www.lsmsa.edu teacher Robert Dalling's Physics Lectures (see also www.ushumans.net)
One-dimensional motion
We want to understand the meanings of time, distance, speed, velocity, and acceleration. We all have some experience with motion but, as for many aspects of daily life, we have a somewhat vague notion of this common event. We have all driven from our home to a friend’s house but we never had any reason to get our rulers and stopwatches to carefully quantify motion beyond our relative words of fast and slow, moving or stopped. As we measure specific aspects of nature, we are always surprised that it is not exactly what we had assumed. We are surprised by the results of measurements. Science means "facts and understandings obtained from repeatable measurements–billions of measurements have been made." These are the most important points of the course.
CQ: Write on a sheet of paper to turn in at the end of the class time. Take 15 seconds to answer each of the following questions. What sort of things have you seen in motion? Which things can you see moving right now? Describe “distance.” What amount of time elapses between heartbeats? Describe something you can do in 1 second. In 10. In 100. For how long does a thrown ball stay in the air? How long does it take you to walk up a flight of stairs? How long does it take a car to traverse a highway on-ramp? How long does it take a rock dropped from shoulder-height to hit the ground? How long will a horizontally shot bullet stay in the air? Do you believe that the dropped rock and fired bullet both take the same amount of time to fall to the ground? Nature surprises us. We will be surprised in every chapter. In this chapter, we will refine our notion of motion.
Imagine for a moment a world in which nothing is moving–no cars, trains, or planes in motion, not even a strolling animal or a bush blowing in the wind. Just 150 years ago–before we developed engines for trains and such–very few objects were in motion besides people and animals. Since it is only living things that can move themselves of their own accord, about the only other objects in motion were the wind, rising smoke, rivers, clouds, and falling rain. Ask tribespeople why clouds move and they might answer “Because that’s what they like to do.” This makes sense to me. When our earliest natural philosophers pondered the nature of motion, they had only the motions of these few objects to explain. Empedocles (490 - 430 BCE) and Aristotle (384 - 322 BCE) explained these motions in terms of things seeking their own level in matter’s hierarchy of water, earth, air, fire, and heavens.
CQ: Do stars, comets, and planets move? Can you think of other things that move? Do they move of their own accord?
They also asked what is it that temporarily keeps a rock moving after you have thrown it. How would you explain it? Our first moving machinery–sailboats and grinding mills, for example–were pushed or pulled by people, animals, wind, or water. They moved only as something else pushed them. Back around the year 1850, the first time people saw a train moving under its own power they typically said that they were astonished because “It moved just like a living thing." Before the train, only living things could move under their own power. We are constantly in motion today, driving around town and flying around the continents, but just a couple centuries ago, a busy city was filled only with foot traffic and some horse- or people-drawn carts. The city contained only motionless buildings. In this Wisconsin Historical Society photo, foot traffic in an otherwise motionless city is seen as people walk to work at a Chicago factory in 1886. Copy the following URL into your browser to see the caption for this photo.
By the way, animals are naturally equipped to detect the motion of other objects because such movement often is of biological consequence. We also quickly learn to ignore those motions which are of no biological consequence, such as the swaying of grass in the wind, though we still might still enjoy such a sight in a mountain meadow. But our animal senses are easily fooled. Akiyoshi Kitaoka from the Ritsumeikan University in Kyoto, Japan demonstrates that sometimes motion is just an illusion. Motion is also art in interactive exhibits, movement, dance, and in kinetic sculptures. Motion can be frozen in time.
Motion is a fundamental aspect of nature, but what exactly is the nature of motion? While driving your car past trees and such, has it ever seemed to you that you are motionless and that instead it is the objects outside the car that are moving past you? From your point of view, are you in motion or are the passing tress and such in motion?
CQ: Are or are not those trees in motion?
Motion involves relative movement. While driving your car for example, you and your passengers are not in motion relative to each other but you are in motion relative to the ground, less so relative to other cars you pass, and about twice so relative to oncoming cars.
Can you be in motion without knowing it? You may have noticed that if you close your eyes while riding in a car, you are unaware of the car’s constant speed but do sense its changes in motion (termed “acceleration”). We feel the acceleration as the car seat pushes us forward. We also feel acceleration as an elevator begins moving and later slows to a stop. But without sensing it at all, we are in motion as the Earth spins in its daily cycle and travels in its annual path around the Sun; at the same time, the Solar System orbits the center of the galaxy, and the galaxies are moving away from each other within the expanding universe. How would you quantify motion?
The motion experiments of Galileo
Galileo was among the first persons to experimentally study motion. His work began to reveal to us the true nature of motion. He is the hero of this chapter. Galileo (1564-1642) and Shakespeare were both born in the same year. (This is also the year that Michelangelo died.) Some 78 years later, Galileo died during the same year that Newton was born (not fig Newton but Isaac Newton, who is the hero of the next chapter).
Around the year 1590, Galileo 2 began to quantify motion by measuring the time needed for rolling balls to travel given distances along tilted pathways or inclined planes. The steeper the incline, the faster the balls will travel. At the time, there were variously named units of distance–cubits and leagues, for example–and the day had been divided into twenty-four hours but no clocks yet existed that counted the tiny intervals we call seconds. The day was divided only into morning, noon, afternoon, and evening and such. Galileo made his measurements of motion before second-hands existed on clocks. He measured time intervals by counting heartbeats, water drops, and pendulum swings. By the way, what causes the oscillatory motion of a pendulum?
Dropped objects take only a fraction of a second to fall to the ground; it is hard to time such a short interval by counting fractions of heartbeats. Galileo used inclined planes to stretch the time duration of the motion. The less steep the incline, the longer will be the motion’s duration.
Speed represents a distance traveled divided by the time taken to travel that distance. For a modern example, driving at 60 mph means that we traverse a distance of 60 miles in one hour. Galileo quantified motion in terms of distance traversed in each of a series of, for example, three-heartbeat time units. That is, what distance was covered in the first three-heartbeat time period and what distance was covered in the second three-heartbeat time period, and the third, and so on. Galileo defined acceleration to be the change in speed per unit time. (While driving our car, we say that we press the accelerator to speed up.)
CQ: Suppose you begin from a stopped position to drive down a highway on-ramp, you traverse a small distance during the first three-heartbeat time unit (write down an estimate of this distance), a larger distance through the next three-heartbeat time unit (write down an estimate of this distance), and a yet larger distance through the next (write down an estimate of this distance) as your speed increases from slow to fast and then faster.
