And yet it moves.
Galileo, the “father of modern physics,” stated a distinction between the cause of motion and the description of motion. Kinematics is that modern description of motion that answers questions such as:
How far does this object travel?
How fast and in what direction does it move?
At what rate does its speed change?
Kinematics are the mathematical tools for describing motion in terms of displacement, velocity, and acceleration.
An object’s position is its location in a certain space. Since it is difficult to describe an object’s location, in mathematics, we typically use a coordinate system to show where an object is located. Using a coordinate system with an origin also helps to determine positive and negative positions. Typically we set the object in question as the origin and relate its surroundings to the object.
Displacement is an object’s change in position. It’s the vector that points from the object’s initial position to its final position, regardless of the path actually taken. Since displacement means change in position, it is generically denoted ∆s, where ∆ denotes change in and s means spatial location. The displacement is only the distance from an object’s final position and initial position. The net distance takes into account the total path taken (meaning if an object went backward and then forward, those distances are included in the total distance). Since a distance is being measured, the SI unit for displacement is the meter [∆s] = m.
Example 1 A rock is thrown straight upward from the edge of a 30 m cliff, rising 10 m, then falling all the way down to the base of the cliff. Find the rock’s displacement.
Solution. Displacement only refers to the object’s initial position and final position, not the details of its journey. Since the rock started on the edge of the cliff and ended up on the ground 30 m below, its displacement is 30 m, downward. Its distance traveled is 50 m; 10 m on the way up and 40 m on the way down.
Example 2 An infant crawls 5 m east, then 3 m north, then 1 m east. Find the magnitude of the infant’s displacement.
Solution. Although the infant crawled a total distance of 5 + 3 + 1 = 9 m, this is not the displacement, which is merely the net distance traveled.
Using the Pythagorean Theorem, we can calculate that the magnitude of the displacement is
Example 3 In a track-and-field event, an athlete runs exactly once around an oval track, a total distance of 500 m. Find the runner’s displacement for the race.
Solution. If the runner returns to the same position from which she left, then her displacement is zero.
The total distance covered is 500 m, but the net distance—the displacement—is 0.
A Note About Notation
∆s is a more general term that works in any direction in space. The term x or “∆x = xf – xi” has a specific meaning that is defined in the x direction. However, to be consistent with AP notation and to avoid confusion between spacial location and speed, from this point on we will use x in our development of the concepts of speed, velocity, and acceleration. x is just a reference and can refer to any direction in terms of the object’s motion (for example straight down or straight up).
LOOKING AT DISTANCE VS. TIME GRAPHS
You should also be able to handle kinematics questions in which information is given graphically. One of the popular graphs in kinematics is the position-versus-time graph. For example, consider an object that’s moving along an axis in such a way that its position x as a function of time t is given by the following position-versus-time graph:
What does this graph tell us? It says that at time t = 0, the object was at position x = 0. Then, in the first second, its position changed from x = 0 to x = 10 m. Then, at time t = 1 s to 3 s, it stopped (that is, it stayed 10 m away from wherever it started). From t = 3 s to t = 6 s, it reversed direction, reaching x = 0 at time t = 5 s, and continued, reaching position x = –5 m at time t = 6 s.
Example 4 What is the total displacement?
Solution. Displacement is just the final position subtracted from the initial position. In this case, that’s –5 – 0 = –5 m (or 5 meters to the left, or negative direction).
What is the total distance traveled?
10 m + 15 m = 25 meters
Segment #1 Segment #3
SPEED AND VELOCITY
When we’re in a moving car, the speedometer tells us how fast we’re going; it gives us our speed. But what does it mean to have a speed of say, 10 m/s? It means that we’re covering a distance of 10 meters every second. By definition, average speed is the ratio of the total distance traveled to the time required to cover that distance:
The car’s speedometer doesn’t care what direction the car is moving. You could be driving north, south, east, west, whatever; the speedometer would make no distinction. 55 miles per hour, north and 55 miles per hour, east register the same on the speedometer: 55 miles per hour. Remember that speed is scalar, so it does not take into account the direction.
However, we will also need to include direction in our descriptions of motion. We just learned about displacement, which takes both distance (net distance) and direction into account. The single concept that embodies both speed and direction is called velocity, and the definition of average velocity is:
units = meter/seconds
Note that the bar over the v means average. Because ∆x is a vector, is also a vector, and because ∆t is a positive scalar, the direction of is the same as the direction of ∆x. The magnitude of the velocity vector is called the object’s speed, and is expressed in units of meters per second (m/s).
