﻿ ﻿Motion in a Straight Line - Review the Knowledge You Need to Score High - 5 Steps to a 5: AP Physics 1: Algebra-Based 2017 (2016)

## 5 Steps to a 5: AP Physics 1: Algebra-Based 2017 (2016)

### Review the Knowledge You Need to Score High

CHAPTER 10 Motion in a Straight Line

CHAPTER 11 Forces and Newton”s Laws

CHAPTER 12 Collisions: Impulse and Momentum

CHAPTER 13 Work and Energy

CHAPTER 14 Rotation

CHAPTER 15 Gravitation

CHAPTER 16 Electricity: Coulomb”s Law and Circuits

CHAPTER 17 Waves and Simple Harmonic Motion

CHAPTER 18 Extra Drills on Difficult but Frequently Tested Topics

### Motion in a Straight Line1

IN THIS CHAPTER

Summary: The entire goal of motion analysis is to describe, calculate, and predict where an object is, how fast it”s moving, and how much its speed is changing. In this chapter you”ll review two separate approaches to make these predictions and descriptions: graphs and algebra.

Definitions

The cart”s position (x ) tells where the cart is on the track.

The cart”s speed (v ) tells how fast the cart is moving. 2

Acceleration (a ) tells how much the object”s speed changes in one second. When an object speeds up, its acceleration is in the direction of its motion; when an object slows down, its acceleration is opposite the direction of its motion.

Displacementx ) tells how far the object ends up away from its starting point, regardless of any motion in between starting and ending positions.

The graphical analysis of motion includes position-time graphs and velocity-time graphs . On a position-time graph, the slope is the object”s speed, and the object”s position is read from the vertical axis. For velocity-time graphs, the speed is read from the vertical axis, and the slope is the object”s acceleration.

The five principal motion variables are:

v 0 initial velocity

v f final velocity

Δx displacement

a acceleration

t time

In any case of accelerated motion, when three of the five principal motion variables are known, the remaining variables can be solved for using the kinematic equations.

Free fall means no forces other than the object”s weight are acting on the object.

A projectile is an object in free fall, but it isn”t falling in a straight vertical line. To approach a projectile problem, make two motion charts: one for vertical motion and one for horizontal motion.

Introduction to Motion in a Straight Line

Pretty much all motion problems can be demonstrated with a cart on a track, like in the diagram above. The motion detector can read the location of the cart up to 50 times each second. This detector can make graphs of position or velocity versus time.

The entire goal of motion analysis is to describe, calculate, and predict where the cart is; how fast it”s moving; and how much its speed is changing. You”ll use two separate approaches to make these predictions and descriptions: graphs and algebra.

Graphical Analysis of Motion

Before you start any analysis, tell yourself which kind of graph you”re looking at. The most common mistake in studying motion graphs is to interpret a velocity-time graph as a position-time graph, or vice versa.

Position-Time Graphs

Example 1: The preceding position-time graph represents the cart on the track. The motion detector is located at position x = 0; the positive direction is to the left.

An AP Exam question could ask all sorts of questions about this cart. How should they be approached? Use these facts, and reason from them.

FACT: In a position-time graph, the object”s position is read from the vertical axis.

Look at the vertical axis in Example 1. At the beginning of the motion, the cart is located 40 cm to the left of the detector. 3 After 2 s, the cart is located 10 cm left of the detector. Therefore, in the first 2 s of its motion, the cart moved 30 cm to the right.

FACT: In a position-time graph, the object”s speed is the slope of the graph. The steeper the slope, the faster the object moves. If the slope is a front slash (/), the movement is in the positive direction; if the slope is a backslash (\), the movement is in the negative direction.

Wait a second, how can I tell the slope of a curved graph? Just look at how the graph is sloped at one specific place on the graph. If I want to know how fast the object is moving after 2 seconds of motion, I take the slope drawn in the following figure.

Mathematically, this means you divide (change in y -value)/(change in x -value) to get a speed that”s a bit more than 6 cm/s. The exact calculation is not something to obsess over. It”s more important to think in terms of comparisons.

You”re not often going to be asked “Calculate the speed of the cart at t = 2 s.” Instead, you”ll be asked to describe the motion of the cart in words and to justify your answer. When you describe motion, use normal language that your grandparents would understand. Avoid technical terms like “acceleration” and “negative.” Justify your answer with direct reference to the facts.