Galileo found that as a ball rolls down an inclined plane, its speed increases linearly in time–it is accelerating–but its position increases as the square of time. Mathematically, these are written v ∝ t and x ∝ t2. Using the letter ‘a’ as the constant of proportionality, these two relations are also written v = at and x = ½ a t2.
CQ: Would you have guessed before making these measurements that nature behaves in this specific, mathematical way? Would you expect nature to behave in a mathematical way in any situation?
Luckily, these speed and acceleration equations hold no matter what object–from wooden spheres to metal cylinders–rolls down the inclined plane. They also hold whether the incline is set nearly horizontal or completely vertical, which then results in “free-fall” motion. In the year 1632, Galileo published his results in Dialogue Concerning Two New Sciences. Before Galileo, nobody had stumbled across the importance of the time rate of change of speed. Galileo quantified the change in speed, which is acceleration, a = ( vf - vi ) / ( tf - ti ) = Δv / Δt = dv / dt. We will use the Greek letter delta for the symbol to represent “change.” It is always the “final minus initial” values. Just a few decades later, Isaac Newton would build upon Galileo’s work to figure out that forces cause accelerations.
What other sorts of natural phenomena might be describable by an equation? The answer is: all of them. We find out more accurately how nature works when we make measurements and discover the equations that describe those measurements. This is the goal of the physicist. Every scientist is seeking the “mother load” as a particular aspect of nature whose understanding clears up previous confusion and opens new doors of inquiry. Remember also that nature nearly always has more imagination then us, we rarely predict ahead of time how nature behaves: we instead go measure nature and are surprised by the result. Science means facts and understandings obtained from repeatable measurements–billions of measurements have been made. These are the most important points of the course.
Lets look one-by-one at distance, time, speed, and acceleration. A goal of this chapter is for us to gain an understanding of the physical meaning of speed and acceleration.
Time is measured in seconds. We sometimes count seconds by counting out loud: “one thousand, two thousand.” The time that elapses between heartbeats is about three-fourths seconds.
Lifetimes typically last for about one billion heart beats, no matter the species or heart rate. In general, the larger the animal, the slower the heart rate. Small animals having heart rates ten times the human rate and have lifetimes that are one-tenth as long as a human’s.
Example:
How many seconds are in a year?
When estimating calculations, a year has about π x 107 seconds.
Example:
How many years are there in a billion seconds? 109 s / 3.15 x 107 sec / year = 31.7 years.
The age of the Earth is 4.5 billion years or 1.5 x 1017 seconds, and the age of the universe is 13 billion years or 4 x 1017 seconds.
CQ: What is the duration of the trajectory of this tossed item? How many seconds are there in 17 years?
Distance is a measure of how far you have traveled and is written with the symbol ‘x’. Distance is measured in meters and sometimes stated in terms of kilometers or miles and such. One mile is 1.6 kilometers. Let’s look at the size of millimeters and centimeters on this meter stick. Let’s look though Jeff Phillips’ list of order-of-magnitude estimates of times, lengths, masses, speeds, accelerations, forces, and energies for various objects. Let’s look at the National High Magnetic Field Laboratory’s visual tour through the powers of ten, from the Universe, to the Milky Way Galaxy, and down to the subatomic universe of electrons and protons. By the way, the NASA Worldwind website lets you zoom from satellite altitude down to any location on the Earth–down to street level for many U.S. cities.
CQ: What is the length of your desk top? What is the width of this room? How far do you live from the classroom?
Speed is a measure of how fast you are traveling past something else and is written with the symbol v. Speed is measured in m/s (pronounced meters per second, just as 6 slices of pie per 2 persons is written 6 pieces/2 persons = 3 pieces/person). The speed of a car is 60 mph = 100 km/h. When a car moves at a constant speed–say 20 m/s–its position is changing by a constant amount–20 meters, in this example–every second, even between the seventh and eighth seconds. By the way, fingernails and hair grow and continents drift at a speed of about 2 cm/year or 2 x 10-9 m/s. Speed is always measured relative to something else, which may or may not be in motion relative to something else. For example, a child might walk across your lap while you are sitting in a car that is passing another car along a row of houses. Suppose that you stand on a moving continent and point one finger in the direction of the continent’s motion and another opposite its motion. With what speed do the fingernails on these two hands then move?
The definition of speed is
v = Δx/Δt = ( Xfinal - Xinitial.) / ( tfinal - tinitial ) = ( xf - xi ) / ( tf - ti ).
Speed = change in position / change in time. It is a measure of how quickly the position of an object is changing.
The change Δ is always “final minus initial.” We often choose subscripts 0 and 1 to represent the two values or drop the “final” subscript on one variable. The numerical value of speed calculated from this equation is independent of the axis origin. One stationary observer might put x = 0 in Los Angeles (Pacific time) while another puts x = 0 in Natchitoches (Central time) but both calculate the same speed. For example, the Natchitoches observer sees a car at xi = 50 miles at time ti = 5 hour and then is at position xf = 150 miles at time tf = 7 hours. Its speed is then
v = Δx/Δt = ( xf - xi ) / ( tf - ti ) = ( 150 - 50 miles) / ( 7 - 5 h) = 50 mph.
The same numerical speed is calculated by the observer sitting in Los Angeles, 2000 miles away using clocks shifted by two hours.
v = Δx/Δt = ( xf - xi ) / ( tf - ti ) = ( 2150 - 2050 miles) / ( 5 - 3 h) = 50 mph.
Physicists require that models of nature be independent of the observers coordinate system and independent of the observer’s system of units (this is gauge theory).
CQ: Use the ruler and stop watch to measure your own walking speed or the speed of a bug or fly. (By the way, how does a fly go about landing on the ceiling? Does it first invert while flying and then land while inverted?)
Example:
To convert speeds from mph to m/s, multiply by 0.44, or to convert m/s to mph multiply by 2.25. For example, 10 m/s = 22.5 mph.
CQ:
Show that 60 mph = 88 ft/s. This means that 600 mph = 880 ft/s.
Example:
What is the speed of blood flowing in your arm? When I plug and release the vein in my arm here, I see blood move 0.05 m in 0.25 s. Its speed is then v = Δx/Δt = 0.05 m / 0.25 s = 0.2 m/s.
CQ: If a dog runs 100 meters in 20 seconds then its speed v = Δx/Δt = 100 m / 20 s = 5 m/s.
Example:
The speed of light is 3 x 108 m/s. At this speed, how many times n in one second would light travel a distance equal to the circumference of the Earth? The radius of the Earth is re = 6.37 x 106 m so its circumference is Ec = 2πr = ( 2 )( 3.14 )( 6.37 x 106 m ) = 4.00 x 107 m. The distance light moves is given by x = vt = ct = nEc. Solve for n = ct/Ec = ( 3 x 108 m/s )( 1 s ) / 4.00 x 107 m = 7.5.