Note the distinction between speed and velocity. Velocity is speed plus direction. An important note: The magnitude of the velocity is the speed. The magnitude of the average velocity is not called the average speed (as we will see in these next two examples).
Example 5 Assume that the runner in Example 3 completes the race in 1 minute and 18 seconds. Find her average speed and the magnitude of her average velocity.
Solution. Average speed is total distance divided by elapsed time. Since the length of the track is 500 m, the runner’s average speed was (500 m)/(78 s) = 6.4 m/s. However, since her displacement was zero, her average velocity was zero also: = ∆x/∆t = (0 m)/(78 s) = 0 m/s.
Example 6 Is it possible to move with constant speed but not constant velocity? Is it possible to move with constant velocity but not constant speed?
Solution. The answer to the first question is yes. For example, if you set your car’s cruise control at 55 miles per hour but turn the steering wheel to follow a curved section of road, then the direction of your velocity changes (which means your velocity is not constant), even though your speed doesn’t change.
The answer to the second question is no. Velocity means speed and direction; if the velocity is constant, then that means both speed and direction are constant. If speed were to change, then the velocity vector’s magnitude would change.
When you step on the gas pedal in your car, the car’s speed increases; step on the brake and the car’s speed decreases. Turn the wheel, and the car’s direction of motion changes. In all of these cases, the velocity changes. To describe this change in velocity, we need a new term: acceleration. In the same way that velocity measures the rate-of-change of an object’s position, acceleration measures the rate-of-change of an object’s velocity. An object’s average acceleration is defined as follows:
units = (meters/second)2 = m/s2
Note that an object can accelerate even if its speed doesn’t change. (Again, it’s a matter of not allowing the everyday usage of the word accelerate to interfere with its technical, physics usage). This is because acceleration depends on ∆v, and the velocity vector v changes if (1) speed changes, or (2) direction changes, or (3) both speed and direction change. For instance, a car traveling around a circular racetrack is constantly accelerating even if the car’s speed is constant, because the direction of the car’s velocity vector is constantly changing.
Example 7 A car is traveling in a straight line along a highway at a constant speed of 80 miles per hour for 10 seconds. Find its acceleration.
Solution. Since the car is traveling at a constant velocity, its acceleration is zero. If there’s no change in velocity, then there’s no acceleration.
Example 8 A car is traveling along a straight highway at a speed of 20 m/s. The driver steps on the gas pedal and, 3 seconds later, the car’s speed is 32 m/s. Find its average acceleration.
Solution. Assuming that the direction of the velocity doesn’t change, it’s simply a matter of dividing the change in velocity, vf − vi, 32 m/s – 20 m/s = 12 m/s, by the time interval during which the change occurred: = ∆v/∆t = (12 m/s)/(3 s) = 4 m/s2.
Example 9 Spotting a police car ahead, the driver of the car in Example 8 slows from 32 m/s to 20 m/s in 2 seconds. Find the car’s average acceleration.
Solution. Dividing the change in velocity, 20 m/s – 32 m/s = –12 m/s, by the time interval during which the change occurred, 2 s, give us = ∆v/∆t = (–12 m/s)/(2 s) = –6 m/s2. The negative sign here means that the direction of the acceleration is opposite the direction of the velocity, which describes slowing down.
Let’s next consider an object moving along a straight axis in such a way that its velocity, v, as a function of time, t, is given by the following velocity-versus-time graph:
What does this graph tell us? It says that, at time t = 0, the object’s velocity was v = 0. Over the first two seconds, its velocity increased steadily to 10 m/s. At time t = 2 s, the velocity then began to decrease (eventually becoming v = 0, at time t = 3 s). The velocity then became negative after t = 3 s, reaching v = –5 m/s at time t = 3.5 s. From t = 3.5 s on, the velocity remained a steady –5 m/s.
Segment #1: What can we ask about this motion? First, the fact that the velocity changed from t = 0 to t = 2 s tells us that the object accelerated. The acceleration during this time was
Note, however, that the ratio that defines the acceleration, ∆v/∆t, also defines the slope of the v versus t graph. Therefore,
The slope of a velocity-versus-time graph gives the acceleration.
