5 Steps to a 5: AP Physics C (2016)
Review the Knowledge You Need to Score High
IN THIS CHAPTER
Summary: As soon as an object”s velocity changes, you need to analyze the problem using kinematics, which deals with aspects of motion separate from considerations of mass and force.
Kinematics problems involve five variables: initial velocity, final velocity, displacement, acceleration, and time interval.
Use the three kinematics equations whenever acceleration is constant.
Average speed is the total distance in a given time divided by the time it takes you to travel that distance.
Velocity is just like speed, except it”s a vector.
Acceleration is the change in velocity divided by a time interval.
Displacement is the vector equivalent of distance.
The key rule of projectile motion is that an object”s motion in one dimension does not affect its motion in any other dimension.
The constant-acceleration kinematics equations, which we refer to as the “star” equations:
The equilibrium problems we saw in the last chapter all had something in common: there was no acceleration. Sure, an object can move at a constant velocity and still be in equilibrium, but as soon as an object”s velocity changes, you need a new set of tricks to analyze the situation. This is where kinematics comes in.
Velocity, Acceleration, and Displacement
We”ll start with a few definitions.
In this definition, Δx means “displacement” and Δt means “time interval.” Average speed is the total displacement you travel in a straight line in a given time divided by the time it takes you to travel that distance. This is different from “instantaneous speed,” which is your speed at any given moment. WARNING: The formula you learned in seventh grade, “speed = distance/time” is ONLY valid for an average speed, or when something is moving with constant speed. If an object speeds up or slows down, and you want to know its speed at some specific moment, don”t use this equation! 1
Questions on the AP exam tend to focus on velocity more than speed, because velocity says more about an object”s motion. (Remember, velocity has both magnitude and direction.)
Acceleration occurs when an object changes velocity.
The symbol Δ means “change in.” So Δv = vf − v 0 , where v f means “final velocity” and v 0 means “initial velocity” and is pronounced “v-naught.” Similarly, Δt is the time interval during which this change in velocity occurred.
Just as velocity is the vector equivalent of speed, displacement is the vector equivalent of distance—it has both magnitude and direction.
So, let”s say that you head out your front door and walk 20 m south. If we define north to be the positive direction, then your displacement was “−20 m.” If we had defined south to be the positive direction, your displacement would have been “+20 m.” Regardless of which direction was positive, the distance you traveled was just “20 m.” (Or consider this: If you walk 20 m north, followed by 5 m back south, your displacement is 15 m, north. Your displacement is not 25 m.)
Constant-Acceleration Kinematics Equations
Putting all of these definitions together, we can come up with some important lists. First, we have our five variables:
Using just these five variables, we can write the three most important kinematics equations. An important note: The following equations are valid ONLY when acceleration is constant. We repeat: ONLY WHEN ACCELERATION IS CONSTANT . Which is most of the time. 2
We call these equations the “star equations.” You don”t need to call them the “star equations,” but just be aware that we”ll refer to the first equation as “* ,” the second as “** ,” and the third as “*** ” throughout this chapter.
These are the only equations you really need to memorize for kinematics problems.
Constant-Acceleration Kinematics Problem-Solving
Step 1 : Write out all five variables in a table. Fill in the known values, and put a “?” next to the unknown values.
Step 2 : Count how many known values you have. If you have three or more, move on to Step 3 . If you don”t, find another way to solve the problem (or to get another known variable).
Step 3 : Choose the “star equation” that contains all three of your known variables. Plug in the known values, and solve.
Step 4 : Glory in your mastery of physics. Feel proud. Put correct units on your answer .
Be sure that you have committed these steps to memory. Now, let”s put them into action.
A rocket-propelled car begins at rest and accelerates at a constant rate up to a velocity of 120 m/s. If it takes 6 s for the car to accelerate from rest to 60 m/s, how long does it take for the car to reach 120 m/s, and how far does it travel in total?
Before we solve this problem—or any problem, for that matter—we should think about the information it provides. The problem states that acceleration is constant, so that means we can use our kinematics equations. Also, it asks us to find two values, a time and a distance. Based on the information in the problem, we know that the time needed for the car to reach 120 m/s is greater than 6 s because it took 6 s for the car just to reach 60 m/s. Moreover, we can estimate that the car will travel several hundred meters in total, because the car”s average velocity must be less than 120 m/s, and it travels for several seconds.
So now let”s solve the problem. We”ll use our four-step method.
Step 1 : Table of variables.
The car begins at rest, so v 0 = 0 m/s. The final velocity of the car is 120 m/s. We”re solving for time and displacement, so those two variables are unknown. And, at least for right now, we don”t know what the acceleration is.
Step 2 : Count variables.
We only have two values in our chart, but we need three values in order to use our kinematics equations. Fortunately, there”s enough information in the problem for us to solve for the car”s acceleration.
Acceleration is defined as a change in velocity divided by the time interval during which that change occurred. The problem states that in the first 6 s, the velocity went from 0 m/s to 60 m/s.
We now have values for three of our variables, so we can move to Step 3 .
Step 3 : Use “star equations” to solve.
All three of our known values can be plugged into *, which will allow us to solve for t .
Now that we know t , we can use either ** or *** to solve for displacement. Let”s use **:
Step 4 : Units.
Always remember units! And make sure that your units are sensible—if you find that an object travels a distance of 8 m/s, you”ve done something screwy. In our case, the answers we found have sensible units. Also, our answers seem reasonable based on the initial estimates we made: It makes sense that the car should travel a bit more than 6 s, and it makes sense that it should go several hundred meters (about half a mile) in that time.
Problems that involve something being thrown off a cliff 3 are great, because vertical acceleration in these problems equals g in just about every case.
g : The acceleration due to gravity near the Earth”s surface; about 10 m/s2
Falling-object problems should be solved using the method we outlined above. However, you have to be really careful about choosing a positive direction and sticking to it. That is, figure out before you solve the problem whether you want “up” to be positive (in which case a equals −10 m/s2 ) or “down” to be positive (where a would therefore equal +10 m/s2 ).
Exam tip from an AP Physics veteran:
You may remember that a more precise value for g is 9.80 m/s2 . That”s correct. But estimating g as 10 m/s2 is encouraged by the AP readers to make calculation quicker.
—Jake, high school junior
Here”s a practice problem:
You are standing on a cliff, 30 m above the valley floor. You throw a watermelon vertically upward at a velocity of 3 m/s. How long does it take until the watermelon hits the valley floor?
Begin by defining the positive direction. We will call “up” positive. Then use the four-step method to solve the problem.
Step 1 : Table of variables.
Why do we always indicate what part of the motion the kinematics chart is for? Well, this problem could be solved instead by making two separate charts: one for the upward motion (where vf would be zero), and one for the downward motion (where v o would be zero). Be crystal clear how much of an object”s motion you are considering with a chart.
Remember that displacement is a vector quantity. Even though the melon goes up before coming back down, the displacement is simply equal to the height at which the melon ends its journey (0 m) minus its initial height (30 m). Another way to think about displacement: In total, the melon ended up 30 m BELOW where it started. Because down is the negative direction, the displacement is −30 m.
Step 2 : Count variables.
Three! We can solve the problem.
Step 3 : Solve.
The rest of this problem is just algebra. Yes, you have to do it right, but setting up the problem correctly and coming up with an answer that”s reasonable is more important than getting the exact right answer. Really! If this part of an AP free-response problem is worth 5 points, you might earn 4 of those points just for setting up the equation and plugging in values correctly, even if your final answer is wrong.
But which equation do you use? We have enough information to use ** (x − x 0 = v 0 t + 1 /2 at 2 ) to solve for t . Note that using ** means that we”ll have to solve a quadratic equation; you can do this with the help of the quadratic formula. 4 Or, if you have a graphing calculator, you can use it to solve. But almost always there”s a way to avoid the quadratic.
Algebra hint: You can avoid quadratics in all constant acceleration kinematics problems by solving in a roundabout way. Try solving for the velocity when the watermelon hits the ground using *** then plug into * (v f = v 0 + at ). This gives you the same answer.
What If Acceleration Isn”t Constant?
A typical Physics C kinematics question asks you to use calculus to find position, velocity, or acceleration functions. Then you can solve a motion problem even if acceleration is not constant. The way to remember what to do is, first and foremost, to understand graphical kinematics as discussed in the section below. Then, we know that the slope of a graph is related to the derivative of a function; the area under a graph is related to the integral of a function. Therefore:
- To find velocity from a position function, take the derivative with respect to time:
- To find acceleration from a velocity function, also take the time derivative:
- To find position from a velocity function, take the integral with respect to time:
- To find velocity from an acceleration function, take the time integral:
Most of the time, even on the Physics C exam, you”ll be able just to use the star equations to solve a kinematics problem. Reserve your use of calculus for those problems that explicitly include an unusual function for position, velocity, or acceleration.
