Hello there! Welcome to lecture 3: linear motion!
Linear motion allows us to start exploring kinematics, which is the study of the motion of objects. The topics in this chapter will allow us to answer the questions of how far something goes, how fast it goes, and how quickly it speeds up or slows down in the process.
These topics will enable us to study the motion of objects traveling at a constant velocity and also objects that are moving under the influence of gravity or some other force. It will set the stage for a more expanded look at forces, momentum, and energy in subsequent lectures.
Each of the following concepts will be discussed in this video: path and displacement, speed and velocity, acceleration, and free-fall.
Path and displacement
The concepts of path and displacement enable us to analyze how far an object moves as it travels. Path describes the total motion of an object as it moves from one point to another. It is a scalar quantity. The symbol that we use could be the lowercase letter d, the lowercase letter x, or the lowercase letter y. We typically use the letter d when discussing the concept of distance. We use the letter x when discussing a horizontal motion. We use the letter y when discussing a vertical motion. Either way, the units we use for path are meters.
Displacement describes the straight-line distance from one point to another. It is a vector quantity, so it must include direction. The symbol that we use could be Delta d, Delta x, or Delta y. The delta symbol makes it clear that we are interested in the change of position. The units used are meters.
First let’s consider a rather straightforward example. In this video, I recorded myself walking at a mostly constant pace from one end of the classroom to the other. I marked on the whiteboard intervals of one meter, to get an idea of the distance. I started at the zero meter mark, and walked to the four meter mark. Therefore, the distance that I walked was four meters. My path describes the distance that I walked, which is equal to four meters.
Let’s look at a more complicated example. Consider driving in a car from Naperville to the College of DuPage. When driving a car, it isn’t always possible to go in a direct straight-line. Instead, somebody driving a car has to follow the roads. This route describes the path of the car. The path from Naperville to COD would therefore include the entire route that is driven. The path, in this specific example, is 15,600 meters, or 15.6 kilometers.
The displacement between Naperville and the college is 11,100 meters, or 11.1 kilometers Northeast. In other words, the displacement is the change in position from start to finish. Regardless of the path that’s taken to move between the two points, the displacement remains the same. It is always equal to the vector pointing from the starting point to the finishing point.
In general, as you drive your car around, you can determine your path by taking a look at your odometer. What distance does it take to drive between two different points?
It’s slightly more difficult to calculate the displacement as you drive from one place to another, unless you are able to drive in a completely straight-line. However, if you know coordinates of the starting and ending location, you can draw a straight line distance between two points. By splitting the straight line into horizontal and vertical components, you can use the Pythagorean theorem to calculate the displacement.
As an example, and a good review for calculating the magnitude of a vector, let’s say that we’re given coordinates of the starting and ending location of an object’s motion. The starting location is (300,400) meters. The ending location is (1200,900) meters. Regardless of the path taken, we can draw a vector from start to finish. That vector denotes the displacement of the object.
To calculate the magnitude of the vector, we can calculate the two components. The horizontal component of the vector tells us the change in position in the horizontal direction, Delta x. Delta x is equal to the final x position minus the initial x position. In this example, Delta x is 1200 meters minus 300 meters, which is 900 meters. We repeat this for the vertical component. The vertical component of the vector tells us the change in position in the vertical direction, Delta y. Delta y is equal to the final y position minus the initial y position. In this example, Delta y is 900 meters minus 400 meters, which is 500 meters.
We can now use the Pythagorean theorem. The magnitude of the vector is equal to the square root of 900 meters squared plus 500 meters squared, which is equal to 1030 meters. Therefore, the magnitude of the displacement is 1030 meters.
Let’s take a look at another example. In this case, I have a piece of plastic tubing that’s coiled up. When I place a BB in the tube, it travels along the length of the tubing until it comes out the other end. The path of the BB considers the length that the BB moves through all of the tubing, which is 1.5 meters. The displacement of the BB only considers the straight-line distance between start and end. The displacement of the BB is -0.73 meters. We use a negative sign in this case because it points down, and displacement is a vector, so we need to include the direction somehow.
Let’s go back to the example of me walking around the physics classroom. This time, I start at the zero meter mark, walk to the four meter mark, and then go back to the two meter mark. My path describes the total distance that I walk, which is six meters. However, my displacement only considers how far I moved relative to my initial position. Displacement is delta x, final position minus initial position. Two meters minus zero is equal to two meters.