Segment #2: What was the acceleration from time t = 2 s to time t = 3.5 s? The slope of the line segment joining the point (t, v) = (2 s, 10 m/s) to the point (t, v) = (3.5 s, –5 m/s) is
Between time t = 2 s to time t = 3.5 s, the object experienced a negative acceleration. Between time t = 2 s to time t = 3 s, the object’s velocity was slowed down till it reached 0 m/s. Between time t = 3 s to time t = 3.5 s, the object began increasing its velocity in the opposite direction.
Segment #3: After time t = 3.5 s, the slope of the graph is zero meaning the object experienced zero acceleration. This, however, does not mean the object did not move since its velocity was constant and in the negative direction.
Let’s look at Example 9 on this page again: How far did the object travel during a particular time interval? For example, let’s figure out the displacement of the object from time t = 4 s to time t = 6 s. During this time interval, the velocity was a constant –5 m/s, so the displacement was ∆x = v∆t = (–5 m/s)(2 s) = –10 m.
Geometrically, we’ve determined the area between the graph and the horizontal axis. After all, the area of a rectangle is base × height and, for the shaded rectangle shown below, the base is ∆t, and the height is v. So, base × height equals ∆t × v, which is displacement.
We say signed area because regions below the horizontal axis are negative quantities (since the object’s velocity is negative, its displacement is negative). Thus, by counting areas above the horizontal axis as positive and areas below the horizontal axis as negative, we can make the following claim:
Given a velocity-versus-time graph, the area between the graph and the t-axis equals the object’s displacement.
What is the object’s displacement from time t = 0 to t = 3 s? Using the fact that displacement is the area bounded by the velocity graph, we figure out the area of the triangle shown below:
Since the area of a triangle is × base × height, we find that Δx = (3 s)(10 m/s) = 15 m.
Example 10 How far did the object travel from time 0 s to time t, given an intial velocity of vo and a final velocity of vf in the graph below?
Let’s consider the relationship between acceleration and velocity in a velocity versus time graph.
If we take an object’s original direction of motion to be positive, then an increase in speed corresponds to positive acceleration. This is indicative of Segment #1 and Segment #3.
A decrease in velocity corresponds to negative acceleration (as indicated in top part of Segment #5).
However, if an object’s original direction is negative, then an increase in speed corresponds to negative acceleration, indicated by the bottom part of Segment #5. Whatever is below the x-axis shows us that you are speeding up backward.
When the velocity and acceleration are in opposite directions, the object slows down, as is indicated in Segment #7. Note that Segments #2, #4, and #6 indicate no acceleration.
UNIFORMLY ACCELERATED MOTION AND THE BIG FIVE
The simplest type of motion to analyze is motion in which the acceleration is constant.
Another restriction that will make our analysis easier is to consider only motion that takes place along a straight line. In these cases, there are only two possible directions of motion. One is positive, and the opposite direction is negative. Most of the quantities we’ve been dealing with—displacement, velocity, and acceleration—are vectors, which means that they include both a magnitude and a direction. With straight-line motion, direction can be specified simply by attaching a + or – sign to the magnitude of the quantity. Therefore, although we will often abandon the use of bold letters to denote the vector quantities of displacement, velocity, and acceleration, the fact that these quantities include direction will still be indicated by a positive or negative sign.
Let’s review the quantities we’ve seen so far. The fundamental quantities are position (x), velocity (v), and acceleration (a). Acceleration is a change in velocity, from an initial velocity (vi or v0) to a final velocity (vf or simply v—with no subscript). And, finally, the motion takes place during some elapsed time interval, ∆t. Also, if we agree to start our clocks at time ti = 0, then ∆t = tf – ti = t – 0 = t, so we can just write t instead of ∆t in the first four equations. This simplification in notation makes these equations a little easier to write down. Therefore, we have five kinematics quantities: ∆x, v0, v, a, and ∆t.
These five quantities are related by a group of five equations that we call the Big Five. They work in cases where acceleration is uniform, which are the cases we’re considering.
| || ||Variable that’s missing|
|Big Five #1:||Δx = (v0 + v)t||a|
|Big Five #2:||v = v0 + at||x|
|Big Five #3:||x = x0 + v0t + at2||v|
|Big Five #4:||x = x0 + v0t − at2||v0|
|Big Five #5:||v2 = v02 + 2a(x − x0)||t|
Big Five #1 is the definition of velocity (this is the area under a velocity versus time graph, which will be covered a little further in this chapter). Big Five #2 is the definition of acceleration (this is the slope at any given moment of a velocity versus time graph).