Referring back to Example 1, because the slope was steeper at earlier times and shallower at later times, the cart must be slowing down. The car is moving to the right the whole time, because the slope is always a backslash.

Velocity-Time Graphs

Example 2: The preceding velocity-time graph represents a different cart on the track. The positive direction is to the left.

FACT: In a velocity-time graph, the object”s speed is read from the vertical axis. The direction of motion is indicated by the sign on the vertical axis.

In Example 2, at the beginning of the motion, the vertical axis reads 1.8 m/s. This means that initially, the cart was moving 1.8 m/s to the left. After one second, the cart was moving about 0.8 m/s. A bit less than two seconds into the motion, the vertical axis reads zero, so the cart stopped.

FACT: In a velocity-time graph, the object”s acceleration is the slope of the graph.

You could do the rise/run calculation to find the amount of the acceleration, or you could use the definition of acceleration to see that the object lost 1 m/s of speed in one second, making the acceleration 1 m/s per second. 4 The cart in Example 2 was slowing down and moving to the left. When an object slows down, its acceleration is opposite the direction of its motion; this cart has an acceleration to the right.

The Mistake

Acceleration is not the same thing as speed or velocity. Speed says how fast something moves; acceleration says how quickly speed changes. Acceleration doesn”t say anything about which way something is moving, unless you know whether the thing is speeding up or is slowing down.

Someone who says, “This car has an acceleration of 4 m/s per second, so it is moving at about a jogging pace,” has made “The Mistake.” The car is speeding up or slowing down by 4 m/s every second; the car could well be an Indy 500 racecar traveling 94 m/s right now, but only 90 m/s a second later.

Someone who says “The cart in Figure 2 has a negative acceleration, so it is moving to the right” has made “The Mistake.” An acceleration to the right means either speeding up and moving right, or slowing down and moving left. While the cart”s acceleration is negative—after all, the slope of the line is a backslash—the car was slowing down, making the velocity”s direction opposite the acceleration”s direction. The acceleration is right, and the velocity is left.

It takes a lot of practice to avoid “The Mistake.” Just continually remind yourself of the meaning of acceleration (how much an object”s speed changes in one second), and you”ll get there.

FACT: The object”s displacement is given by the area between the graph and the horizontal axis. The location of the object can”t be determined from a velocity-time graph; only how far the object ended up from its starting point can be determined.

To find how far the cart in Example 2 moved, take the area of the triangle in the graph, 5 giving about 1.6 m. Since the cart”s velocity as read from the vertical axis was positive that whole time, and the positive direction is left, the cart ended up 1.6 m left of where it started. But exactly where it started, no one knows.

Algebraic Analysis of Motion

Example 3: A model rocket is launched straight upward with an initial speed of 50 m/s. It speeds up with a constant upward acceleration of 2.0 m/s per second until its engines stop at an altitude of 150 m.

Sometimes you”ll be asked to analyze motion from a description and not a graph. Start your analysis by defining a positive direction and clearly stating the start and the end of the motion you”re considering. For example, take the upward direction as positive, 6 and consider from the launch to when the engines stop.

Next, make a chart giving the values of the five principal motion variables. Include a plus or minus sign on every one (except time—a negative time value means you”re in a Star Trek–style movie). If a variable isn”t given in the problem, leave that variable blank.

The Five Principal Motion Variables for Your Chart

v 0 initial velocity

v f final velocity

Δ x displacement

a acceleration

t time

For Example 3, the chart looks like this:

The acceleration is positive because the rocket was speeding up; therefore, acceleration is in the same direction as the motion, which was upward. Upward was defined as the positive direction here.

FACT: In any case of accelerated motion when three of the five principal motion variables are known, the remaining variables can be solved for using the kinematic equations.

In Example 3, we know three of the five motion variables; therefore, we can find the others, and the physics is done .

Whoa there . Um, how is the physics “done”? Don”t we have to plug the numbers in to the kinematic equations, which incidentally you haven”t mentioned yet?

Well, remember the AP Physics 1 revolution: While you will occasionally be asked, say, to calculate how much time the engines run for, you”ll just as often be asked something that doesn”t involve calculation. For example, “Is it possible to determine the running time of the engines?” Or, “When the engines have run for half of their total run time, is the rocket at a height greater than, less than, or equal to 75 m?” 7

More to the point, actually doing the math here is, well, a math skill not a physics skill. As long as your answers are reasonable—a model rocket will likely burn for a few seconds, not a few thousands of seconds—the exam is likely to award close to full, or sometimes even full, credit for a correct chart and for recognizing the correct equation to use.