Now, if I start at the zero meter mark, walk to the four meter mark, and then go back to the zero meter mark, let’s calculate path and displacement again. My path is the total distance I walk, eight meters. My displacement is final position minus initial position. Because my final and initial positions are at the same location, my displacement is zero. Whenever the starting and ending location are in the same spot, the displacement will be zero, regardless of how large a number the path is.
Speed and velocity
The concepts of speed and velocity enable us to analyze how fast (or slow) an object moves as it travels. Speed is the scalar quantity that describes the rate at which an object changes its position. The symbol used can be the lowercase letter s, or the lowercase letter v in absolute value sign, indicating that we are looking at the magnitude of the velocity. The units of speed are meters per second.
Velocity is the vector quantity that describes the rate at which an object changes its position. Because velocity is a vector, it’s either denoted in bold font or with an arrow over the top. The symbol is the lowercase letter v. The units of velocity are meters per second.
Both speed and velocity can be expressed as averages or as instantaneous values. That is, I can express how fast an object moves on average from the beginning to end of its motion. That would be the average speed or velocity. Average speed has an equation of path divided by time. Average velocity has an equation of displacement divided by time.
The instantaneous value tells us, how fast is an object going at any given snapshot in time? This might be the same as the average value, or it might be different. Instantaneous speed and velocity can either be measured, or we can use an equation. In this section of the lecture, we’ll talk about how to measure the instantaneous speed. We’ll talk about the equation to use to calculate instantaneous speed and velocity when we discuss acceleration in several minutes.
Let’s use the examples from the previous section of this lecture to gain a better understanding of speed and velocity.
Let’s reconsider the example where I walked at a mostly constant pace from one end of the classroom to the other. Because speed is scalar, it relates to path, which is also scalar. My path was four meters. It took me 6.2 seconds to walk that distance. Therefore, my average speed is equal to path divided by time. This is 0.65 meters per second. Because I walked at a constant speed, my instantaneous speed at any moment during my walk would also be equal to 0.65 meters per second.
Now, let’s re-consider the drive from Naperville to COD. The average speed is going to be equal to the path divided by the amount of time it took to drive that distance. I drove this distance in my car and recorded the time at 20 minutes, or one-third of an hour. The average speed is therefore 15.6 kilometers divided by one third of an hour, or 46.8 kilometers per hour. If you’re not comfortable with units of KPH, to give you some context, this is equal to an average speed of 29 miles per hour.
The average speed seems a little slow if we’re familiar with some of these roads, where the speed limit is much higher than 30 miles per hour! What’s the deal? Well, average speed accounts for all of the times I’m stopped at red lights or stop signs. It also accounts for times I’m speeding up from a stop, or slowing down to stop or to turn. It accounts for all of the variations in my motion as I drive along the path.
This is where we see a big difference between average speed and instantaneous speed. As I drive along this path, my instantaneous speed is given by the readout on my car’s speedometer at any given moment. At times it could be as high as 72 kilometers per hour, and at times it could be as low as 0 kilometers per hour. The average and instantaneous speeds are not necessarily going to be equal!
Now that we know speed, let’s calculate velocity. The displacement of the drive from Naperville to COD was 11.1 kilometers northeast. The amount of time it took me to drive between the two points was 20 minutes. Therefore, the average velocity of this trip was 11.1 kilometers northeast divided by one third of an hour, which is 33.3 kilometers per hour northeast.
Average velocity tells us, on average, what the velocity was throughout the motion of an object. The displacement is always going to be less than or equal to the path of an object, as the shortest distance between two points is a straight line. Therefore, the average velocity will always be less than or equal to the average speed of an object. That hopefully makes sense intuitively, as the fastest way I can get from one point to another would be to travel in a straight line. Any other way I travel, that’s not a direct path, is going to add extra time to my motion and slow me down.
The average velocity for the trip from Naperville to COD is the average over all of the motion. At any given snapshot in time, the instantaneous velocity as I drive to the college would be indicated by my current speed and direction. I don’t spend the entire time traveling at 33.3 kilometers per hour northeast! Sometimes I’m stopped and have an instantaneous velocity of zero. Sometimes I might be going faster or pointed in a different direction.