Big Five #1 and #2 are simply the definitions of and ā written in forms that don’t involve fractions. The other Big Five equations can be derived from these two definitions and the equation = (v0 + v), using a bit of algebra.
Example 11 An object with an initial velocity of 4 m/s moves along a straight axis under constant acceleration. Three seconds later, its velocity is 14 m/s. How far did it travel during this time?
Solution. We’re given v0, ∆t, and v, and we’re asked for x. So a is missing; it isn’t given and it isn’t asked for, and we use Big Five #1:
x = t = (v0 + v)t = (4 m/s + 14 m/s)(3 s) = 27 m
Structured Example Method to Solve Problems
v(initial) = 4 m/s
v(final) = 14 m/s
t = 3 s
Unknown (What we are solving for):
Big Five #1
It’s okay to leave off the units in the middle of the calculation as long as you remember to include them in your final answer. Leaving units off your final answer will cost you points on the AP Exam.
Example 12 A car that’s initially traveling at 10 m/s accelerates uniformly for 4 seconds at a rate of 2 m/s2, in a straight line. How far does the car travel during this time?
Solution. We’re given v0, t, and a, and we’re asked for x. So, v is missing; it isn’t given, and it isn’t asked for. Therefore, we use Big Five #3:
x = x0 + v0Δt + a(Δt)2 = (10 m/s)(4s) + (2 m/s2)(4s)2 = 56 m
Example 13 A rock is dropped off a cliff that’s 80 m high. If it strikes the ground with an impact velocity of 40 m/s, what acceleration did it experience during its descent?
Solution. If something is dropped, then that means it has no initial velocity: v0 = 0. So, we’re given v0, ∆x, and v, and we’re asked for a. Since t is missing, we use Big Five #5:
Note that since a has the same sign as (x – x0), the acceleration vector points in the same direction as the displacement vector. This makes sense here, since the object moves downward, and the acceleration it experiences is due to gravity, which also points downward.
ADDITIONAL KINEMATIC GRAPHICAL ASPECTS
Not all graphs have nice straight lines as shown so far. Straight line segments represent constant slopes and therefore constant velocities, or accelerations, depending on the type of graph. What happens as an object changes its velocity in a position versus time graph? The “lines” become “curves.” Let’s look at a typical question that might be asked about such a curve.
a) What is the average velocity from 0 to 10 seconds?
b) What is the average velocity from 10 to 20 seconds?
c) What is the average velocity from 0 to 20 seconds?
Solution. This is familiar territory. To find the average velocity, use
The instantaneous velocity is the velocity at a given moment in time. When you drive in a car and look down at the speedometer, you see the magnitude of your instantaneous velocity at that time.
To find the instantaneous velocity at 10 seconds, we will need to approximate. The velocity from 9–10 seconds is close to the velocity at 10 seconds, but it is still a bit too slow. The velocity from 10–11 seconds is also close to the velocity at 10 seconds, but it is still a bit too fast. You can find the middle ground between these two ideas, or the slope of the line that connects the point before and the point after 10 seconds. This is very close to the instantaneous velocity. A true tangent line touches the curve at only one point, but this line is close enough for our purposes.
Example 15 Using the graph below, what is the instantaneous velocity at 10 seconds?
Solution. Draw a tangent line. Find the slope of the tangent by picking any two points on the tangent line. It usually helps if the points are kind of far apart and it also helps if points can be found at “easy spots” such as (0, 5) and (15, 1,000) and not (6.27, 113) and (14.7, 983)
Beyond all the math, you are at a clear advantage when you start to recognize that position-versus-time and velocity-versus-time graphs have a few basic shapes, and that all the graphs you will see will be some form of these basic shapes. Having a feel for these building blocks will go a long way toward understanding kinematics graphs in physics.
Either of the following two graphs represents something that is not moving.
• no change in position
• zero velocity
• zero acceleration
Either of the following two graphs represents an object moving at a constant velocity in the positive direction.
• positive change in position
• constant velocity
• zero acceleration
Either of the following two graphs represents an object moving at a constant velocity in the negative direction.
• negative change in position
• constant velocity
• zero acceleration
Either of the following two graphs represents an object speeding up in the positive direction.
