Motion Along a Straight Line
The world, and everything in it, moves. Even things that seem stationary, such as a roadway, move with Earth’s rotation. The Earth orbits around the Sun. The Sun orbits around the center of the Milky Way, and the galaxy moves relative to other galaxies.
This post explores the elements needed to understand motion along a straight line.
We’ll start with basic concepts such as position, displacement, and average velocity. Eventually we’ll speed up and explore the definition of acceleration, and from there, a whole new way to understand rectilinear motion will unveil.
If you’ve been studying this topic in physics and struggle to understand the concepts, I invite you to explore this page and see what I have to share, since I’ve made this not only user-friendly, but engaging and inviting too.
In order to understand motion, we have to first understand no-motion. Life has it this way and I love it. It takes you to different places to get where you wanted to be in the first place.
What I just said may sound nonsense for some, and perhaps profound for others. All I’m trying to do is give you some perspective to approach this concept, the concept of position.
Position is the equivalent to frames in animations. Animations seem to move, to have motion, yet they are just a sequence of frames moving a certain speed to create the illusion of motion. Each frame represents a still image of the animated object. This still image is needed to perceive the effect of motion in the animation, and without it, the animation could not be understood.
In order to see an object moving, we need to know where is moving. We need to give it a reference, so that when we see the object move, it is easy to say it moved from here to there.
The here and the there, are spots that show that in deed, an object moved. In physics, the here and the there have a name, and we call them, positions. To be more precise, we call them position 1 and position 2, or initial and final positions.
While it may seem simple, the concept of position is fundamental to understand the motion of an object or a particle along a straight line. It is through a series of positions, that we can study the motion of an object.
In short, the position of a moving object, is like a frame of an animation, because both help build the concept of motion.
Since this motion is along a straight line, we give that straight line a name: axis. An axis is just that, a straight line we arbitrarily picked to helps us describe the motion of the object we want to study. The axis is sliced in segments of equal distance one of each other, that will serve as reference to tell the position along the axis of the moving object.
To better understand the concept of an axis, think of a football field with the marks of yards. Now, take a very thin section of it, still with the yard marks. This would be a perfect example of what an axis is.
For simplicity, we often use the horizontal axis and describe the motion on its positive direction, that is, from left to right. We’ll talk more about positive and negative directions later on the post.
Also, the horizontal axis is called, “x” axis, and from now on we’ll be using it a lot!
Displacement is the concept that describes motion itself. Going back to our frames example, a series of frames would describe displacement or motion.
We talk about the displacement of an object when we observe a difference in position. That is, when you say a thing was here and then was there, that thing has displaced. And going back to the x axis, when an object has initial and final positions, we can determine the displacement of this object by comparing both positions along the x axis and telling the difference between both positions.
The net change in position is called displacement, and we often use the Greek and uppercase symbol ∆ (Delta) to represent a change in a quantity, thus, you’ll see ∆x used to represent displacement, as ∆x means: change in the x axis position, or displacement along the x axis.
It is important to say that the axis letter could be also “y” or “z”. In most cases, “x” is used for simplicity, but ∆x, ∆y, or ∆z are commonly seen in when studying physics.
Just like everything man uses to describe the world around him, the concept of average velocity is also arbitrary. Yet this does not mean is not useful or incorrect. In fact, it is very useful and helps to describe, analyze, predict, and plan the motion of an object.
Average velocity is just that, the average velocity of a moving object between two different positions.
Velocity is a concept we have not explored, but velocity is simply how fast something moves, that is, displacement timed. It is by the introduction of the variable time, that we can tell how fast something moves. The less time it takes for something to move from its initial to final position, the faster it will move.
If you take the velocities along the different positions of an object, you’ll know its average velocity.
If you divide the displacement of an object by the time it took the object to make that displacement, you’ll find average velocity as well.
Lastly, speed and velocity are not the same. Speed is the magnitude of the quantity velocity, while velocity is the magnitude and the direction of the quantity velocity.