Distinguish average from instantaneous speed: get into your car, turn it on, speed up, drive 10 miles at 60 mph, passing at 70 mph and slowing to 40 mph for animals, then stop and get out. Picture this motion by putting yourself into that car making that trip. You will have an "average speed = total distance traveled /total time" but at any instant your "instantaneous speed" might have been any number between 0 and 70 mph.
We have a speed if our location is changing, otherwise we are not moving. What is the difference between speed and acceleration? We have an acceleration if our speed is changing.
Acceleration is a measure of how quickly a speed is changing. It is written with the symbol ‘a’ and is measured in m/s2 (pronounced meters per second squared). Moving at constant acceleration means your speed increases by a constant amount per unit of time. The definition of acceleration is
a = Δv/Δt.
Acceleration = change in speed / change in time. It is a measure of how quickly the speed of an object is changing.
Here are some typical values of acceleration
Example:
A person jumps off a platform and experiences a vertical increase in speed of 4.9 m/s in 0.5s. The acceleration of the person is
a = Δv/Δt = ( vf - vi ) / ( tf - ti ) = ( 4.9 m/s - 0 m/s ) / 0.5s = 9.8 m/s 2. We use the symbol ’g’ to represent this acceleration due to gravity.
Example:
It takes the Shuttle 50 seconds to reach a speed of 750 mph, 1 minute 47 seconds to reach a speed of 2600 mph, and 8.5 minutes to reach 17,000 mph. What is its average acceleration at each of these three points? We multiply 750, 2600, and 17,000 mph by 0.44 to obtain 333.3, 1156, and 7560 m/s. The acceleration is then a = Δv/Δt = ( 333.3 m/s ) / 50 s = 6.7 m/s2, which is 6.7/9.8 = 0.7g, a = Δv/Δt = ( 1156 m/s ) / (60 + 47 s) = 10.8 m/s2, which is 10.8/9.8 = 1.1 g, and a = Δv/Δt = ( 7560 m/s ) / (8.5 x 60 s ) = 14.8 m/s2, which is 14.8/9.8 = 1.5g. For a period just before the engines are cutoff, its acceleration grows to 3g.
How does this acceleration feel? Listen.
CQ: During take off, the speed of a 727 jetliner goes from a speed of 0 to 88 m/s (200 mph) in 30 seconds. What is its acceleration? a = Δv/Δt = (88 m/s) / 30 s = 2.9 m/s2. Divide that acceleration by g (9.8 m/s2) to show that the acceleration of the airplane calculate is 2.9/9.8 = 0.3 g.
We have no sensation of speed but we feel acceleration in an elevator as it starts or stops. As you push the “accelerator pedal” to speed up to get onto the highway, you see the speedometer reading increase and you feel the acceleration as your back is pressed into the car seat, making you feel “heavier.” (On page 44 of The Sciences, An Integrated Approach (James Trefil and Robert M. Hazen, 2004, John Wiley & Sons, Inc., Trefil describes his experience of g-forces as being similar to having multiple persons sitting on you all at once. Trefil and jet pilots say that a 5g acceleration feels like five people sitting on top of you.) If this press stays constant then the acceleration is constant. If you press and release the peddle several times, you are thrown into and out of the seat and will feel the acceleration change in an irritating way. Can acceleration change? Yes, but not in this course. We want to understand the meaning of acceleration and not bury it for no reason with complicated mathematics. If acceleration changes then the motion equations given below do not apply; instead, calculus must be used to obtain other forms of these equations. The time rate of change of acceleration has been aptly named “jerk.” This is of interest in the design of public transportation vehicles but very few natural phenomenon involve the rate of change of acceleration. While driving our cars, we might accelerate gently and feel a gentle push, or press the pedal to the floor to produce a high 'a' and feel strongly pressed into the chair of the car. By the way, the time rate of change of a is called jerk--because we do feel a jerking motion. You'll feel da/dt if you quickly press and release the gas pedal several times.
Suppose that, as you walked along, water dripped out of a bottle tied to your waste. (Or, as you walked along the chalkboard and marked dots in equal-time intervals.) If the water dripped once every second, you could look back and see a sort of time-record of where you had been. If you walk at constant speed then you would traverse equal distances every second and the water drops would be equally spaced, as here •--•--•--•--•--•. If you then speed up to begin running, adjacent drops will begin to have greater separations, as here •---•----•-----•-------•--------•. You might then continue running for a while at that higher, constant speed. The entire record of water drops would show that you walked, sped up, and then steadily ran. The drops would produce the following series of dots.
•-•-•-•-•-•-•-•--•---•----•-----•------•------•------•------•------•------•------•------•
walking (equally spaced) speeding up (spreading out) running (equally spaced but wider)
The same dot-records might represent a car moving at constant speed, speeding up, and then continuing to move at that higher speed. See other ticker tapes.
CQ: Can you think of another object whose motion would be described by this sequence of water drops?
Look at the picture frames. The camera sits at one place while a person runs past. A strobe flash is used to show equal amounts of time elapsed between each successive picture.
Here is a person running at constant speed.
O---O---O---O---O (Person running)
/\----/\---/\----/\---/\
Here is a representation of this motion using ticker tape or water dripping at equally-spaced moments in time.
●----●----●----●----● (Ticker tape)
Here is a coordinate system with its origin at the left edge of the motion. The ticks in the coordinate system have an equal distance between adjacent numbers while the strobe flash has equal amounts of time between marks.
012345678901234567890 (coordinate system)
Here is a person who is speeding up or accelerating.
O-O---O---O-----O-----O
/\--/\---/\----/\-----/\------/\
●-●---●----●-----●------●
Here is a person who is slowing down or accelerating at a negative rate. We say that “the acceleration is negative” rather than saying “it is decelerating.”
O------O-----O----O---O-O
/\-------/\-----/\---- /\--- /\-/\
●-------●-----●----●---●-●
Let’s draw a ticker-tape representation of a jet that is initially sitting at the beginning of a runway and then accelerates at a constant rate to attain lift-off speed.