• positive change in position
• increasing velocity
• positive acceleration
Either of the following two graphs represents an object slowing down in the positive direction.
• positive change in position
• decreasing positive velocity
• negative acceleration
Either of the following two graphs represents an object slowing down in the negative direction.
• negative change in position
• decreasing negative velocity
• positive acceleration
Either of the following two graphs represents an object speeding up in the negative direction.
• negative change in position
• increasing negative velocity
• negative acceleration
Example 16 Below is a position-versus-time graph. Describe in words the motion of the object and sketch the corresponding velocity-versus-time graph.
Solution. Part A is a constant speed moving away from the origin, part B is at rest, part C is speeding up moving away from the origin, part D is slowing down still moving away from the origin, part E is speeding up moving back toward the origin, and part F is slowing down moving back toward the origin.
The area under an acceleration-versus-time graph gives the average velocity.
Example 17 The velocity of an object as a function of time is given by the following graph:
At which point (A, B, C, D, or E) is the magnitude of the acceleration the greatest?
Solution. The acceleration is the slope of the velocity-versus-time graph. Although this graph is not composed of straight lines, the concept of slope still applies; at each point, the slope of the curve is the slope of the tangent line to the curve. The slope is essentially zero at Points A and D (where the curve is flat), small and positive at B, and small and negative at E. The slope at Point C is large and positive, so this is where the object’s acceleration is the greatest.
The simplest real-life example of motion under pretty constant acceleration is the motion of objects in Earth’s gravitational field, near the surface of the Earth and ignoring any effects due to the air (mainly air resistance). With these effects ignored, an object can fall freely. That is, it can fall experiencing only acceleration due to gravity. Near the surface of the Earth, the gravitational acceleration has a constant magnitude of about 9.8 m/s2; this quantity is denoted g (for gravitational acceleration). On the AP Physics 1 Exam, you may use g = 10 m/s2 as a simple approximation to g = 9.8 m/s2. Even though you can use the exact value on your calculator, this approximation helps to save time, so in this book, we will always use g = 10 m/s2. And, of course, the gravitational acceleration vector, g, points downward.
Since gravity points downward, for the sake of keeping things consistent throughout the rest of this chapter, we will set gravity as g = –10 m/s2.
Example 18 A rock is dropped from a cliff 80 m above the ground. How long does it take to reach the ground?
Solution. We are given vo, and asked for t. Unless you specifically see words to the contrary (such as, “you are on the moon where the acceleration due to gravity is…”), assume you are also given a = –10 m/s2. Because v is missing, and it isn’t asked for, we can use Big Five equation #3.
y = y0 + v0t + at2 ⇒ y = at2
If we set the origin y0 = 0 at the base of the cliff and v0 = 0, we get:
Note: The negative in front of the 80 is inserted because the rock fell in the down direction.
Example 19 A baseball is thrown straight upward with an initial speed of 20 m/s. How high will it go?
Solution. We are given v0, a = –10 m/s2 is implied, and we are asked for y. Now, neither t nor v is expressly given; however, we know the vertical velocity at the top is 0 (otherwise the baseball would still rise). Consequently, we use Big Five equation #5.
v2 = v02 + 2a(y − y0) ⇒ −2ay = v02
We set y0 = 0 and we know that v = 0, so that leaves us with:
Example 20 One second after being thrown straight down, an object is falling with a speed of 20 m/s. How fast will it be falling 2 seconds later?
Solution. We’re given v0, a, and t, and asked for v. Since x is missing, we use Big Five #2:
v = v0 + at = (–20 m/s) + (–10 m/s2)(2 s) = –40 m/s
The negative sign in front of the 40 simply indicates that the object is traveling in the down direction.
Example 21 If an object is thrown straight upward with an initial speed of 8 m/s and takes 3 seconds to strike the ground, from what height was the object thrown?
Solution. We’re given a, v0, and t, and we need to find y0. Because v is missing, we use Big Five #3:
In general, an object that moves near the surface of the Earth will not follow a straight-line path (for example, a baseball hit by a bat, a golf ball struck by a club, or a tennis ball hit from the baseline). If we launch an object at an angle other than straight upward and consider only the effect of acceleration due to gravity, then the object will travel along a parabolic trajectory.