FACT: To calculate the missing values in a motion chart, use the three kinematic equations listed as follows. Choose whichever equation works mathematically. Never solve a quadratic equation. If the math becomes overly complicated, try solving for a different missing variable first.

Kinematic Equations

1. vf= v 0 + at

Continuing with Example 3, we can use equation (3) to solve for the final velocity of the rocket 8 ; this is about 56 m/s. 9 Then we can use equation (1) to get the time before the engines shut off, which is 3 seconds. If you had tried to use equation (2) to solve for time, you would have gotten a quadratic; that”s why I said to use equation (3) and then (1).

Objects in Free Fall

FACT: When an object is in free fall, its acceleration is 10 m/s per second 10 toward the ground. “Free fall” means no forces other than the object”s weight are acting on the object.

Let”s do more with the rocket in Example 3. When the engines stop, the rocket is moving upward at 56 m/s. The rocket doesn”t just stop on a dime. It keeps moving upward, but it slows down, losing 10 m/s of speed every second.

Try making a chart for the motion from when the engines stop to when the rocket reaches the peak of its flight. We”ll keep the positive direction as upward.

Three of the five variables are known so the physics is done .

Now, be careful that you keep grounded in what”s physically happening, not in the algebra. For example, you might be asked for the maximum height that the rocket in Example 3 reaches.

Good job recognizing that you need equation (3) to solve for Δ x . Dropping the units during the calculation, that gives 02 = 562 + 2(−10)(Δ x ). Solving with a calculator you get about 160 m for Δ x .

Wait a second! Think what the 160 m answer means—that”s the distance the rocket goes between when the engines stop and when the rocket reaches its highest height. That”s not the height above the ground, just the additional height after the engines stop! The actual maximum height is this 160 m, plus the 150 m that the rocket gained with its engines on, for a total of 310 m.

If you were blindly plugging numbers into equations, you would have totally missed the meaning behind these different distances. The AP Physics 1 Exam will repeatedly ask targeted questions that check to see whether you understand physical meaning. Calculation? Pah. In comparison, it”s not so important.

Projectile Motion

A projectile is defined as an object in free fall. But this object doesn”t have to be moving in a straight line. What if the object were launched at an angle? Then you treat the horizontal and vertical components of its motion separately.

Example 4: A ball is shot out of a cannon pointed at an angle of 30° above the horizontal. The ball”s initial speed is 25 m/s. The ball lands on ground that is level with the cannon.

FACT: A projectile has no horizontal acceleration and so moves at constant speed horizontally. A projectile is in free fall, so its vertical acceleration is 10 m/s per second downward.

To approach a projectile problem, make two motion charts: one for vertical motion and one for horizontal motion.

FACT: To find the vertical component of a velocity at an angle, multiply the speed by the sine of the angle. To find the horizontal component of a velocity at an angle, multiply the speed by the cosine of the angle. This always works, as long as the angle is measured from the horizontal.

Here are the two charts for this ball”s motion in Example 4. Consider up and right to be the positive directions. Let”s consider the motion while the ball is in free fall—that means, starting right after the ball was shot, and ending right before the ball hits the ground. Note that the initial vertical velocity is (25 m/s)(sin 30°) = 13 m/s. The initial horizontal velocity is (25 m/s)(cos 30°) = 22 m/s. You needed to use your calculator to get these values.

Two entries here are tricky. Remember that displacement only means the distance traveled start to end, regardless of what happens in between. Well, this ball landed on “level ground.” That means that the ball ends up at the same vertical height from which it was shot; it didn”t end up any higher or lower than it started. Thus, vertical displacement is zero.

Second, the final vertical velocity is unknown, not zero. Sure, once the ball hits the ground it stops; but then it”s not in free fall anymore. The “final” velocity here is the velocity in the instant before the ball hits the ground.

FACT: The horizontal and vertical motion charts for a projectile must use the same value for time.

The vertical chart is completely solvable, because three of the five variables are identified. Once the time of flight is calculated from the vertical chart, that time can be plugged into the horizontal chart, and voila , we have three of five horizontal variables identified; the chart can be completed.

Practice Problems

Note: Extra drills on describing motion based on graphs can be found in Chapter 18 .