●---●-----●-------●---------●----------------●-------------------------●
Our representation of this motion shows the jet’s position at equal intervals of time (not distance). Place the origin at the left. The dots are located at 1, 4, 9, 16, and then 25 distance units. These numbers are increasing quadratically in time because greater distances are being covered in successive time intervals. Between successive time intervals, the jet has traveled 1, 3, 5, 7, and 9 distance units. These differences are 1 - 0 = 1, 4 - 1 = 3, 9 - 4 = 5, 16 - 9 = 7, and then 25 - 16 = 9 units. This means that successive speeds have values 1, 3, 5, 7, and 9. Notice that these speeds or differences in distances are increasing linearly in time. Speed is determined from the separation between successive dots and is increasing linearly here because the separation between successive dots is increasing linearly. The difference of successive velocities is always two here because 3 - 1 = 2, 5 - 3 = 2, 7 - 5 = 2, and 9 - 7 = 2. Acceleration is determined from the difference of successive speeds, which is constant here. In this example, acceleration is positive and constant. We see that acceleration is the difference in successive speeds which are the differences in successive positions. In this example, distances grew quadratically as 12, 22, 32, 42, and 52 while speed, which is the difference in successive positions, grew linearly as 1, 3, 5, 7, and 9. Acceleration, which is the difference between successive speeds, remained a constant.
When we write down the letter ‘a’ in our homework, we are not doing math, except in that it is a short hand notation for physics. While writing 'a' we are thinking acceleration and internally feeling what occurs when we speed up or slow down in our car or in an elevator or as being pressed back into the chair of our car.
Moving at constant acceleration means your speed is increasing by a constant amount (the same amount) per unit of time, during the 1st or 51st second, but you are covering greater and greater distances with each passing second.
Example:
A car goes from a speed of zero to 20 m/s (20x2.25=45mph) in 10 seconds. Its acceleration is change in speed / elapsed time = 20 m/s / 10 s = 2 m/s2. While accelerating for just 3 of those 10 seconds, its speed was v = at = 2 m/s2 x 3 s = 6 m/s and it had traveled a distance of
x = ½at2 = 0.5 x 2m/s2 x (3s)2 = 9 m.
Example:
A car speeds up from 30 to 60 mph in 30 seconds while getting onto the hwy by using the on-ramp, what is its acceleration? a= change in speed / elapsed time = (60 - 30 mph) / 30 s = 1 mph/s. Does this seem like a quick or slow increase in speed? Do you take 2, 5, 10, 20, or 30 seconds to speed up by that amount?
CQ: While driving along at a constant speed of 25 mph, what is your acceleration? No equation needed here.
Sandia National Labs has accelerated a small plate from a speed of zero to 76,000 mph in 3.4 x 10-8 s. What is the acceleration?
We have a = Δv/Δt = ( 76,000 mph )( 0.44 m/s /mph ) / 3.4 x 10-8 s = 9.9 x 1011 m/s2 = 1 x 1011 g. This is an acceleration of 10 billion g.
The motion equations for constant acceleration
For constant acceleration, the average speed occurs at the midpoint of a time interval
vave = ( v + vo )/2.
(This is seen in the graph of a straight line. We use this fact in labs involving spark tape).
Let’s set tf = 0 and drop the subscript on ti so that vave = Δx/Δt = ( xf - xi ) / ( tf - ti ) becomes
vave = ( xf - xi ) / t,
which can be written as xf = xi + vave t or, since one subscript is easier than two,
x = xo + vave t.
Let’s set tf = 0 and drop the subscript on ti so that a = Δv/Δt = ( vf - vi ) / ( tf - ti ) becomes a = ( vf - vi ) / t, which can be written as vf = vi + at or, since one subscript is easier than two,
v = vo + at.
(Notice that we can solve this for t to get t = ( v - vo ) / a.)
Substitute this v in the equation x = xo + vave t = xo + ( v + vo )t/2 to get
x = xo + [ vo + at + vo ]t/2 = xo + [ 2vo + at ]t/2 = xo + vot + ½at2.
If instead of substituting v in the above equation, we can instead substitute t as noted to get
x = xo + ( v + vo )t/2 = xo + ( v + vo )( v - vo ) / 2a or
v2 = vo2 + 2a(x-xo).
Galileo’s measurements of motion down inclined planes resulted in the following so-called motion equations that we use to describe motion under constant acceleration.
(1) x = xo + vot + ½at2 (notice that this is a quadratic equation in t and so has two solutions)
(2) v = vo + at
(3) v2 = vo2 + 2a(x-xo)
(4) x = xo + ½( v + vo )t
(5)
= ( vi + vf ) / 2 (for linear relations, the average occurs at the midpoint of an interval).
Notice that when xo = 0 and vo = 0 these equations become x = ½at2, v = at, and v2 = 2ax.
This looks like five equations but it is really only two independent equations plus three rearrangements. For these two equations, we can have only two unknowns.
It does not matter what is the object–from falling coins to rocket ships–the motion of any object moving with constant acceleration can be described by these motion equations. In these equations, x represents the distance an object has moved in time t from its starting place, xo. The object’s speed is represented by v and its initial speed by vo. Speed is visualized in terms of how quickly an object’s position is changing. Notice that x represents how far an object has moved while v represents how fast it is moving or how quickly the object’s position is changing while acceleration represents how quickly its speed is changing in time. Speed is the time rate of change of position while acceleration is the time rate of change of speed. In symbols, speed v=Δx/Δt, where the change in position is Δx = Xfinal - Xinitial., and a = Δv/Δt. (In calculus, differential changes are used, making v = dx/dt and a = dv/dt = d2x/dt2.)
Example:
Suppose a car is accelerated from a stopped position to a speed of 20 m/s (45 mph) in 10 seconds. Its acceleration is a = Δv/Δt = ( 20 m/s ) / 10 s = 2 m/s2 and it will have traveled a distance x = ½at2 = ( ½ )( 2 m/s2 )( 10s )2 = 100 m (109 yards). Compared to your own driving style, does this seem like a high or low rate of acceleration? Do you typically accelerate from 0 to 20 m/s (45 mph) in 10 seconds while traveling just 100 meters?
For differential changes in speed, a = dv/dt or v = ∫a dt = a ∫dt = at + c. The integration constant is determined from the condition that v(t=0) = vo, giving v = vo + at. From v = dx/dt or x = ∫v dt = ∫ (vo + at ) dt, integrating gives x = vo t + ½ a t2 + c. The constant is found from x(t=0) = xo. Writing a = dv/dt = dx/dt dv/dx = v dv/dx and integrating we get ∫v dv = ∫a dx or v2 = vo2 + 2a(x-xo)
Suppose an object travels through a first distance at a constant speed (x = vt), then travels through a second and equal distance at a second speed, and then a third and so on. The total distance traveled is
Δx = v1 Δt + v2 Δt + v3 Δt + v4 Δt + v5 Δt + v6 Δt + ... + vn Δt = ∑ vi Δt = ∫v dt.