In general, as you drive or ride around in a car, you can determine your instantaneous speed by taking a look at your speedometer. If you want to know what your instantaneous velocity is, look at your speedometer and a compass to determine what direction you’re moving in.
Let’s calculate the average speed and average velocity of the BB in the tube Recall that the path was 1.5 meters. I used a stopwatch to measure the time it took for the BB to travel the length of the tube, which was 1.7 seconds. Path divided by time is 1.5 meters divided by 1.7 seconds, giving us an average speed of 0.88 meters per second.
The displacement, the straight-line distance from start to finish was -0.73 meters. Because it took 1.7 seconds to go from start to finish, the average velocity is -0.43 meters per second. Recall that the negative sign means that the vector points down.
To determine the instantaneous value of speed or velocity at any given moment in the BB’s motion, we would need to use some kind of motion detector that is capable of measuring that value.
Let’s reconsider the example where I walk around the physics classroom, starting at the zero meter mark, walking to the four meter mark, and then back to the zero meter mark, which was my starting location. We had calculated the path to be eight meters, and the displacement to be zero meters.
The average speed is eight meters divided by 14.2 seconds, which is 0.56 meters per second. Because I walked at a constant pace the entire time, my instantaneous speed is equal to my average speed.
The average velocity is displacement divided by time, which gives an average velocity of zero meters per second. Regardless of how far I walk, if I end in the same position as I started, then my displacement and average velocity will be zero.
I can calculate instantaneous velocity on my walk out, and my instantaneous velocity for my walk back. I walked at a constant pace to the right at four meters divided by 7.1 seconds, giving an instantaneous velocity throughout the first half of my motion at 0.56 meters per second. On the way back, I walked a constant pace to the left at four meters divided by 7.1 seconds, giving an instantaneous velocity throughout the second half of my motion at negative 0.56 meters per second. Note that the negative sign reflects the fact that I moved to the left.
Acceleration quantifies the rate at which an object changes its velocity. The symbol for acceleration is the lowercase letter a. The units for acceleration are meters per second-squared. The acceleration of an object is equal to the change in velocity divided by time. Acceleration equals delta v divided by time. In other words, acceleration equals final velocity minus initial velocity divided by time.
We have to be very careful when using the word acceleration in a physics class, as it differs from the definition of “acceleration” that you may be used to using in the English language. The physics definition of acceleration can mean three things: speeding up, slowing down, and changing direction.
I could therefore look at something that is slowing down, and say “that object is accelerating,” and I would be correct. Whereas in the English language, people tend to use the word acceleration only to refer to objects that are speeding up. We have to be careful and precise with our words in physics. Try to get out of the habit of saying “accelerate” when you really mean to say “speed up.”
Because it’s so important, I’m going to repeat what is meant by acceleration in physics: that an object is either speeding up, slowing down, or changing direction.
For the most part, this class will only deal with situations where the acceleration is constant throughout an object’s motion. That means that instantaneous and average acceleration will be equal to each other. I’d like to point out that instantaneous and average acceleration are not equal in some of the scenarios we looked at previously in this lecture: the BB traveling through the tube, and the car driving from Naperville to COD. Those two examples are great for discussing concepts of “how far” and “how fast”, but would make for complicated analysis of acceleration. So we’ll consider simpler examples when we talk about acceleration.
Each of the demos in this section use a motion detector to record the position, velocity, and acceleration of a cart as it moves along a low friction track. In this way, we can look at how position, velocity, and acceleration all relate to each other.
In this example, a cart travels along a flat low friction track. Because there are no forces acting on the cart to change its speed, it has a constant speed throughout its motion. The position vs time graph shows a straight line pointing upward, indicating that the cart increases its position at a steady rate. The velocity vs time graph is a straight horizontal line, indicating that the velocity stays the same throughout the cart’s motion. The speed at the start is 0.5 meters per second, and the speed at the end is 0.5 meters per second. The acceleration is therefore zero meters per second-squared. This is confirmed by the acceleration vs time graph, which shows a flat line at zero, showing an acceleration of zero meters per second-squared.