1 . The velocity-time graphs represent the motion of two cars, Car A and Car B. Justify all answers thoroughly.

(a) Which car is moving faster at the start of the motion?

(b) Which car ends up farther from its starting point?

(c) Which car experiences a greater magnitude of acceleration?

2 . The following questions refer to the preceding position-time graph, which is the readout of an eastward-pointing motion detector. Justify all answers thoroughly.

(a) Rank the speed of the object in each of the four labeled regions of the graph, from fastest to slowest. If the object has the same speed in two or more regions, indicate so in your ranking.

(b) What total distance did the object travel in the 12 s, including all parts of the motion?

(c) How far from the object”s starting point did the object end up after the 12 s?

(d) Which of the following objects could reasonably perform this motion?

(A) A baby crawling

(B) A sprinter

(C) A car on the freeway

(D) A jet airplane during takeoff

(E) An amoeba in a petri dish

3 . A ball is dropped from rest near Earth. Neglect air resistance. 11 Justify all answers thoroughly.

(a) About how far will the ball fall in 3 s?

(b) The same ball is dropped from rest by an astronaut on the Moon, where the free-fall acceleration is one-sixth that on Earth. In 3 s, will the ball on the moon fall:

(A) One-sixth as far as the ball on Earth

(B) One-36th as far as the ball on Earth

(C) The same distance as the ball on Earth

(D) Six times as far as the ball on Earth

(E) Thirty-six times as far as the ball on Earth

4 . Four projectiles, A, B, C, and D, were launched from and returned to level ground. The preceding data table shows the initial horizontal speed, initial vertical speed, and time of flight for each projectile. Justify all answers thoroughly.

(a) Rank the projectiles by the horizontal distance traveled while in the air.

(b) Rank the projectiles by the maximum vertical height reached.

(c) Rank the projectiles by the magnitude of their acceleration while in the air.

Solutions to Practice Problems

1 . (a) On a velocity-time graph, speed is read off of the vertical axis. At time = 0, Car A has a higher vertical axis reading than Car B, so Car A is moving faster.

(b) Displacement is determined by the area under a velocity-time graph. Car A”s graph is a small triangle; Car B”s graph is a trapezoid of obviously larger area. Both cars” graphs are always above the horizontal axis, so both cars move in the same direction the whole time; Car B moves farther away from its starting point.

(c) Acceleration is the slope of the velocity-time graph. Car A”s graph is steeper, so its acceleration is larger. [Sure, the slope of Car A”s graph is negative, but that just means acceleration is in the negative direction, whatever that is; the question asks for the “magnitude” of the acceleration, meaning the amount, regardless of direction.]

2 . (a) IV > I = III > II. Speed is the steepness on a position-time graph, without reference to direction (i.e., whether the slope is positive or negative). Segment IV is steepest. Segments I and III seem to be the same steepness. Segment II has 0 slope, so it represents an object that doesn”t move.

(b) It”s a position-time graph, so read the vertical axis to figure out where the object is at any time. The object travels from its original position at x = 0 m to x = 40 m, then backtracks another 20 m. The total distance traveled is 60 m.

(c) It”s a position-time graph, so read the vertical axis to figure out where the object is at the beginning and after 12 s. At the beginning the object was at x = 0 m; after 12 s, the object was at position x = 20 m. The object traveled 20 m. (If your justification didn”t explicitly mention that the object started at x = 0 m, or that you must find the difference between the final and initial positions, then it”s incomplete.)

(d) At its top speed in segment IV, the object travels 20 m in about 2 s. That”s a speed of 10 m/s. If you”re familiar with track and field, you”ll know that the best sprinters run the 100-m dash in somewhere in the neighborhood of 10 s, so the sprinter is an obvious choice. It might be easier 12 to approximate a conversion to miles per hour. The result of 1 m/s is a bit more than 2 miles per hour. This object goes between 20 and 25 miles per hour. This is the speed of a car on a neighborhood street. There is no way a baby or an amoeba can keep up; takeoff speeds for most airplanes are at least in the high tens of miles per hour; and you”d be a danger to yourself and others if you drove on the freeway at 25 miles per hour.

3 . (a) Use the equation with v 0 = 0 and a = 10 m/s per second. You should get about 45 m.

(b) In the equation we use in (a), the time of 3 s is still the same, as is v 0 . The only difference is the acceleration a , which is in the numerator and is neither squared nor square rooted. Therefore, reducing a by one-sixth also reduces the distance fallen by one-sixth. That”s choice A. (By the way, if your answer is A but your justification included “setting up a proportion” or anything without specific reference to this equation, your answer is incorrect.)

4 . (a) Horizontal speed remains constant throughout a projectile”s flight. Use horizontally with the acceleration term equal to zero. That means you”re multiplying the horizontal speed by the time of flight. This gives D > C > A = B.

(b) Regardless of the time of flight, the vertical speed is directly related to the maximum height reached. Why? Use vertically with v f = 0 and a = 10 m/s per second. The bigger the v 0 , the bigger the Δx . So A > C > B = D.

(c) Easy—all objects in free fall have a downward acceleration of 10 m/s per second. A = B = C = D.

Rapid Review

• In a position-time graph, the object”s position is read from the vertical axis.
• In a position-time graph, the object”s speed is the slope of the graph. The steeper the slope, the faster the object moves. If the slope is a front slash (/), the movement is in the positive direction; if the slope is a backslash (\), the movement is in the negative direction.
• In a velocity-time graph, the object”s speed is read from the vertical axis. The direction of motion is indicated by the sign on the vertical axis.
• In a velocity-time graph, the object”s acceleration is the slope of the graph.
• In a velocity-time graph, the object”s displacement is given by the area between the graph and the horizontal axis. The location of the object can”t be determined from a velocity-time graph; only how far it ended up from its starting point can be determined.
• In any case of accelerated motion when three of the five principal motion variables are known, the remaining variables can be solved for using the kinematic equations.
• To calculate the missing values in a motion chart, use the three kinematic equations listed below. Choose whichever equation works mathematically.

(1) v f = v 0 + at

(2)

(3)

When an object is in free fall, its acceleration is 10 m/s per second10 toward the ground. “Free fall” means no forces other than the object”s weight are acting on the object.

A projectile has no horizontal acceleration, and so it moves at constant speed horizontally. A projectile is in free fall, so its vertical acceleration is 10 m/s per second downward.

To find the vertical component of a velocity at an angle, multiply the speed by the sine of the angle. To find the horizontal component of a velocity at an angle, multiply the speed by the cosine of the angle. This always works, as long as the angle is measured from the horizontal.

The horizontal and vertical motion charts for a projectile must use the same value for time.

1 We”re discussing motion with constant acceleration; that covers pretty much all motion in a straight line in AP Physics 1 except things attached to springs.

2 Strictly speaking, speed is the magnitude—the amount—of the velocity vector. Velocity tells how fast something moves, as well as in which direction it moves. I tend to use “speed” and “velocity” interchangeably, especially in this unit; the distinction between the two is not important here.

3 Why not 40 cm to the right of the detector? Because the position value is +40 cm and the positive direction is left.

4 Textbooks and problems on the AP Exam will write this as 1 m/s2 . Well, that”s silly—what the heck is a “second squared,” anyway? When you see that notation, read it as “meters per second per second.” I suggest you always write the units of acceleration as “m/s per second.” Then you”ll be far less likely to make “The Mistake.”

5 The area of a triangle is (1/2) base × height.

6 Could I have called the downward direction positive? Sure. Then signs of displacement, velocity, and acceleration would all be switched.

7 The answer is less than 75 m. The rocket is speeding up throughout the time when the engines burn. In the second half of the burn time, the rocket is (on average) moving faster, and so it covers more distance.

8 This calculation requires a calculator, of course: v f 2 = (50 m/s)2 + 2(2 m/s per second)(150 m). See, here”s another reason you”re probably not going to have to actually carry out this math—you”re not likely to need a calculator more than a few times on the entire exam.

9 Not 55.67764363 m/s. Don”t make me ask your chemistry teacher to talk to you about significant figures again. Just use two or three figures on all values, and we”ll all be happy.

10 No, stop it with the 9.8 m/s per second. The CollegeBoard is very clear that you can and should use a free-fall acceleration of 10 m/s per second. Really. Limited calculator use, remember?

11 Always neglect air resistance, unless it is extremely, abundantly, and unambiguously clear from the problem”s context that air resistance is important (i.e., talking about “terminal velocity”).

12 For an American, who doesn”t usually know from meters per second (m/s), anyway. If you”d prefer km/hr, multiply speeds in m/s by about 4 to get km/hr.

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