3 m/s west and 6 m/s west is 4.5 m/s west. This average velocity can then be used to calculate the distance you travelled during your acceleration period, which was 10 seconds long. The distance is simply the average velocity multiplied by the time interval, so 45 m.
Similar to the way the average velocity is related to the displacement, the average acceleration is related to the change in velocity: the average acceleration is the change in velocity over the time interval (in this case a change in velocity of 3 m/s in a time interval of 10 seconds). The instantaneous acceleration is given by:
As with the instantaneous velocity, the time interval is very small (unless the acceleration is constant, and then the time interval can be as big as we feel like making it).
On the way out, you travelled at a constant velocity, so your acceleration was zero. On the trip back your instantaneous acceleration was 0.3 m/s2 for the first 10 seconds, and then zero after that as you maintained your top speed. Just as you arrived back at your front door, your instantaneous acceleration would be negative, because your velocity drops from 6 m/s west to zero in a small time interval. If you took 2 seconds to come to a stop, your acceleration is -6 / 2 = -3 m/s2.
When the acceleration of an object is constant, calculations of the distance travelled by an object, the velocity it's travelling at a particular time, and/or the time it takes to reach a particular velocity or go a particular distance, are simplified. There are four equations that can be used to relate the different variables, so that knowing some of the variables allows the others to be determined.
Note that the equations apply under these conditions:
The equations are:
Doing a sample problem is probably the best way to see how you would use the kinematics equations. Let's say you're driving in your car, and approaching a red light on Commonwealth Avenue. A black Porsche is stopped at the light in the right lane, but there's no-one in the left lane, so you pull into the left lane. You're travelling at 40 km/hr, and when you're 15 meters from the stop line the light turns green. You sail through the green light at a constant speed of 40 km/hr, and pass the Porsche, which accelerated from rest, at a constant rate of 3 m/s2, starting at the moment the light turned green.
(a) How far from the stop line do you pass the Porsche?
(b) When does the Porsche pass you?
(c) If a Boston police officer happens to get you and the Porsche on the radar gun at the instant the Porsche passes you, will either of you be pulled over for speeding? Assume the speed limit is 50 km/hr.
Step 1 - Write down everything you know. Define an origin - the stop line is a good choice in this problem. Then choose a positive direction. In this case, let's take the positive direction to be the direction you're traveling. Decide on a system of units...meters and seconds is a good choice here, so convert your speed to m/s from km/hr. Drawing a diagram is also a good idea.
Origin = stop line
Step 2 - Figure out what you need to solve for. At the instant you pass the Porsche, the x values (yours and the Porsche's) have to be equal. You're both the same distance from the stop line, in other words. Write out the expression for your x-value and the Porsche's. We'll use the equation:
For you : x = -15 + 11.11 t + 0
For the Porsche : x = 0 + 0 + 1/2 (3) t2 = 1.5 t2
At some time t, when you pass the Porsche, these x values will be the same. So, we can set the equations equal to one another and solve for time, and then plug the time back in to either x equation to get the distance from the stop line. Doing this gives:
-15 + 11.11 t = 1.5 t2
Bringing everything to one side gives:
1.5 t2 - 11.11 t + 15 = 0 This is a quadratic equation, which we can solve using the quadratic formula:
where a = 1.5, b = -11.11, and c = 15
This gives two values for t, t = 1.776 s and t = 5.631 s.
What do these two values mean? In many cases only one answer will be relevant, and you'll have to figure out which. In this case both are relevant. The smaller value is when you pass the Porsche, while the larger one is when the Porsche passes you back.
To get the answer to question (a), plug t = 1.776 into either of your x expressions. They should both give you the same value for x, so you can use one as a check.
For you, at t = 1.776, x = 4.73 m.
For the Porsche, at t = 1.776 s, x = 4.73 m.
We've actually already calculated the answer to (b), when the Porsche passes you, which is at t = 5.6 s.
To get the answer to part (c), we already know that you're travelling at a constant speed of 40 km/hr, which is under the speed limit. To figure out how fast the Porsche is going at t = 5.631 seconds, use:
v = vo + a t = 0 + (3) (5.631) = 16.893 m/s.
Converting this to km/hr gives a speed of 60.8 km/hr, so the driver of the Porsche is in danger of getting a speeding ticket.
Objects falling straight down under the influence of gravity are excellent examples of objects travelling at constant acceleration in one dimension. This also applies to anything you throw straight up in the air which, because of the constant acceleration downwards, will rise until the velocity drops to zero and then will fall back down again.
The acceleration experienced by a dropped or thrown object while it is in flight comes from the gravitational force exerted on the object by the Earth. If we're dealing with objects at the Earth's surface, which we usually are, we call this acceleration g, which has a value of 9.8 m/s2. This value is determined by three things: the mass of the Earth, the radius of the Earth, and a number called the universal gravitational constant. We'll be dealing with all that later in the semester, though, so don't worry about it yet. For now, all you need to remember is that g is 9.8 m/s2 at the surface of the Earth, directed down.
A typical one-dimensional free fall question (free fall meaning that the only acceleration we have to worry about is g) might go like this.
You throw a ball straight up. It leaves your hand at 12.0 m/s.
(a) How high does it go?
(b) If, when the ball is on the way down, you catch it at the same height at which you let it go, how long was it in flight?
(c) How fast is it travelling when you catch it?
Origin = height at which it leaves your hand
Positive direction = up
(a) At the very top of its flight, the ball has an instantaneous velocity of zero. We can plug v = 0 into the equation:
This gives:
0 = 144 + 2 (-9.8) x
Solving for x gives x = 7.35 m, so the ball goes 7.35 m high.
(b) To analyze the rest of the problem, it's helpful to remember that the down half of the trip is a mirror image of the up half. In other words, if, while going up, the ball passes through a particular height at a particular velocity (2 m/s up, for example), on its way down it will pass through that height at the same speed, with its velocity directed down rather than up. This means that the up half of the trip takes the same time as the down half of the trip, so we could just figure out how long it takes to reach its maximum height, and then double that to get the total time.
Another way to do it is simply to plug x = 0 into the equation:
This gives 0 = 0 + 12 t - 4.9 t2
A factor of t can be cancelled out of both terms, leaving:
0 = 12 - 4.9 t, which gives a time of t = 12 / 4.9 = 2.45 s.
(c) The answer for part (c) has to be 12 m/s down, because of the mirror-image relationship between the up half of the flight and the down half. We could also figure it out using the equation:
v = vo + a t
which gives:
v = 12 - 9.8 (2.45) = -12 m/s.