INDEPENDENCE OF MOTION
TRUE VELOCITY AND DISPLACEMENT
RELATIVE VELOCITY
PROJECTILE MOTION
The previous chapter introduced three kinematic equations for objects moving along one-dimensional lines. In one-dimensional kinematics, objects typically move along either the x-axis or the y-axis. This chapter will consider objects that move along the x-axis and the y-axis simultaneously: two-dimensional motion. The kinematic equations must now be used separately. You must analyze the x-motion and y-motions independently. This chapter focuses on the following objectives:
■Define and discuss the independence of motion.
■Determine the true velocity and displacement of objects.
■Solve projectile motion problems.
Table 4.1 lists the variables that will be used in this chapter and their units.
Table 4.1 Variables used in Two-Dimensional Kinematics
New Variables |
Units |
|
m/s (meters per second) |
|
m/s (meters per second) |
|
m (meters) |
vix = initial velocity in the x-direction |
m/s (meters per second) |
viy = initial velocity in the y-direction |
m/s (meters per second) |
vfx = final velocity in the x-direction |
m/s (meters per second) |
vfy = final velocity in the y-direction |
m/s (meters per second) |
Chapter 3 demonstrated how kinematic equations are used determine the position, velocity, and acceleration of an object moving along a one-dimensional line. Consider an example of an astronaut in space throwing a ball horizontally with an initial velocity of 
i. If the astronaut is very far from Earth, where gravity is negligibly small, the ball will continue to move in a straight line. The diagram below indicates the instantaneous velocity vectors on the ball at four different locations during its motion.
Figure 4.1. Velocity vectors
IF YOU SEE two-dimensional motion
Solve the kinematic equations for x and y independently.
The motion in the x- and y-directions may be completely different.
Notice that all the velocity vectors have equal magnitudes and directions. If no external forces are acting on the ball after its release, the ball will continue moving with its initial velocity, i, indefinitely.
Consider if someone throws the same ball horizontally near the surface of Earth. If the ball is given the same initial velocity, i, its path will resemble a parabola.
Figure 4.2. Projectile velocity vectors
The ball will continue to move in the x-direction at velocity i. However, it will also experience a vertical acceleration due to gravity. As a result, the vertical velocity, vy, will increase uniformly. The resulting path of the ball is a parabola.
Kinematic equations can be used for motion only along a straight line. Therefore, separate kinematic equations must be employed for x-variables and for y-variables. The resulting motion is described by two kinematic equations in combination. Adding x- and y-subscripts to the kinematic variables allows you to distinguish between similar variables acting in different directions. Table 4.2 compares the kinematic equations in one and in two dimensions.
Table 4.2 Kinematic Equations in One and Two Dimensions
Although the kinematic equations can never contain a mixture of x- and y-variables, the individual x- and y-equations do share one very important variable—time, t. The mathematically independent x- and y-motions take place simultaneously and share the same time, t.
IF YOU SEE two-dimensional motion
Time is key.
Time is the common variable shared by the x- and y-equations.
TRUE VELOCITY AND DISPLACEMENT
The kinematic equations solve for x- and y-direction velocities and displacements. However, in two-dimensional motion problems, the path followed by the object does not lie solely along either the x- or y-axis. The ball in Figure 4.3 follows a parabolic path.
Figure 4.3. Projectile motion velocity vectors and their components
At any instant during this motion, there is the true velocity, , tangent to the motion of the object. The true velocity is found using vector mathematics as described in Chapter 3. The x- and y-velocities, calculated by the separate one-dimensional kinematic equations, are the components of the true velocity. You can find the true velocity of the object by using the Pythagorean theorem.
The true displacement of an object can be found in a similar manner. The kinematic equations solve separately for the x- and y-displacements. These displacements are also the vector components of the true displacement of an object. Use the Pythagorean theorem to find the true displacement.
The motion of an object, as described by two observers, may differ depending on the location of the observers. For example, a car reported as moving at 30 meters per second by a stationary observer will appear to be moving at 10 meters per second as observed by a driver in a car traveling alongside at 20 meters per second. Problems involving multiple velocities are known as relative velocity problems. Common two-dimensional relative velocity problems involve a boat moving across a river or an airplane flying through the air. In these problems, the velocities in both the x- and y-directions are constant. Therefore, the acceleration in the x-direction, ax, and the acceleration in the y-direction, ay, are both equal to zero, as shown in Table 4.3. This greatly simplifies the kinematic equations.
Table 4.3 Relative Velocity Equations
Kinematic Equation Used for Relative Velocity |
Modified for x-direction when ax = 0 |
Modified for y-direction when ay = 0 |
x = vit + |
x = vixt |
y = viyt |
The velocities of river currents and the wind change the true velocity of boats and airplanes. Relative velocity problems require you to understand both vector addition and independence of motion.
EXAMPLE 4.1
Determining True Velocity
An airplane is heading due north at 400 kilometers per hour when it encounters a wind from the west moving at 300 kilometers per hour, as shown in the following diagram.
(A) Determine the magnitude of the true velocity of the plane with respect to an observer on the ground.
WHAT’S THE TRICK?
An observer on the ground will see the true velocity of the airplane. Mathematically, this is the resultant vector created by adding the airplane and wind velocity vectors tip to tail, as shown below, and using the Pythagorean theorem.
These vectors form a 3-4-5 right triangle. The airplane has a speed of 500 kilometers per hour relative to the ground.
(B) Which vector, of the choices given below, describes the direction the pilot must aim the plane in order for the plane to have a true velocity that points directly north?
WHAT’S THE TRICK?
The velocity vectors for the plane and the wind must add together to create a true velocity that points north. Since the wind is blowing out of the west, the plane must have a component that moves toward the west to cancel the effect of the wind.
Answer (i) is the correct heading for the plane. When the plane is aimed northwest, it will actually move with a true velocity directly north.
EXAMPLE 4.2
Determining True Displacement
A boat capable of moving at 6 meters per second attempts to cross a 60-meter-wide river as shown in the diagram above. The river flows downstream at 3 meters per second. The boat begins at point P and aims for point Q, a point directly across the river. How far downstream from point Q will the boat drift?
WHAT’S THE TRICK?
The boat and the river both move at constant velocity. The motion of the boat and the river are mathematically independent but take place simultaneously. You should include subscripts to distinguish between the two velocities. The velocity of the boat, vb, produces a cross-stream displacement, x. The velocity of the river, vr, produces a downstream displacement, y.
IF YOU SEE relative velocity problems
x = vt
Solve this equation in both the x- and y-directions.
The variable shared by both equations is time. Simply substitute the known values into both equations. You can then solve one equation, followed by the next equation.
Solve the first equation for time.
t = 10 s
Now solve the second equation.
y = (3 m/s)(10 s) = 30 m
The boat will drift 30 m downstream from point Q.
Projectile motion describes an object that is thrown, or shot, in the presence of a gravity field. An object is considered a projectile only when it is no longer in contact with the person, or device, that has thrown it and before it has come into contact with any surfaces. As a result, the downward acceleration of gravity is the only acceleration acting on a projectile during its flight.
One of the most important aspects of projectile motion is the type of motion experienced in each direction. If the vertical acceleration of gravity is the only acceleration acting on a projectile, the horizontal speed of a projectile cannot change. Therefore, the horizontal component of velocity must always remain constant. Both the vertical velocity and the vertical displacement will be affected by the acceleration of gravity. When calculating the vertical portion of projectile motion, you must use the complete kinematic equations, as shown in Table 4.4. You should include additional subscripts to distinguish the x- and y-velocities from each other and from the true velocity, .
Table 4.4 Kinematic Equations with Gravity
The first step in projectile motion problems involves determining the initial velocity in the x-direction, vix, and the initial velocity in the y-direction, viy. These velocities are the components of the initial launch velocity, i. Table 4.5 shows the velocities of two typical launches.
IF YOU SEE projectile motion
Horizontal velocity is constant.
This is true for all projectiles regardless of launch angle.
Table 4.5 Velocity of Two Typical Launches
The next step in projectile motion usually involves determining time, t. As with all two-dimensional motion problems, time is the only variable common to both the x- and y-directions. You should note that the time of flight, t, depends on y-direction variables, not on x-direction variables. Therefore, when the time of flight is not given, most problems begin by solving one of the three y-direction kinematic equations.
IF YOU SEE projectile motion
Acceleration of gravity acts in the y-direction.
The y-velocity decreases by 10 m/s2 every second.
Horizontally Launched Projectiles
Horizontally launched projectiles are the most common projectile motion problems encountered on introductory physics exams. As with any kinematics problem, identifying variables (especially hidden variables) is extremely important. In horizontal launches, the initial velocity in the y-direction is zero as shown in Table 4.6. This simplifies the y-direction equations.
Table 4.6 Kinematic Equations for Horizontal Launches
IF YOU SEE projectile motion
Time is controlled by y-variables.
Time is usually solved using y-component equations.
EXAMPLE 4.3
Horizontally Launched Projectiles
A ball is thrown horizontally at 15 meters per second from the top of a 5-meter-tall platform, as shown in the diagram above. Determine the horizontal distance traveled by the ball.
WHAT’S THE TRICK?
First determine the x- and y-components of the initial velocity. This is easy for horizontal launches. All of the initial velocity is directed horizontally, and none is directed vertically.
The velocity in the y-direction is a hidden zero, which simplifies the y-equations.
If time is unknown, solve for time using y-direction equations. The vertical displacement of 5 meters is given. Use an equation containing both displacement and time.
Finally, time is the one variable common to motion in any direction. Now use time in the horizontal equation to determine the horizontal distance.
x = vix t
x = (15 m/s)(1 s) = 15 m
Projectiles Launched at an Angle
Projectiles launched at angles are more difficult to solve mathematically. As a result, complex calculations for these projectiles may not appear on the SAT Subject Test in Physics. However, projectiles launched at upward angles do have several unique characteristics that will be tested conceptually.
Figure 4.4 depicts the flight path of a projectile launched with an initial speed of i = 50 m/s at an upward launch angle. The projectile lands at the same height from which it was launched, y = 0, and has a final speed vf = 50 m/s. The projectile is shown every second during its flight. In each of these positions, the instantaneous horizontal component of velocity, vx, and the instantaneous vertical component of velocity, vy, are shown.
Figure 4.4. Projectile motion vectors
Examination of the diagram reveals four key facts about projectiles launched at angles.
1. The horizontal component of velocity, vx, remains constant.
2. When the projectile is moving upward, its vertical speed decreases by 10 meters per second every second until the projectile reaches an instantaneous vertical speed of zero at maximum height. The decreasing velocity then results in a changing downward speed that increases by 10 meters per second every second.
3. The projectile passes through each height twice, once on the way up and once on the way down (except the single point at maximum height). At points with equal height, the magnitude of the vertical velocity on the way up equals the magnitude of the vertical velocity on the way down.
4. The time the projectile rises equals the time the projectile falls, as long as the final height equals the initial height.
Retaining a mental image of the above diagram in your memory will help you answer conceptual problems for upwardly launched projectiles. The most common questions tend to focus on two key locations during the flight: the very top of the flight path (maximum height) and the landing point, as shown in Figure 4.5.
Figure 4.5. Two key instants during a projectile flight
Remember the following facts.
■The vertical component of velocity at maximum height is zero. Therefore, the true velocity at maximum height equals the horizontal component of the launch velocity: max height = vix.
■Although the vertical component of velocity at maximum height becomes zero, the acceleration in the vertical direction remains a constant −10 meters per second squared.
■The maximum height, y, can be solved using 
. At maximum height vfy = 0.
■When a projectile lands at the same height as its launch point, the vertical displacement is zero: y = 0. As a result, the final speed of the projectile equals the launch speed: vf = vi.
■For a projectile landing at its initial launch height, the time to maximum height is half the total time of flight.
■To reach maximum range, xmax, a projectile must have a launch angle of 45 degrees.
■Any two launch angles totaling 90 degrees will have the same range. For example, if a projectile is launched at 30 degrees, it will reach the same impact point if it is launched at 60 degrees.
1. INDEPENDENCE OF MOTION. The horizontal and vertical forces and motions on an object are independent of each other. Although one dimension does not affect the other, when taken together, they describe the overall two-dimensional motion of the object. Both motions occur simultaneously during the same time, t.
2. TRUE VELOCITY, 
. This is also referred to as the net velocity or resultant velocity. The true velocity is the velocity of the object as seen by a stationary observer. It is found by adding all the velocity vectors acting on an object by using vector addition.
3. PROJECTILE MOTION. In the horizontal direction, a projectile will always move at constant velocity. In the vertical direction, a projectile will accelerate at –10 meters per second squared due to gravity. The resulting motion of a projectile is a parabola. Kinematic equations can be used to describe the motion of the projectile as long as x- and y-variables are not used together in the same kinematic equation. One set of equations is used to calculate x-variables, and another separate set is used to calculate y-variables. However, time—t, is the one variable common to both x- and y-equations.
4. RESOLVING VELOCITY VECTORS. The initial velocity of a projectile thrown at an angle above the horizontal is the hypotenuse of a right triangle. The x- and y-component vectors are the two legs of the right triangle. The x-component velocity vector remains constant throughout the flight of the projectile. The y-component continually changes due to the acceleration of gravity.
If You See |
Try |
Keep in Mind |
Any type of two-dimensional motion |
Solve the x- and y-kinematic equations separately. |
Time is the one variable shared by x- and y-variables. |
Relative velocity |
Use the constant velocity formula in both the x- and y-directions. x = vxt y = vyt |
Solve one equation first, followed by the other. |
Projectile motion involving a horizontal launch |
Solve for time using a simplified y-direction equation. y = Then use time in the x-direction equation to solve for range, x. x = vixt |
viy = 0 is the hidden zero in horizontal launches. The y-variables control time. The motion in the x-direction is constant velocity. |
A projectile launched at an angle greater than horizontal |
Resolve the launch velocity into its components, vix and viy. There is no vertical velocity at the top of the flight path, vfy = 0. Therefore, the true velocity at the top of the flight path is equal to the initial x-velocity.
When solving for maximum altitude, try vfy2 = viy2 + 2gy where vfy = 0. If a projectile returns to its original launch height, its speed, vf, equals the launch speed, vi. vf = vi |
The velocity in the x-direction, vx, remains constant. The velocity in the y-direction, vy, decreases by 10 m/s every second on the way up, is zero at the top of the flight, and increases by 10 m/s every second on the way down. |
1.In the diagram above, six cars are shown driving on a four-lane highway. Two observers are riding in car X. When compared with the observers, which car appears to have a velocity of 10 meters per second in the opposite direction?
(A)A
(B)B
(C)C
(D)D
(E)E
Questions 2−4
A boat with a speed of 4 meters per second will cross a 40-meter-wide river that has a current of 3 meters per second. The boat aims directly across the river, as shown in the diagram below.
2.How long will the boat take to cross the river?
(A)10 s
(B)20 s
(C)30 s
(D)40 s
(E)50 s
3.How far downstream will the boat drift as a result of the river current?
(A)0 m
(B)15 m
(C)20 m
(D)30 m
(E)45 m
4.You want to cross the 40-meter-wide river, as shown, in the least amount of time. The location on the other side of the river is not important, only minimizing the time to cross the river matters. You should swim
(A)completely upstream
(B)slightly upstream
(C)directly across the stream
(D)slightly downstream
(E)completely downstream
5.Which variable(s) remain(s) constant during the entire flight of a projectile?
(A)Horizontal component of velocity only
(B)Vertical component of velocity only
(C)Acceleration only
(D)Both A and C
(E)Both B and C
6.A ball thrown horizontally from the edge of a 20-meter-tall structure lands 30 meters from the base of the structure as shown above. Determine the speed at which the ball was thrown.
(A)5 m/s
(B)10 m/s
(C)15 m/s
(D)20 m/s
(E)25 m/s
7.A convertible car with an open top is driving at constant velocity when a ball is thrown straight upward by one of the passengers. If there is no air resistance, where will the ball land?
(A)In front of the car
(B)In the car
(C)Behind the car
(D)The answer depends on the speed of the projectile.
(E)The answer depends on the speed of the car.
Questions 8−10
The diagram below depicts a projectile launched from point A with a speed v at angle of θ, above the horizontal. The projectile reaches its maximum height, h, at point B. The projectile impacts the ground at point C, achieving a final range of x. Points A and C are at the same height. The total time of flight from point A to point C is t seconds.
8.Determine the speed of the projectile at point B.
(A)zero
(B)v
(C)
v
(D)v cos θ
(E)v sin θ
9.Determine the speed of the projectile at point C.
(A)zero
(B)v
(C)v
(D)v cos θ
(E)v sin θ
10.Which of the statements below is true regarding the acceleration acting on the projectile during its flight from point A to point C?
(A)The acceleration decreases from A to B and increases from B to C.
(B)The acceleration increases from A to B and decreases from B to C.
(C)At B, the magnitude of the acceleration equals the x-component of the magnitude at A.
(D)At C, the magnitude of the acceleration is the same as at A but the direction is opposite.
(E)The magnitude and direction of the acceleration remain constant from A to C.