Positives and Negatives
Eventually you’ll see that positions, displacements, and velocities come with a positive or negative sign. This positive or negative refers the direction in which the motion is happening along the axis.
- A negative displacement means the object moved from right to left, and vice versa for a positive displacement.
- A negative velocity means the object moved from right to left, and vice versa for a positive velocity.
Later when we see the concept of acceleration, we’ll have to talk about the signs again.
Also, you must know that both displacement and velocity, are vector quantities, which means they have direction and magnitude. We’ll talk about vectors on a different post, but for now, that is enough.
Average velocity and RC Truck Sample Problem
Imagine you drive an RC truck along a straight road for 0.7 km at 20 km/h. Here, your truck runs out of battery and stops. Over the next 30 min, you walk 2.0 km farther along the road to a hobby store.
- What is your total displacement from the beginning of your drive to your arrival at the hobby store?
- What is the time interval ∆t from the beginning of your drive to your arrival at the hobby store?
- What is your average velocity vavg from the beginning of your drive to your arrival at the hobby store? Express your answer both with numbers and a graphic.
- Suppose you get a new (charged) battery, and walk back to the RC truck for another 30 min. What is your average speed from the beginning of your drive to your return to the RC truck with the battery?
Instantaneous Velocity and Speed
More than often, when we talk about how fast something moves, we refer to its velocity at a specific point in time. That is, its instantaneous velocity or speed. So far we’ve talked about average velocity and we know how to calculate that for a moving object in a rectilinear motion, but the instantaneous velocity or speed is something different.
The instantaneous velocity is nothing more than the velocity at given instant. A given instant is a fraction of time close to (but not) 0 (zero).
In fact, the way to calculate instant velocity and speed is to approach ∆t to 0. What we do is to find the limit of v(t) when ∆t tends gets close to zero. We say v(t) tends to a given value as ∆t gets close to zero.
When a particle’s velocity changes, we say the particle accelerates (or experiments acceleration). For motion along an axis, the average acceleration aavg over a time interval ∆t is
where the particle has velocity v1 at time t1 and velocity v2 at time t2. The instantaneous acceleration, or simply acceleration is
To simplify, the acceleration of a particle at any instant is the rate at which its velocity is changing at that instant. Graphically, the acceleration at any point is the slope of the curve of v(t) at that point.
Also, the acceleration of a particle at any instant is the second derivative of its position x(t) with respect to time. Acceleration has both magnitude and direction (it is yet another vector quantity). Its algebraic sign represents its direction on an axis just as for displacement and velocity; that is, acceleration with a positive value is in the positive direction of an axis, and acceleration with a negative value is in the negative direction.
The sign of an acceleration has a nonscientific meaning: positive acceleration means that the speed of an object is increasing, and negative acceleration means that the speed is decreasing (the object is decelerating).
For example, if a car with an initial velocity v = -25 m/s brakes to a stop in 5.0 s, then aavg = +5.0 m/s . The acceleration is positive, but the car’s speed has decreased. Since the direction of the acceleration is opposite to the direction of the velocity, the signs are different.
If the signs of velocity and acceleration are the same, the speed increases. If the signs are opposite, the speed decreases.
Acceleration and dv/dt Sample Problem
- Because position x depends on time t, the marble must be moving. Find the marble’s velocity function v(t) and acceleration function a(t).
- Is there ever a time when v = 0?
- Describe the marble’s motion for t ≥ 0.
- To get the function of velocity v(t), we get the first derivative of the position function x(t) with respect to time. Then we get the acceleration function a(t) with the first derivative of the v(t) function or the second derivative of the x(t) function.
- v = -27 + 3t2
- a= + 6t
- To know when v = 0, all we do is use the v(t) equation and make it equal to 0. Then we solve for t and we’ll know when v = 0
- 0 = -27 + 3t2
- 27 = 3t2
- t =√9
- t = ±3 s
To describe the motion of the marble, we need to understand what happens with its position, velocity and acceleration.
If we evaluate the position, velocity and acceleration expressions at t=0, the marble’s x(0) = +4 m. Its v(0) = -27 m/s, which means it moves in the negative direction of the x axis, and its a(0) = 0 because just at that moment, its velocity is not changing.
From 0 to 3 seconds, its position is now on the negative side of x axis. Its velocity is still on the negative direction of the x axis, and as t approaches to 3s, it becomes zero, as per our calculations before.
Its acceleration is zero when t = 0 s, but after that it grows and it is always positive. Since the velocity and acceleration have different signs, between 0 s an 3 s, the marble must be slowing down.
After 3 s, position is still negative, until a bit more than 5 s, its position changes from negative to positive.
We already know that at 3 s, velocity is zero, but after that, velocity becomes positive, and starts to increase. This means the marble moves in the positive direction of the x axis.
As for the acceleration, this was always positive, and since velocity and acceleration after 3 s now have positive signs, acceleration acts in the same direction than velocity, which means the marble accelerates.
Remember that if velocity and acceleration have equal signs, that means that velocity increases, it can be either in the positive or negative direction of the x axis, but the object speeds up.
On the other hand, if velocity and acceleration have different signs, this means the object slows down, and the motion is in the direction of the velocity.
It is fairly common that in many types of motion, that acceleration is either constant or approximately so. For instance, you might accelerate a car at an approximately constant rate when a traffic light turns from red to green.
Then, the graphs of your position, velocity, and acceleration would resemble those below.
Note that a(t) is constant, which requires v(t) to have a constant slope. Later when you brake the car to a stop, the acceleration (or deceleration in common language) might also be approximately constant.
Such cases are so common that a special set of equations has been derived for dealing with them. One approach to the derivation of these equations is given in this section. A second approach is given in the next section.
Throughout both sections and later when you work on the homework problems, keep in mind that these equations are valid only for constant acceleration (or situations in which you can approximate the acceleration as being constant).
First Basic Equation. When the acceleration is constant, the average acceleration and instantaneous acceleration are equal and thus we can derive a new equation like this:
Second Basic Equation. Just like we did before, we can get the average velocity equation and rewrite it like this:
Third Basic Equation. If we solve for tf in our 1st basic equation, and then substitute that in our 2nd basic equation we get:
Fourth Basic Equation. If we solve for a in our 1st basic equation, and then substitute that in our 2nd basic equation we get:
Constant Acceleration Problem Sample 1
Constant Acceleration Problem Sample 2
Free Fall Acceleration
It was Galileo Galilei who questioned the widely accepted idea that if two objects fell from the same height and at the same time, the heavier one would reach the floor sooner. This happened in the 16th century, and he even conducted several experiments to show the first idea was wrong.
In the 4th century B.C., Aristotle theorized the first idea, and believe it or not, it was the standard for almost 2000 years. If you ask me, this proves that we could be wrong in many things we have believed to be true in the last 2000 years as well.
This portion of the post deals with free-fall acceleration. And one of the central ideas to keep in mind is what Galileo discovered while conducting his experiments:
Regardless of their mass, shape, or density, two objects will reach the floor at the same time, when free-falling from the same height at the same time.
Here is a short animation with the basic idea.
In the everyday world, you can’t replicate this experiment with a feather. And this is because we live in a world filled with air. There is no vacuum like in the previous example. But, if you make the experiment with an object that is heavier than a feather, say a bowling ball and a football, chances are you’ll be able to see Galileo’s prediction.
In this case, the falling objects accelerate at a rate of 9.8 m/s2. Since both objects accelerate at the same rate, with the same initial velocity and from the same initial position, the equations of constant acceleration developed before predict and prove that both objects will reach its final position (the floor) at the same time.
In fact, the equations of rectilinear motion apply the same way for vertically or horizontally moving objects, and the only thing that changes is the value of the acceleration.
In this case, a = g, which is the value of gravitational acceleration for planet Earth close to its surface.
Again, the equations of motion along a straight line continue to work if the object is falling down or falling up, that is, if the vertical motion is down or up.