Things don”t always move in a straight line. When an object moves in two dimensions, we look at vector components.
The super-duper-important general rule is this: An object”s motion in one dimension does not affect its motion in any other dimension .
The most common kind of two-dimensional motion you will encounter is projectile motion. The typical form of projectile-motion problems is the following:
“A ball is shot at a velocity v from a cannon pointed at an angle è above the horizontal …”
No matter what the problem looks like, remember these rules:
- The vertical component of velocity,vy , equals v (sin θ ).
- The horizontal component of velocity,vx , equals v (cos θ ) when θ is measured relative to the horizontal.
- Horizontal velocity is constant.
- Vertical acceleration isg , directed downward.
Here”s a problem that combines all of these rules:
A ball is shot at a velocity 25 m/s from a cannon pointed at an angle θ = 30° above the horizontal. How far does it travel before hitting the level ground?
We begin by defining “up” to be positive and writing our tables of variables, one for horizontal motion and one for vertical motion.
Note that because horizontal velocity is constant, on the horizontal table, v f = v 0 , and a = 0. Also, because the ball lands at essentially the same height it was launched from, Δx = 0 on the vertical table. You should notice, too, that we rounded values in the tables to two significant figures (for example, we said that v 0 in the vertical table equals 13 m/s, instead of 12.5 m/s). We can do this because the problem is stated using only two significant figures for all values, so rounding to two digits is acceptable, and it makes doing the math easier for us.
We know that t is the same in both tables—the ball stops moving horizontally at the same time that it stops moving vertically (when it hits the ground). We have enough information in the vertical table to solve for t by using equation **.
Using this value for t , we can solve for x − x 0 in the horizontal direction, again using **.
The cannonball traveled 57 m, about half the length of a football field.
You may have learned in your physics class that the range of a projectile (which is what we just solved for) is
If you feel up to it, you can plug into this equation and show that you get the same answer we just got. There”s no need to memorize the range equation, but it”s good to know the conceptual consequences of it: the range of a projectile on level earth depends only on the initial speed and angle, and the maximum range is when the angle is 45°.
A Final Word About Kinematics Charts
The more you practice kinematics problems using our table method, the better you”ll get at it, and the quicker you”ll be able to solve these problems. Speed is important on the AP exam, and you can only gain speed through practice. So use this method on all your homework problems, and when you feel comfortable with it, you might want to use it on quizzes and tests. The other benefit to the table method, besides speed, is consistency; it forces you to set up every kinematics problem the same way, every time. This is a time-tested strategy for success on the AP exam.
You may see some graphs that relate to kinematics on the AP test. They often look like those in Figure 11.1 . We call these graphs by the names of their axes: For example, the top graph in Figure 11.1 is a “position–time graph” and the second one is a “velocity–time graph.”
Figure 11.1 Typical motion graphs. (As an excercise, you may want to describe the motion these represent; answers are at the end of this section.)
Here are some rules to live by:
- The slope of a position–time graph at any point is the velocity of the object at that point in time.
- The slope of a velocity–time graph at any point is the acceleration of the object at that point in time.
- The area under a velocity–time graph between two times is the displacement of the object during that time interval.
It”s sometimes confusing what is meant by the area “under” a graph. In the velocity–time graphs below, the velocity takes on both positive and negative values. To find the object”s displacement, we first find the area above the t -axis; this is positive displacement. Then we subtract the area below the t -axis, which represents negative displacement. The correct area to measure is shown graphically in Figure 11.2a . Whatever you do, don”t find the area as shown in Figure 11.2b ! When we say “area,” we measure that area to the t -axis only.
Figure 11.2a : Do this.
Figure 11.2b : Don”t do this.
Problems involving graphical analysis can be tricky because they require you to think abstractly about an object”s motion. For practice, let”s consider one of the most common velocity–time graphs you”ll see:
A ball”s velocity as a function of time is graphed below. Describe with words the ball”s motion. (The positive direction is up.)
Whenever you have to describe motion in words, do so in everyday language, not physics-speak. Don”t say the word “it”; instead, give the object some specificity. Never say “positive” or “negative”; + and − merely represent directions, so name these directions. 5
In this case, let”s consider a ball going up (positive) and down (negative). Here”s how we”d answer the question:
“At first the ball is moving upward pretty fast, but the ball is slowing down while going upward. (I know this because the speed is getting closer to zero in the first part of the graph.) The ball stops for an instant (because the v–t graph crosses the horizontal axis); then the ball begins to speed up again, but this time moving downward.”
Now, no numerical values were given in the graph. But would you care to hazard a guess as to the likely slope of the graph”s line if values were given? 6
Figure 11.1 Graphs
The position–time graph has a changing slope, so the speed of the object is changing. The object starts moving one way, then stops briefly (where the graph reaches its minimum, the slope, and thus the speed is zero). The object then speeds up in the other direction. How did I know the object”s velocity changed direction? The position was at first approaching the origin, but then was getting farther away from the origin.
The second graph is a velocity–time graph, and it must be analyzed differently. The object starts from rest, but speeds up; the second part of the motion is just like the example shown before, in which the object slows to a brief stop, turns around, and speeds up.
Air Resistance and the First-Order Differential Equation
The force of air resistance is usually negligible in kinematics problems. You probably don”t believe me, though. After all, unless you”re on the moon, or unless your teacher is using a vacuum chamber for demonstrations, air is all around us. And if I dropped a piece of paper simultaneously and from the same height as a lead weight, the weight would hit the ground first—and by a huge margin. Certainly. Obviously.
Why don”t you try it? But be sure you crinkle up the paper first and drop it from about waist height. Notice how the lead weight hits the ground WAY before …
The weight and the paper hit the ground essentially at the same time. If the weight did hit first, the difference wasn”t anything you could reliably time or even be sure enough to gamble on.
Conclusion: As long as we”re not throwing objects out of our car on the freeway, air resistance is not important in kinematics.
The most common questions asked:
When is air resistance important? And how should it be dealt with?
Air resistance should only be considered when the problem explicitly says so. Usually, a problem will suggest that the force of air acts opposite to an object”s velocity, and is equal to a constant times the velocity: F air = bv . 7
- Find the terminal speed.Terminal speed means that, after a long time, the object”s speed becomes constant. To find that terminal speed, do an equilibrium problem: Free body, set up forces equal down forces, and left forces equal right forces. If something”s falling straight down with no other forces, usually you”ll get bv = mg . Then solve for v . That”s the terminal speed.
- Sketch a graph of the speed of the object as a function of time.Perhaps the problem will say the object was dropped from rest, or give an initial velocity. Well, you can plot that point at time t = 0. Then you can find the terminal velocity using the method above—the terminal speed is the constant velocity after a long time. Plot a horizontal line for the terminal velocity near the right-hand side of the t -axis.
In between the initial velocity and the terminal velocity, just know that the velocity function will look like an exponential function, changing rapidly at first, and changing less rapidly as time goes on. Sketch a curve in between the point and the line you drew. Done.
- Describe the motion in words, including what”s happening to the acceleration and/or the velocity.Be sure to use a free-body diagram, and to separate the motion if necessary into parts when the object is moving up, and moving down. When the speed is zero, the force of air resistance is zero. This doesn”t mean no acceleration, of course. When the speed is not zero, use a free body to figure out the amount and direction of the net force. Remind yourself of the basics of kinematics—the net force is the direction of acceleration. If the acceleration is in the direction of motion, the object speeds up; if the acceleration is opposite motion, the object slows down.
- Solve a differential equation to find an expression for the velocity as a function of time.Again, start with a free-body diagram, and write F net = ma . Now, though, you”ll need to do some calculus: acceleration . Perhaps your Newton”s second law equation might say something like mg – bv = ma .
Okay, solve for a and write the calculus expression for . This type of equation is called a differential equation, where a derivative of a function is proportional to the function itself. Specifically, since the first derivative is involved, this is called a “first order” differential equation.
Your calculus class may well have taught you how to solve this equation by a technique known as separation of variables: put all the v terms on one side, the t terms on the other, and integrate. If you know how to do that, great; if not, it”s complex enough not in any way to be worth learning in order to possibly—possibly —earn yourself one or two points. Everyone, though, should be able to recognize and write the answer using the knowledge that the solution to a first-order differential equation will involve an exponential function . You can use facts about the initial velocity and the terminal velocity to write this function without an algorithmic solution.
Imagine that a ball was dropped from rest in the presence of air resistance F air = bv . Writing the second law gives you the equation shown above. What”s the solution? Well, the initial velocity is zero; the terminal velocity can be found by setting acceleration to zero, meaning . The answer will always be something like v (t ) = Ae – kt or v (t ) = A (1 − e − kt ). Start by figuring out which: does the speed start large, and get small? If so, use the first expression. Or, does the speed start small, and get larger? If so, use the second expression. In this case, the speed starts from zero and ends up faster. So we use v (t ) = A (1 − e − kt ). Generally, the k term is going to be whatever”s multiplying the v in the original equation. In this case, then, k = b/m .
Now, look at the initial and final conditions to find the value of A . At time t = 0, v = 0; that works no matter the value of A , because e 0 = 1. But after a long time, we know the terminal velocity is mg /b . And in the equation, e −ktgoes to zero for large t . Meaning: after a long time, the velocity function equals A . This A must be the terminal velocity!
So our final equation looks like: .
1 . A firework is shot straight up in the air with an initial speed of 50 m/s. What is the maximum height it reaches?
(A) 12.5 m
(B) 25 m
(C) 125 m
(D) 250 m
(E) 1250 m
2 . On a strange, airless planet, a ball is thrown downward from a height of 17 m. The ball initially travels at 15 m/s. If the ball hits the ground in 1 s, what is this planet”s gravitational acceleration?
(A) 2 m/s2
(B) 4 m/s2
(C) 6 m/s2
(D) 8 m/s2
(E) 10 m/s2
3 . An object moves such that its position is given by the function x (t ) = 3t 2 − 4t + 1. The units of t are seconds, and the units of x are meters. After 6 s, how fast and in what direction is this object moving?
(A) 32 m/s in the original direction of motion
(B) 16 m/s in the original direction of motion
(C) 0 m/s
(D) 16 m/s opposite the original direction of motion
(E) 32 m/s opposite the original direction of motion
4 . An airplane attempts to drop a bomb on a target. When the bomb is released, the plane is flying upward at an angle of 30° above the horizontal at a speed of 200 m/s, as shown below. At the point of release, the plane”s altitude is 2.0 km. The bomb hits the target.
(a) Determine the magnitude and direction of the vertical component of the bomb”s velocity at the point of release.
(b) Determine the magnitude and direction of the horizontal component of the bomb”s velocity at the point when the bomb contacts the target.
(c) Determine how much time it takes for the bomb to hit the target after it is released.
(d) At the point of release, what angle below the horizontal does the pilot have to look in order to see the target?
Solutions to Practice Problems
1 . Call “up” the positive direction, and set up a chart. We know that v f = 0 because, at its maximum height, the firework stops for an instant.
Solve for Δx using equation ***: . The answer is (C) 125 m, or about skyscraper height.
2 . Call “down” positive, and set up a chart:
Plug straight into ** (Δx = v 0 t + 1 /2 at 2 ) and you have the answer. This is NOT a quadratic, because this time t is a known quantity. The answer is (B) 4 m/s2 , less than half of Earth”s gravitational field, but close to Mars”s gravitational field.
3 . First find the velocity function by taking the derivative of the position function: v (t ) = 6t − 4. Now plug in t = 6 to get the velocity after 6 s; you get 32 m/s. Note that this velocity is positive. Was the object originally moving in the positive direction? Plug in t = 0 to the velocity formula to find out … you find the initial velocity to be −4 m/s, so the object was originally moving in the negative direction, and has reversed direction after 6 s. The answer is (E).
4 . (a) Because the angle 30° is measured to the horizontal, the magnitude of the vertical component of the velocity vector is just (200 m/s) (sin 30°), which is 100 m/s. The direction is “up,” because the plane is flying up.
(b) The horizontal velocity of a projectile is constant. Thus, the horizontal velocity when the bomb hits the target is the same as the horizontal velocity at release, or (200 m/s)(cos 30°) = 170 m/s, to the right.
(c) Let”s call “up” the positive direction. We can solve this projectile motion problem by our table method.
Don”t forget to convert to meters, and be careful about directions in the vertical chart.
The horizontal chart cannot be solved for time; however, the vertical chart can. Though you could use the quadratic formula or your fancy calculator to solve x − x 0 = v 0 t + 1 /2 at 2 , it”s much easier to start with ***, v f 2 = v 0 2 + 2a (x − x 0 ), to find that vf vertically is −220 m/s (this velocity must have a negative sign because the bomb is moving down when it hits the ground). Then, plug in to *(v f = v 0 + at ) to find that the bomb took 32 s to hit the ground.
(d) Start by finding how far the bomb went horizontally. Because horizontal velocity is constant, we can say distance = velocity × time. Plugging in values from the table, distance = (170 m/s)(32 s) = 5400 m. Okay, now look at a triangle:
By geometry, tan θ = 2000 m/5400 m. The pilot has to look down at an angle of 20°.
- Average speed is total distance divided by total time. Instantaneous speed is your speed at a particular moment.
- Velocity is the vector equivalent of speed.
- Acceleration is a change in velocity divided by the time during which that change occurred.
- Displacement is the vector equivalent of distance.
- The three “star equations” are valid only when acceleration is constant.
- To solve any constant-acceleration kinematics problem, follow these four steps:
° Write out a table containing all five variables—v 0 , v f , x − x 0 , a, t —and fill in whatever values are known.
° Count variables. If you have three known values, you can solve the problem.
° Use the “star equation” that contains your known variables.
° Check for correct units.
- When an object falls (in the absence of air resistance), it experiences an acceleration ofg , about 10 m/s2 . It”s particularly important for problems that involve falling objects to define a positive direction before solving the problem.
- An object”s motion in one dimension does not affect its motion in any other dimension.
- Projectile motion problems are usually easier to solve if you break the object”s motion into “horizontal” and “vertical” vector components.
- The slope of a distance–time graph is velocity.
- The slope of a velocity–time graph is acceleration.
- The area under a velocity–time graph is displacement.
1 Use the “star equations,” which we will address in detail momentarily.
2 When can”t you use kinematics, you ask? The most common situations are when a mass is attached to a spring, when a roller coaster travels on a curvy track, or when a charge is moving in a non-uniform electric field produced by other charges. To approach these problems, use conservation of energy, as discussed in Chapter 14 .
3 The writers of the AP exam love to throw things off cliffs.
5 Why shouldn”t I say “positive” and “negative,” you ask? Well, how do these directions to the store sound: “Define north as positive. Start from zero, and go positive 20 constantly; then come back at –20, also constantly.” You”d never say that! But, this is what you”ll sound like unless you use common language.
6 −10m/s2 , if we”re on Earth.
7 Occasionally you might see a different form, F air = bv 2 . In this case, you might be asked about the terminal speed, but you will NOT have to solve a differential equation.
1 . A cannonball is shot straight up in the air and then falls back to Earth. The force of air resistance is given by the equation Fair = -bv 2 . If air resistance cannot be neglected, the acceleration of the cannonball is
(A) constant while traveling upward.
(B) increasing while traveling upward.
(C) greater while traveling upward than while traveling downward.
(D) equal to 9.8 m/s2 while traveling downward.
(E) constant while traveling downward.
2 . A shortstop throws a baseball to the first baseman. Which graphs correctly depict the horizontal velocity and horizontal acceleration of the baseball while in flight, assuming air resistance can be ignored?
3 . You drop a baseball, then after 1.0 s, you drop a second baseball. Assuming air resistance can be neglected, which of the following statements is correct?
(A) The difference in velocity between the two baseballs decreases as they fall.
(B) The difference in velocity between the two baseballs remains constant as they fall.
(C) The difference in velocity between the two baseballs increases as they fall.
(D) The distance between the baseballs decreases as they fall.
(E) The distance between the baseballs stays constant as they fall.
4 . An object begins at the origin and has a velocity as shown in the figure. When does the object return to the origin?
(A) 2 s
(B) 3 s
(C) Between 3 and 4 s
(D) 4 s
(E) The object never returns to the origin.
1 . C —On the way up, the acceleration of the cannonball is . On the way down, the acceleration is . Therefore, the acceleration on the way up is larger in magnitude because both gravity and drag are acting in the same direction. The acceleration is not constant, as the velocity changes during the cannonball”s flight. The velocity decreases on the upward journey, so the acceleration will also decrease in magnitude. Finally, the acceleration is only g when the velocity is zero at the very top of the flight.
2 . A —The horizontal velocity is constant, and there is no horizontal acceleration in the absence of air resistance.
3 . B —Both balls start from rest and pick up speed at the same constant rate of g = 9.8 m/s2 . The first ball simply starts 1.0 s ahead of the second ball and will remain a constant 9.8 m/s ahead. Since the first ball is always 9.8 m/s faster than the second, the distance between the balls will always increase as they fall.
4 . C —The displacement of the object is given by the area of the graph. The positive area from 0 to 2 s is 3 m. The negative area from 2 to 3 s is 1 m, which is not enough to return to the origin. The negative area from 2 to 4 s is 4 m, which is too large. Therefore, the object will pass through the origin between 3 and 4 s as it travels in the negative direction.