Graphically, this is a sawtooth. For a time-varying speed, one first computes the total distance using small time intervals, say Δ = 0.02 seconds, and then continues re-computing with intervals that are half as large as the previous sum until the total distance stops changing. Mathematically, this is an integration but in all but the rarest, simple cases, a computer is used to compute sums of increasingly finer time intervals, looking for a total answer that is independent of further decreases in the time interval. When you see an integral, consider it to be shorthand for the process of “adding up the little pieces.”
Notice that the rectangles of varying height vi but constant width Δt cover the entire area below the plotted curve. The area always has a physical meaning. In the v-t plot, the area under the curve is equal to the total distance traveled.
Example:
Plot the first quarter of a circle of radius 1 such that both x and y go from 0 to ½. Consider this to be a plot of speed versus time. Obtaining vi from the center of the ith time interval, compute the total distance x = ∑ vi Δt using a time interval of 0.25. Then repeat the process using increasingly fine time intervals by continually cutting the previous intervals in half–0.12, 0.06, 0.03, and 0.015 seconds, and so on–until the total distance stops changing. What is the value to which the total is converging? Do you find that value to be π/4? We expect that value since this plot is the area of one-quarter of an entire circle, which has area πr2. Archimedes suggested that this procedure be used to obtain a formula for the volume of a sphere by adding up the area of all the little circles of varying radii that comprise the sphere.
Calculus consists mostly of two ideas: the slope of a curve–for example, the steepness of a hill–and the process of summing up the little pieces that comprise a whole. Calculus was invented in about 1650 by Liebnitz and separately by Newton (not fig newton but Isaac Newton, who will be the hero of the next chapter). Calculus is used to add a string of numbers or things varying in magnitude. In algebra things can only be constant. Algebra was invented in the year 800 by a Persian (Iranian) named al-Khwarizami. Engineers and scientists use calculus every minute to build today's civilization. Algebra helped build only medieval civilization. Everyone should learn calculus.
Several websites have data for motion having non-constant acceleration, including a = kt-1.3 for fish.
Did you believe that the steeper Galileo sets his incline, the greater the resulting acceleration? Do you believe the acceleration would be greatest when the inclined plane is vertical so that the object falls straight down? As something falls, just how is its speed changing in time? Does its speed stay the same as it falls? Does a light object fall as fast as one that is heavier? When an object is dropped, gravity causes it to undergo a constant acceleration of g = 9.8 m/s2. When a car moves at a constant speed of 20 m/s, its position is changing by 20 meters every second–for example, between the seventh and eighth seconds. Falling at constant gravitational acceleration of 9.8 m/s2 means that the speed of an object increases by 9.8 m/s every second–for example, between the seventh and eighth seconds. That is, the speed after, say, 20 seconds will be 9.8 m/s greater than it was after 19 seconds, and the speed after 38 seconds will be 9.8 m/s greater than it had been after falling for 37 seconds. Suppose a small rock is dropped from a window that is 4.9 meters above the ground. Its initial speed is zero but after one second, the rock will be falling at a speed of 9.8 m/s and will have fallen the entire distance of 4.9 meters. It will have taken one second for it to hit the ground. If the window was instead four times as high, which is 19.8 meters, then the ball would take only twice as long to hit the ground, which is two seconds, and just before hitting the ground, its speed will be twice as fast–19.8 m/s–as when it had fallen 4.9 m, see Table 1. By the way, you may have heard of Galileo’s claim that a large and a small rock will both hit the ground at the same time when simultaneously dropped from the Leaning Tower of Pisa. This experiment was repeated on the Moon by Apollo astronauts who simultaneously dropped a feather and a hammer.
When falling, v = gt, v up by ten m/s with each second, down by 10 when thrown upward. After having released the thrown straight-up ball, is anything pushing it up in the air? It wants to keep moving at constant speed forever but the downward force of gravity is slowing it down. When falling, the object still wants to keep moving at constant speed but it is being accelerated by gravity. A constant change in speed per unit time. Hewitt suggests we trade dimes for each gain or loss of speed. Downward speed when returned from ascent and passing release point same as if thrown down. Table shows distance is x=1/2 a t2, use your calculator to verify.
Table 1. Falling at g = 9.8 m/s2. Speed grows linearly in time, increasing by 9.8 m/s every second, but distance increases as the square of time.
time (s) speed (m/s) speed change total distance fallen (m) change in distance (m)
0 0 0
1 9.8 9.8 4.9 4.9
2 19.6 9.8 19.6 14.7
3 29.4 9.8 44.1 24.5
4 39.2 9.8 78.4 34.3
Demonstration:
Drop Hewitt’s bolts that are tied on a string. Tie equally spaced bolts to a string and drop the string over a metal pan. The bolts hit increasingly frequent as those dropped from increasing heights reach higher speeds. The time between pan-hits is heard to decrease. If the nuts are instead spaced 1, 3, 5, 7, and 9 distance units apart then the hits are heard to be evenly spaced in time.
CQ: During the first second in which an object is falling its speed increases by 9.8 m/s. The object is falling with constant acceleration of 9.8 m/s. By how much does its speed increase between the ninth and tenth seconds? Did the object travel a larger distance between the second and third seconds or between the third and fourth seconds?
CQ2:
If an object falls with constant acceleration, its speed must
a) be constant also
b) continually increase by the same amount each second
c) continually change by varying amounts depending on its speed
d) continually decrease
e) none of these
Demonstration:
Here is a tear-drop-shaped lump of water falling toward the ground.
Example:
A ball is thrown upward with a speed of 29.4 m/s. Its speed will then decrease by 9.8 m/s each second. At t = 1, v = 19.6. At t = 2, v = 9.8, and at t = 3, v = 0. Notice that momentarily, v = 0 at the top of the trajectory. This time at which v = 0 can be found from the equation v = vo - gt, which gives 0 = 30 - 9.8t or t = 3. How high will the object rise? Since we know that the time will be 3 s at the top of the path, we use x = xo + vot + ½at2 we get x =0 + (29.4)(3) - (½)(9.8)(3)2 = 1/2 x 10 x 32 = 44m. Notice that it moves 44 meters in the three seconds before reaching the top and also during the three seconds after reaching the top. If we don’t know the time at which the object reaches the top of its trajectory, then we determine the maximum height by noticing that the speed is momentarily zero at the highest point in the trajectory. We set v = 0 in v2 = vo2 + 2a(x-xo) and then solve for x. How long does it take the object to return to its initial level? A total of 6 seconds, which is 3 upward and 3 downward. What is the speed at t = 6 seconds? It is -29.4 m/s, which is equal but opposite to its initial upward speed when thrown from this same level.
Example:
Drop a rock. When free falling under gravity, a = g = 9.8 m/s2 throughout the motion–at the beginning, for an instant at the peak, at the end, and everywhere in between. Its acceleration is constant throughout its motion even if it is initially thrown upward or downward. If the rock is equipped with an accelerometer, the meter will show a = g = constant.
What if you stand at the top of a cliff and use a gun to shoot a bullet straight down. What is its acceleration? 9.8 m/s2 toward the ground. What if you shoot the bullet straight up? Answer: a = 9.8 m/s2 toward the ground. Do you believe it? We get to refine our notion of motion by studying some details we had never had any reason to look at before.
Solving many problems helps us to refine our notion of motion; that is the purpose of the effort. For example, you may have noticed that a ball thrown upward will have a speed of zero for an instant at the top of its trajectory before again falling downward. But you may not have noticed that the object’s acceleration has the constant downward value of g = 9.8 m/s2 throughout its upward and downward motion, including that moment of zero speed.
CQ: When a rock thrown straight upwards gets to the exact top of its path, its
a) speed is zero and its acceleration is zero
b) speed is zero and its acceleration is about 10 m/s2
c) speed is about 10 m/s and its acceleration is zero
d) speed is about 10 m/s and its acceleration is about 10 m/s2
e) none of these
Procedure for solving motion problems
We use the following procedure to solve problems involving the constant-acceleration motion equations. Breaking a problem into these manageable steps enables success while trying to do the entire problem in your head, writing a final arithmetic statement, does not.
At first it looks like we are using five motion equations but since there are only two independent equations, a problem can have only two unknowns. Of the six variables–x, xo, v, vo, a, and t–involved in the motion equations, four will be given in the problem; the remaining two will be unknown, and hence the goal of your solution. Begin by making a sketch of the problem. In the sketch, try to show the meaning of the variables and also indicate where you’ve chosen to place the x- and y-axis so that the origin and positive and negative directions remain consistent throughout the solution. List the six variables and fill in the values of the four given quantities. Look at the equations to see which contains a single unkown, solve it for that unknown variable, plug in values for the knowns, and then calculate a value for the unknown. Check that your answer has a reasonable value and has the correct positive or negative sign. Then use one of the other motion equations to obtain a value for the second unknown.
Example.
To passing a slow moving vehicle, a car goes from a speed of 20 m/s (45 mph) to 25 m/s (56 mph) in 5 seconds. What is its acceleration and how far will it travel during its acceleration?
First we make a sketch showing the axis and indicating the meaning of the variables. Place the origin wherever it will make more variables have a zero value. We’ll choose to place the origin of the x-axis at the place where the car begins accelerating.
v=20 m/s v=25 m/s
----------|-------------------------|--------------> x
x=0 x=?
The problem states that
x = ?
xo = 0
v = 25 m/s
vo = 20 m/s
a = ?
t = 5 s
It never fails that in the statement of the problem, we are given four of the six variables and we will have two unknowns. We see that in this problem, the two unknowns are x and t. We next look at equations 1 to 3 to find one containing a single unknown. We can not use equation 1 because it contains the two unknown quantities, x and a. We’ll use equation 2 because it contains the single unknown, a. Solving equation 2 for a gives,
a = ( v - vo ) / t = [ (25 m/s) - (20 m/s) ] / 5s = 1 m/s2.
The units in the middle expression are (m/s)/s or m/s2. Notice that the calculated acceleration has a positive value, indicating that the acceleration is in what we chose to be the positive direction of the axis. Now that we know a value for a, we can use either equation 1 or 3 to obtain a value for x. Equation 1 gives
x = xo + vot + ½at2 = 0 + (20 m/s)(5s) + (½)(1 m/s2)(5s)2 = 112.5 m
This means that the car traveled 112.5 meters while accelerating at 1 m/s2 for 5 seconds. Does this distance and time seem reasonable, given your past driving experience?
Example:
While launching, the Space Shuttle accelerates at a rate of 1.5 g for 4 minutes. What is its final speed and how far could it travel in a straight line during that acceleration?
v = at = ( 1.5 )( 9.8 m/s2 )( 4 min )( 60 sec/min ) = 3,500 m/s.
x = ½at2 = (0.5)( 1.5 )( 9.8 m/s2 )[ ( 4 min )( 60 sec/min ) ]2 = 423 km.
What does that acceleration feel like? Trefil would say that it feels like 1.5 persons sitting on you for four minutes.
CQ: Airplane Flight Data Recorder or “Black Boxes” are built to withstand accelerations of 3400 g. What acceleration occurs when the speed of an airplane decreases from 270 knots = 311 mph = 137 m/s down to 0 m/s while traversing a distance of 0.45 meters? List the six variables, fill them in, and solve a motion equation for each of the two unknowns.
x = 0.45 m
xo = 0
v = 0
vo = 270 knots = 311 mph = 137 m/s
a = ?
t = ?
From v2 = vo2 + 2a(x-xo) we get a = ( v2 - vo2 ) / [ 2(x-xo) ] = ( 0 - 1372 ) / [ 2(0.45 - 0) ] = -20,850 m/s2 = 2100g, which is less than the 3400 g rating. (One comedian asked why we don’t make the entire airplane out of Black Box material.)
From v = vo + at we get t = (v-vo)/a = (0 - 137 m/s ) / ( -20,850 m/s2 ) = 0.06 seconds.
Demonstration:
Launch the high pressure water bottle upwards and measure the maximum height of its trajectory. What was its initial speed?
Example:
In the 1940s and 50s, Colonel Stapp, an M.D., conducted rocket-powered sled experiments to determine the effects of acceleration on jet pilots–usually testing on himself. The sled moved along two miles of railroad track that ran right into lake water, which was used to quickly stop the sled. Other experiments used numerous brakes to stop the sled. After some test with dummies, the team used chimpanzees. The first test chimp was placed into the seat and was handed a banana just when the rockets were fired. Knowing the resulting effect of the banana, when the chimp was placed into the seat for a second run, he wisely refused the banana. Colonel Stapp made dozens of runs of varying durations and accelerations, sometimes facing backwards.
During the run shown in this video clip, the sled was slowed from a speed of 632 mph to zero in 1.4 seconds. What was the acceleration and how far did Stapp travel while braking?
x = ?
xo = 0
v = 0
vo = 632 mph = 281 m/s
a = ?
t = 1.4 s
From v = vo + at we get a = (v-vo)/t = (0 - 281 m/s ) / ( 1.4 s ) = -200 m/s2, or 200/9.8 = 20.5 g.