To simplify the analysis of parabolic motion, we analyze the horizontal and vertical motions separately, using the Big Five. This is the key to doing projectile motion problems.
||∆x = v0xt
||y = y0 + v0yt − gt2
||vx = v0x (constant!)
||vy = v0y − gt
||ax = 0
||ay = −g = −10 m/s2
The quantity v0x, which is the horizontal (or x) component of the initial velocity, is equal to v0 cos θ0, where θ0 is the launch angle, the angle that the initial velocity vector, v0, makes with the horizontal. Similarly, the quantity v0y, the vertical (or y) component of the initial velocity, is equal to v0 sin θ0.
Example 22 An object is thrown horizontally with an initial speed of 10 m/s. It hits the ground 4 seconds later. How far did it drop in 4 seconds?
Solution. The first step is to decide whether this is a horizontal question or a vertical question, since you must consider these motions separately. The question How far did it drop? is a vertical question, so the set of equations we will consider are those listed above under vertical motion. Next, How far…? implies that we will use the first of the vertical-motion equations, the one that gives vertical displacement, ∆y.
Now, since the object is thrown horizontally, there is no vertical component to its initial velocity vector v0; that is, v0y = 0. Therefore,
The fact that ∆y is negative means that the displacement is down. Also, notice that the information given about v0x is irrelevant to the question.
Example 23 From a height of 100 m, a ball is thrown horizontally with an initial speed of 15 m/s. How far does it travel horizontally in the first 2 seconds?
Solution. The question, How far does it travel horizontally…?, immediately tells us that we should use the first of the horizontal-motion equations:
∆x = v0xt = (15 m/s)(2 s) = 30 m
The information that the initial vertical position is 100 m above the ground is irrelevant (except for the fact that it’s high enough that the ball doesn’t strike the ground before the two seconds have elapsed).
Example 24 A projectile is traveling in a parabolic path for a total of 6 seconds. How does its horizontal velocity 1 s after launch compare to its horizontal velocity 4 s after launch?
Solution. The only acceleration experienced by the projectile is due to gravity, which is purely vertical, so that there is no horizontal acceleration. If there’s no horizontal acceleration, then the horizontal velocity cannot change during flight, and the projectile’s horizontal velocity 1 s after it’s launched is the same as its horizontal velocity 3 s later.
Example 25 An object is projected upward with a 30° launch angle and an initial speed of 40 m/s. How long will it take for the object to reach the top of its trajectory? How high is this?
Solution. When the projectile reaches the top of its trajectory, its velocity vector is momentarily horizontal; that is, vy = 0. Using the vertical-motion equation for vy, we can set it equal to 0 and solve for t:
At this time, the projectile’s vertical displacement is
Example 26 An object is projected upward with a 30° launch angle from the ground and an initial speed of 60 m/s. For how many seconds will it be in the air? How far will it travel horizontally? Assume it returns to its original height.
Solution. The total time the object spends in the air is equal to twice the time required to reach the top of the trajectory (because the parabola is symmetrical). So, as we did in the previous example, we find the time required to reach the top by setting vy equal to 0, and now double that amount of time:
Therefore, the total flight time (that is, up and down) is tt = 2t = 2 × (3 s) = 6 s.
Now, using the first horizontal-motion equation, we can calculate the horizontal displacement after 6 seconds:
Δx = v0xtt = (v0 cos θ)tt = [(60 m/s)cos 30°](6 s) = 312 m
By the way, assuming it lands back at its original height, the full horizontal displacement of a projectile is called the projectile’s range.
Chapter 4 Review Questions
Answers and Explanations can be found in Chapter 13.
Section I: Multiple-Choice
1. An object that’s moving with constant speed travels once around a circular path. Which of the following statements are true concerning this motion? Select two answers.
(A) The displacement is zero.
(B) The average speed is zero.
(C) The acceleration is zero.
(D) The velocity is changing.
In section 5 of the velocity-time graph, the object is
(A) speeding up moving in the positive direction
(B) slowing down moving in the positive direction
(C) speeding up moving in the negative direction
(D) slowing down moving in the negative direction
3. Which of the following statements are true about uniformly accelerated motion? Select two answers.
(A) If an object’s acceleration is constant, then it must move in a straight line.
(B) If an object’s acceleration is zero, then it’s speed must remain constant.
(C) If an object’s speed remains constant, then its acceleration must be zero.
(D) If the object’s direction of motion is changing, then its acceleration is not zero.