Now, the track has been tilted so that the cart moves downward. The cart is initially at rest, at which point I let go of the cart and it travels down the track. In this case, the cart does not travel at a constant velocity any more. The position vs time graph is not a straight line. Instead, it slopes upward in a curved manner, indicating that the cart is changing its position at an increasing rate as the cart moves down the ramp.
The velocity vs time graph is a straight line pointing upward, showing that the speed increases at a constant rate. The speed when the cart starts moving without my hand influencing it is 0.11 meters per second, and the speed 1.9 seconds later is 1.09 meters per second. The acceleration can be calculated as 1.09 meters per second minus 0.11 meters per second divided by 1.9 seconds, which is 0.52 meters per second-squared. In fact, the acceleration vs. time graph shows a straight horizontal line at 0.52 meters per second-squared, confirming that the acceleration is a constant value throughout the motion of the cart.
It’s important to point out here that, in situations like this where there is an acceleration, it is not possible to use velocity equals distance divided by time to calculate instantaneous velocity. This is because the distance does not change in a linear manner. Therefore this equation does not work in situations where there is acceleration present.
The equations we can use for distance and velocity are different in cases where there’s acceleration. In fact, these equations can be used all the time, even when acceleration is zero. But these equations must be used when the acceleration is nonzero. Distance is equal to one-half acceleration times time squared, plus initial velocity times time.
Let’s use that equation to calculate the distance moved by the cart, and compare with the results from the graph. We calculated the acceleration to be 0.52 meters per second-squared, and the amount of time was 1.9 seconds. The initial velocity was 0.11 meters per second. One half times 0.52 meters per second-squared times 1.9 seconds, squared, plus 0.11 meters per second times 1.9 seconds equals 1.15 meters. This agrees exactly with the readout from the motion detector, which recorded an initial position of 0.2 meters and a final position of 1.35 meters.
Let’s consider one more example that shows all three aspects of acceleration: speeding up, slowing down, and changing direction. The cart is going to be traveling on the same ramp, but now it starts from the bottom of the ramp. I push it up the ramp, giving it an initial velocity of negative 0.96 meters per second. The cart slows down as it moves up the ramp, briefly stops at the top, then changes direction and rolls down the ramp and speeds up as it moves downward again.
The position vs time graph is a curved trend, resembling a kind of bowl shape, called a parabola. This indicates that the cart changes its position at a decreasing rate as it moves up the ramp, then changes its position at an increasing rate as it moves down the ramp.
The velocity vs time graph is a straight line pointing upward, showing that the velocity increases at a constant rate. This time, however, the velocity starts out as negative numbers, increases in value, becomes zero briefly when the cart is at the top of the ramp, and then increases as positive values. The speed at the start, after being pushed, is negative 0.96 meters per second, and the speed after 3.3 seconds is 0.93 meters per second. The acceleration can be calculated as 0.93 minus negative 0.96 divided by 3.3, which is 0.57 meters per second-squared. The acceleration vs. time graph shows a straight horizontal line at 0.57 meters per second-squared, confirming that the acceleration is a constant value throughout the motion of the cart.
The position, velocity, and acceleration graphs tell us a lot about the motion of the cart. The first part, where velocity is negative and acceleration is positive, shows us the period of time when the cart is moving up the ramp and slowing down. An object will slow down when the velocity and acceleration vectors point in opposing directions.
For a brief instant of time, the cart stops. This is because the cart must change direction before it rolls back down the hill. The velocity is zero, but the acceleration is still a constant value of 0.57 meters per second.
For the remainder of the graph, the velocity is positive and the acceleration is positive. This tracks the time when the cart is rolling down the ramp and speeding up. An object will speed up when the velocity and acceleration vectors point in the same direction.
The most prevalent source of acceleration in our lives is gravity. Gravity is a force that pulls us down toward the surface of the Earth. That pull represents an acceleration due to gravity, which is equal to 9.8 meters per second-squared pointing downward. This acceleration is a constant value along the surface of the Earth. It never changes.
Free-fall is defined as a situation where gravity is the only force acting on an object. For example, if an object were dropped off the top of a very tall building, free-fall would define its motion assuming that there are no other forces such as friction and air drag to slow it down. Free-fall is not always a completely accurate way to discuss the motion of an object, but it does a good job in some scenarios, giving us a straightforward way to analyze position, velocity, and acceleration.
When analyzing a free-fall situation, we use the equations for distance and velocity explained previously in this lecture. The value we use for acceleration is negative 9.8 meters per second-squared, as that is the value that comes from gravity.
In this demo, a ballistic cart launches a ball directly upward at an initial speed of 3.6 meters per second. Gravity is the only force acting on the ball. Gravity points downward, and the acceleration due to gravity has a value of negative 9.8 meters per second-squared throughout the motion of the ball.
At the start, the acceleration points downward and the velocity points upward. Because the two vectors are pointing in opposite directions, the ball will slow down. The ball slows down until it reaches the top of its motion, where it stops for a very brief instant in time, before moving downward again. Now the acceleration and velocity vectors both point in the same direction, downward. Because the two vectors point in the same direction, the ball will speed up. The ball speeds up until it is captured by the cart again and ends its motion.
We can use the equations of accelerated motion to calculate properties of the ball’s trajectory. There are three positions in the ball’s motion at which we know the velocity, making those ideal times to analyze the motion of the ball. We know that the ball starts with a speed of 3.6 meters per second as it’s launched from the cart. We know that the ball stops very briefly at the top of its motion, meaning that the velocity is zero at that point. Because the ball is in free-fall, it will have a velocity of negative 3.6 meters per second at the instant right before it falls back into the cart again.
Let’s consider the ball at the very beginning of its motion, and again at the top of the trajectory. By analyzing those two points, we can learn about how long it takes the ball to reach the top, and how high the ball goes. The two equations available to use are the distance and velocity equations. Both of them will work. But when I look at the distance equation, I see that there are two unknowns: time and distance. So I cannot use that equation yet! Let’s use the velocity equation to calculate how long it takes the ball to reach the top of its motion.
The equation is final velocity equals initial velocity plus acceleration times time. We can do some algebra to solve the equation for time. We get that time equals final velocity minus initial velocity divided by acceleration. The final velocity is the velocity at the top of the motion, zero meters per second. The initial velocity is the velocity when the ball is shot out of the cart, 3.6 meters per second. The acceleration is from gravity, negative 9.8 meters per second-squared. Zero minus 3.6 divided by negative 9.8 equals 0.37 seconds. Note that the negative sign on the acceleration can’t be ignored, otherwise we would get a result of negative 0.37 seconds, which doesn’t make any sense.
Now that we know how long the ball takes to travel to the top of its peak, we can use the distance equation to solve for the height. Distance equals one half times negative 9.8 meters per second-squared times 0.37 seconds squared plus 3.6 meters per second times 0.37 seconds. The ballistic cart shoots the ball to a height of 0.66 meters.
If we want, we can now choose two different points to analyze the motion of the ball at a different time. Let’s choose the starting and ending positions to analyze. We can use the velocity equation to calculate how long it takes the ball to make a complete up and down motion. As we saw, time equals final velocity minus initial velocity divided by acceleration. Negative 3.6 meters per second minus 3.6 meters per second divided by negative 9.8 meters per second squared equals 0.73 seconds. It takes equal amounts of time for the ball to go up as it does for the ball to return again.
Now that we know how long it takes for the ball to complete its path, we can use the distance and velocity equations to determine height or velocity at any time throughout the ball’s motion. For example, we can now ask how high the ball is above the base after 0.5 seconds, and what the ball’s velocity is at that time.
Distance equals one half times negative 9.8 meters per second-squared times 0.5 seconds squared plus 3.6 meters per second times 0.5 seconds. This is a height of 0.575 meters above the base of the cart.
Velocity equals 3.6 meters per second plus negative 9.8 meters per second-squared times 0.5 seconds, which is -1.3 meters per second. The minus sign indicates that the ball is moving downward at that time, which makes sense considering we know that after 0.37 seconds, the ball was on its downward motion back to the ballistic cart.
There’s one last thing I want to point out in free-fall situations. I mentioned it before, but it bears repeating. Gravity is constant, and it never turns off. The ball may instantaneously stop at the top of its motion, but that’s because gravity is pushing down on the ball, and as it’s moving upward it slows down and must change its direction before coming back down to the Earth. The acceleration due to gravity is negative 9.8 meters per second-squared the entire time the ball is moving. It is a constant value that never changes.
Thanks for taking the time to learn about linear motion! Until next time, stay well.