Then x = xo + vot + ½at2 = 0 + (281 m/s )(1.4 s) + (0.5)(-200 m/s2 )(1.4s )2 = 197 m, which is the length of two football fields.
Stapp experienced an acceleration of 20.5 g here, but as much as 46.2 g in other runs. During the run discussed above, Stapp felt as if 20 persons were sitting on him.
How did Stapp describe his decelerating experience? "On entry into the waterbrakes, vision became a shimmering salmon, followed by a sensation in the eyes somewhat like the extraction of a molar without an anaesthetic. I was left with two black eyes, which lasted the usual length of time, but vision returned in about eight and a half minutes. There was no fuzziness of vision or sensations of retinal spasms as had been experienced following a run at Edwards in which a retinal haemorrhage occurred. Aside from congestion of the passages and blocking of paranasal sinuses, hoarseness and occasional coughing from congestion of the larynx and the usual burning from strap abrasions, there was only a feeling of relief and elation in completing the run and in knowing that vision was unimpaired."
By the way, today’s jet-fighter pilots routinely experience 5-10 g and so wear “g-suits” that squeeze arms and legs to keep blood from gathering in those appendages.
Stapp demonstrated that not much harm occurs to people when decelerated at 25 gs. Car crashes occur at one-tenth the speed and one-tenth the impact duration, resulting in similar accelerations of about 25g. This is the non-deadly deceleration experienced by seat-belted persons during car crashes. Un-belted persons are hurt or killed because they suffer twice this deceleration as their bodies collide with the windshield or dash board of the car (see the video clip “seat-belted egg”). Every person should learn this. The National Highway Traffic Safety Administration reports that about 40,000 people die every year in six million U.S. car accidents. (How many traffic deaths would occur if people instead rode inside spherical, metal enclosures equipped with race-car style seatbelts? Race drivers routinely walk away from dramatic crashes and rollovers. Is it worth 40,000 lives per year to construct roadways that control the movement of cars?) As the bumper sticker says, it is a good idea to wear seatbelts because “they make it harder for aliens to snatch you out of your seat.” Many cars are now equipped with event data recorders. This report includes measurements of a 7g acceleration such as occurs in very low speed collisions that hardly dent cars. Surprisingly, school buses do not have seat belts or even an on-board supervisor who maintains order while the driver operates the vehicle. This report includes measured accelerations of belted and un-belted crash-test dummies during school bus collisions.
Example:
Estimate the acceleration one experiences when plopping down from a standing height onto cushioned or wooden seats or when landing on the floor after jumping from a chair.
Example:
A car goes from a speed of zero to 20 m/s (20x2.25=45mph) in 10 seconds. What is its average acceleration? It is the change in speed / elapsed time = 20 m/s / 10 s = 2 m/s2. What was its speed after 3 seconds? Its speed was v = at = ( 2 m/s2 )( 3 s ) = 6 m/s. How far had it traveled after 3 seconds? It had traveled a distance of x = ½ at2 = ( 0.5 )( 2m/s2 )( 3 s )2 = 9 m.
Example:
A car speeds up from 30 to 60 mph in 30 seconds while getting onto the hwy by using the on-ramp, what is its acceleration? a= change in speed / elapsed time = (60 - 30 mph) / 30 s = 1 mph/s. Does this seem like a quick or slow increase in speed? Do you take 2, 5, 10, 20, or 30 seconds to speed up by that amount?
Example:
A car travels on the highway at constant speed. What is its acceleration. Zero. A parachuter falls through the air at constant speed. What is its acceleration? Zero. We don't have to put numbers into an equation because the concept of acceleration is "changing speed."
Example.
A blue car sits at a stop light. Just as the light turns green the blue car begins accelerating at 8 kph/s and a yellow car passes at a speed of 80 kph (km/hr). In how many seconds will the blue, accelerating car pass the yellow car that is moving at a constant speed? (Visit here, choose "Kinematics" then "Catch Up.”) The graph of x vs t is a straight line for the yellow car. This line will intersect the parabolic curve of the accelerating car. The two cars reach their point of intersection at the same time t and after having traveled the same distance x. For the yellow car, x = vt. For the blue car, x = ½ at2. Equating these two and solving for t gives t = 2v/a = (2)(80 kph) / (8 kph/s ) = 20 seconds. Each car will then have traveled a distance of x = vt = (80 km/h)(1000 m / km)(1 hour / 3600 s)(20 s) = 440 m.
Discuss the following reports.
Athletes concerned about acceleration may use the Valsalva Acceleration Technique.
John Bentley explains the braking system of trains and gives a numerical example. He says that the air pressure signal that engages the brakes of each car takes several seconds to travel the length of a mile-long train. While braking, the wheels do not stop spinning so the stopping-distance is not independent of the mass of the train.
Discuss the article Affects of Speed and Acceleration on Car Gas Mileage by the Southwest Region University Transportation Center.
Discuss the article Speed and acceleration of lizards on inclines by Irschick DJ and Jayne BC of the Department of Biological Sciences, University of Cincinnati.
NASA is looking into ways to accelerate materials at 10,000 m/s2 using electromagnetic mass drivers.
NASA is looking into ways to accelerate spacecraft using magnetic levitation tracks.
One experimental system accelerates a 10-pound mass from a speed of 0 to 57 mph while traversing 22 feet of track. What is the acceleration?
xo = 0
v = 57 mph = 25 m/s
vo = 0
a = ?
t = ?
From v2 = vo2 + 2a(x-xo) we get a = ( v2 - vo2 ) / [ 2(x-xo) ] = ( 252 - 0 ) / [ 2(6.7 - 0) ] = 48 m/s2 = 4.9g.
We might accelerate spaceships using antimatter, fusion power, or ion propulsion.
Adam Summer wrote Spore Launchers, Ferns and fungi that explosively reproduce in the December 2005–January 2006 edition of the Natural History Magazine. He says that an internal pressure of five-atmospheres builds inside a plant and then accelerates spores so that they attain a speed of 80 miles per hour by the end of a one-quarter-inch path. What is the acceleration?
x = ( 0.25 inch )( 2.54 cm / inch )( 1 m / 100 cm ) = 0.00635 m
xo = 0
v = 80 mph = 35.6 m/s
vo = 0
a = ?
t = ?
From v2 = vo2 + 2a(x-xo) we get a = ( v2 - vo2 ) / [ 2(x-xo) ] = ( 35.62 - 0 ) / [ 2(0.00635 - 0) ] = 100,000 m/s2 = 10,000 g.
Example:
The acceleration due to gravity is 1.796 m/s2 at the surface of Saturn’s moon Io. One of Io’s several volcanoes, ejects sulphur dioxide directly upward to an altitude of 300 km. What is the initial speed of this material? When the origin is at the ground, we have these values.
x = 300,000 m
xo = 0
v = 0
vo = ?
a = -1.796 m/s2
t = ?
From v2 = vo2 + 2a(x-xo) we get vo2 = v2 - 2a( x - xo ) = 0 - (2)( -1.796 m/s2 )( 300,000 m - 0) or vo = 1040 m/s, which is about 1 km/s.
ActivPhysics On-line has animations that explain motion, including graphs, ticker tape, vectors, and problem solving.
Plot speed and acceleration from the following two video clips.
www.audiounlimited.biz/movie/acceleration.mpg
http://undergroundrace.turboblog.fr/voiture/files/Acceleration.MPG
Here is a video clip showing the acceleration of a SmartCar, which gets 70 mpg with a non-hybrid engine.
Mark Verboom connected a computer to his car’s electronics and recorded the speed of the car through time. Plot his measurements.
The Cosmology Research Group, UCB Physics Department, and Emery USD explain how to measure distances on a map of your home town and then use stopwatches and speedometers to measure speed and acceleration as one travels selected paths in town
Garrett Brown describes a machine that monitors the acceleration of an elderly person to detect falls in his article An Accelerometer Based Fall Detector: Development, Experimentation, and Analysis.
Ride along with Andy Green as he accelerates a car to Mach 1, see thrust.avi. Measure the elapsed time and record the speeds he calls out, and then plot them on a graph. From M.S.Cramer’s collection at Virginia Tech.
Jay A. Nelson of Towson University describes the measured acceleration of fish through time in A Laser-beam Detection System for Measuring Burst Swimming Performance of Fish.
While working with Teachers Experiencing Antarctica and the Arctic, Karina Leppik measures the acceleration due to gravity at the Amundson-Scott South Pole Station.
AH Smith describes life for chickens growing and living under permanent conditions of 2.5g in his report Physiological changes associated with long-term increases in acceleration. The bones, heart, and muscles of the chickens become a few times stronger than normal.
Let’s look at graphs of x, v, and a vs t during various scenarios. These help convince us that x represents “how far” or “changing position," speed v represents “how fast," and acceleration represents “changing speeds.”
I move my arm up by 5 distance units in 5 seconds. It’s a history of the motion of my arm. To spread out the record in time, we pull the chalkboard left (or I walk to the right) while I move my arm up. Time is along the horizontal axis. The position of my hand is recorded along the vertical position. If I move my arm faster upwards, the drawn line is steeper, getting farther away from the origin more quickly. Next, if I move my arm more slowly upwards, the drawn line is less steep. The equation of the line in the form “y = mx + b” is x = xo + vt. The slope of x-t graph is speed–even while accelerating. The slope line (rise/run) is tangent to the curve at any instant, even for a wiggly line. This illustrates the meaning of instantaneous speed.
for v=0, which means a=0, too:
x vs t is a straight line at x=Xo
v vs t is y=0
a vs t is y=0
for v > 0 but constant (which still means a = 0)
animated ticker tape and graph
x vs t is a straight line with positive slope, x = vt ( v = physical meaning of slope )
v vs t is y=constant (positive)
a vs t is y=0
for v < 0 but constant (which still means a = 0)
animated ticker tape and graph
x vs t is a straight line with negative slope, x = vt ( v=physical meaning of slope )
v vs t is y = constant (less than zero)
a vs t is y=0
Make one x-t plot showing lines for x = 10 + 5t, x = -5 + 20 t, and x = 5 - 10t. Mention steep, not so steep, and negative slopes.
CQ: make an x-t plot (in meters and seconds) with a range of 0 to 50 on each axis: a person starts at x=10 at t=0, then moves at constant speed to get to x = 50 at t=50. Write the "X=Xo+vt" equation for this line by writing the numerical values for Xo and v.
for a>0 and constant
animated ticker tape and graph
x vs t is a parabola x=1/2 a t2
v vs t is a straight line, v=vt
slope of x vs t is speed, speed>0 (<0) when slope pos (neg)
a vs t is a positive constant y=constant
animated ticker tape and graph
x vs t is a downward parabola x = ½at2
v vs t is a straight line of negative slope, v = at
a vs t is a negative constant y=constant
CQ: Make v-t plot with constant a. Make a v-t plot (in meters and seconds) with a range of 0 to 50 on each axis: a person's initial speed is v=10 at t=0, then accelerates at reach the speed of v = 50 at t=50. Write the "V=Vo+at" equation for this line by writing the numerical values for Vo and a.
Try Mixed, x and v vs t for a>0 a=0 a<0
Plot x, v, and a graphs for the case of starting your car, accelerating to a constant speed, driving a block at constant speed, stopping, reversing direction, accelerating, moving at constant speed toward home, stopping.
A person in a car starts at x=0 (in meters) at t=0 (in seconds) and accelerates to get to x=20 at t=10 then moves at constant speed to get to x = 30 at t = 20, then slows down to get to x = 35 at t = 30, there it slams on the breaks and stops to stay there until t = 40 then backs up or reverses direction and moves at constant speed to get to x = 30 at t = 50. Draw graph. Where was the person at t=15, 25, 35, and 45? What was the speed at t=5, 15, 25, 35, and 45?
The BBC’s educational website uses graphs to explain speed and acceleration.
Student:
Run the Phet Moving Man simulation and then write a paragraph explaining what it does, what you have learned from it, and how it helps you understand this physical phenomenon. Include numerical examples. What else might be added to it and how could it be improved?
Student:
Run one of the ActivPhysics motion-graphers and then write a paragraph explaining what it does, what you have learned from it, and how it helps you understand this physical phenomenon. Include numerical examples.
Purpose of the chapter is to refine our notion of motion. We want to distinguish speed from acceleration and to see that during free fall, acceleration is constant but speed is not. We measured motion in the lab and found the equations for motion under constant acceleration. In symbols and in your own words, define speed and acceleration, note units and typical values, indicate meaning of pos, neg, zero. Moving at constant speed means your total distance traveled increases by a constant amount per unit of time and that your speed does not change. a = change in speed / elapsed time. Moving at constant acceleration means your speed increases by a constant amount per unit of time. In words and symbols, explain the use of the motion equations.
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