4. A baseball is thrown straight upward. What is the ball’s acceleration at its highest point?
(A) g, downward
(B) g, downward
(C) g, upward
(D) g, upward
5. How long would it take a car, starting from rest and accelerating uniformly in a straight line at 5 m/s2, to cover a distance of 200 m ?
(A) 9.0 s
(B) 10.5 s
(C) 12.0 s
(D) 15.5 s
6. A rock is dropped off a cliff and strikes the ground with an impact velocity of 30 m/s. How high was the cliff?
(A) 20 m
(B) 30 m
(C) 45 m
(D) 60 m
7. A stone is thrown horizontally with an initial speed of 10 m/s from a bridge. Assuming that air resistance is negligible, how long would it take the stone to strike the water 80 m below the bridge?
(A) 1 s
(B) 2 s
(C) 4 s
(D) 8 s
8. A soccer ball, at rest on the ground, is kicked with an initial velocity of 10 m/s at a launch angle of 30°. Calculate its total flight time, assuming that air resistance is negligible.
(A) 0.5 s
(B) 1 s
(C) 2 s
(D) 4 s
9. A stone is thrown horizontally with an initial speed of 30 m/s from a bridge. Find the stone’s total speed when it enters the water 4 seconds later, assuming that air resistance is negligible.
(A) 30 m/s
(B) 40 m/s
(C) 50 m/s
(D) 60 m/s
10. Which one of the following statements is true concerning the motion of an ideal projectile launched at an angle of 45° to the horizontal?
(A) The acceleration vector points opposite to the velocity vector on the way up and in the same direction as the velocity vector on the way down.
(B) The speed at the top of the trajectory is zero.
(C) The object’s total speed remains constant during the entire flight.
(D) The vertical speed decreases on the way up and increases on the way down.
11. A stone is thrown vertically upwards with an initial speed of 5 m/s. What is the velocity of the stone 3 seconds later?
(A) 25 m/s, upward
(B) 25 m/s, downward
(C) 35 m/s, upward
(D) 35 m/s, downward
12. A car travelling at a speed of v0 applies its brakes, skidding to a stop over a distance of x m. Assuming that the deceleration due to the brakes is constant, what would be the skidding distance of the same car if it were traveling with twice the initial speed?
(A) 2x m
(B) 3x m
(C) 4x m
(D) 8x m
Section II: Free-Response
1. This question concerns the motion of a car on a straight track; the car’s velocity as a function of time is plotted below.
(a) Describe what happened to the car at time t = 1 s.
(b) How does the car’s average velocity between time t = 0 and t = 1 s compare to its average velocity between times t = 1 s and t = 5 s?
(c) What is the displacement of the car from time t = 0 to time t = 7 s?
(d) Plot the car’s acceleration during this interval as a function of time.
(e) Make a sketch of the object’s position during this interval as a function of time. Assume that the car begins at x = 0.
2. Consider a projectile moving in a parabolic trajectory under constant gravitational acceleration. Its initial velocity has magnitude v0, and its launch angle (with the horizontal) is θ0.
(a) Calculate the maximum height, H, of the projectile.
(b) Calculate the (horizontal) range, R, of the projectile.
(c) For what value of θ0 will the range be maximized?
(d) If 0 < h < H, compute the time that elapses between passing through the horizontal line of height h in both directions (ascending and descending); that is, compute the time required for the projectile to pass through the two points shown in this figure:
3. A cannonball is shot with an initial speed of 50 m/s at a launch angle of 40° toward a castle wall 220 m away. The height of the wall is 30 m. Assume that effects due to the air are negligible. (For this problem, use g = 9.8 m/s2.)
(a) Show that the cannonball will strike the castle wall.
(b) How long will it take for the cannonball to strike the wall?
(c) At what height above the base of the wall will the cannonball strike?
4. A cannonball is fired with an initial speed of 40 m/s and a launch angle of 30º from a cliff that is 25 m tall.
(a) What is the flight time of the cannonball?
(b) What is the range of the cannonball?
(c) In the two graphs below, plot the horizontal speed of the projectile versus time and the vertical speed of the projectile versus time (from the initial launch of the projectile to the instant it strikes the ground).
Graphs are very useful tools to help visualize the motion of an object. They can also help you solve problems once you learn how to translate from one graph to another. As you work with graphs, keep the following things in mind: