3.2

Equations of Kinematics in two Dimensions

To understand how displacement, velocity, and acceleration are applied to two-dimensional motion, consider a spacecraft equipped with two engines that are mounted perpendicular to each other. These engines produce the only forces that the craft experiences, and the spacecraft is assumed to be at the coordinate origin when t0 = 0 s, so that . At a later time t, the spacecraft’s displacement is . Relative to the x and y axes, the displacement has vector components of and , respectively.
In Figure 3-3 only the engine oriented along the x direction is firing, and the vehicle accelerates along this direction. It is assumed that the velocity in the y direction is zero, and it remains zero, since the y engine is turned off. The motion of the spacecraft along the x direction is described by the five kinematic variables x, ax, vx, v0x, and t. Here the symbol “x” reminds us that we are dealing with the x components of the displacement, velocity, and acceleration vectors. (See Sections 1.7 and 1.8 for a review of vector components.) The variables x, ax, vx, and v0x are scalar components (or “components,” for short). As Section 1.7 discusses, these components are positive or negative numbers (with units), depending on whether the associated vector components point along the +x or the −x axis. If the spacecraft has a constant acceleration along the x direction, the motion is exactly like that described in Chapter 2, and the equations of kinematics can be used. For convenience, these equations are written in the left column of Table 3-1.

Figure 3-3 The spacecraft is moving with a constant acceleration ax parallel to the x axis. There is no motion in the y direction, and the y engine is turned off.

Table 3-1 Equations of Kinematics for Constant Acceleration in Two-Dimensional Motion

x Component | Variable | y Component
x | Displacement | y
ax | Acceleration | ay
vx | Final velocity | vy
v0x | Initial velocity | v0y
t | Elapsed time | t

| (3.3a)

| (3.3b)

| (3.4a)

| (3.4b)

| (3.5a)

| (3.5b)

| (3.6a)

| (3.6b)

Figure 3-4 is analogous to Figure 3-3, except that now only the y engine is firing, and the spacecraft accelerates along the y direction. Such a motion can be described in terms of the kinematic variables y, ay, vy, v0y, and t. And if the acceleration along the y direction is constant, these variables are related by the equations of kinematics, as written in the right column of Table 3-1. Like their counterparts in the x direction, the scalar components, y, ay, vy, and v0y, may be positive (+) or negative (−) numbers (with units).

Figure 3-4 The spacecraft is moving with a constant acceleration ay parallel to the y axis. There is no motion in the x direction, and the x engine is turned off.

Need more practice?

Interactive LearningWare3.1
A fox runs 85 m due south in 18 s. He starts from rest and stops for a negligible amount of time at the end of the run. Then he takes off again and runs 62 m due east in 21 s. During this second run, his acceleration is constant. For the entire 39-s interval, find the magnitude and direction of the fox’s (a) average velocity and (b) average acceleration. Express the directions relative to due south.
Related Homework: Problem 10

Problem solving insight

If both engines of the spacecraft are firing at the same time, the resulting motion takes place in part along the x axis and in part along the y axis, as Figure 3-5 illustrates. The thrust of each engine gives the vehicle a corresponding acceleration component. The x engine accelerates the ship in the x direction and causes a change in the x component of the velocity. Likewise, the y engine causes a change in the y component of the velocity. It is important to realize that the x part of the motion occurs exactly as it would if the y part did not occur at all. Similarly, the y part of the motion occurs exactly as it would if the x part of the motion did not exist. In other words, the x and y motions are independent of each other.

Figure 3-5 The two-dimensional motion of the spacecraft can be viewed as the combination of the separate x and y motions.

CONCEPTS AT A GLANCE The independence of the x and y motions lies at the heart of two-dimensional kinematics. It allows us to treat two-dimensional motion as two distinct one-dimensional motions, one for the x direction and one for the y direction. As the Concepts-at-a-Glance chart in Figure 3-6 illustrates, everything that we have learned in Chapter 2 about kinematics in one dimension will now be applied separately to each of the two directions. In so doing, we will be able to describe the x and y variables separately and then bring these descriptions together to understand the two-dimensional picture. Example 1 and Example 2 take this approach in dealing with a moving spacecraft.

Figure 3-6 CONCEPTS AT A GLANCE In two dimensions, motion along the x direction and motion along the y direction are independent of each other. As a result, each can be analyzed separately according to the procedures for one-dimensional kinematics discussed in Chapter 2. On the space shuttle Challenger, motion in perpendicular directions is controlled by thrusters. The photographs show the Challenger in orbit with different thrusters activated. (Courtesy NASA).

| Example 1 | | | The Displacement of a Spacecraft

In Figure 3-5, the directions to the right and upward are the positive directions. In the x direction, the spacecraft has an initial velocity component of and an acceleration component of . In the y direction, the analogous quantities are and . At a time of , find the x and y components of the spacecraft’s displacement.
Reasoning The motion in the x direction and the motion in the y direction can be treated separately, each as a one-dimensional motion subject to the equations of kinematics for constant acceleration (see Table 3-1). By following this procedure we will be able to determine x and y, which specify the spacecraft’s location after an elapsed time of 7.0 s.
Solution The data for the motion in the x direction are listed in the following table:

x-Direction Data
x | ax | vx | v0x | t
? | +24 m/s2 |  | +22 m/s | 7.0 s

Problem solving insight When the motion is two-dimensional, the time variable t has the same value for both the x and y directions.

The x component of the craft’s displacement can be found by using Equation 3.5a:

The data for the motion in the y direction are listed in the following table:

y-Direction Data
y | ay | vy | v0y | t
? | +12 m/s2 |  | +14 m/s | 7.0 s

The y component of the craft’s displacement can be found by using Equation 3.5b:

After 7.0 s, the spacecraft is 740 m to the right and 390 m above the origin, as in Figure 3-5.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 2 | | | The Velocity of a Spacecraft

This example also deals with the spacecraft in Figure 3-5. As in Example 1, the x components of the craft’s initial velocity and acceleration are and , respectively. The corresponding y components are and . At a time of , find the spacecraft’s final velocity (magnitude and direction).
Reasoning Figure 3-7 shows the final velocity vector, which has components vx and vy and a magnitude v. The final velocity is directed at an angle above the +x axis. The vector and its components form a right triangle, the hypotenuse being the magnitude of the velocity and the components being the other two sides. Thus, we can use the Pythagorean theorem to determine the magnitude v from values for the components vx and vy We can also use trigonometry to determine the directional angle .

Figure 3-7 The velocity vector has components vx and vy and a magnitude v. The magnitude gives the speed of the spacecraft, and the angle gives the direction of travel relative to the positive x direction.

Knowns and Unknowns The data for this problem are listed in the table that follows:

Description | Symbol | Value | Comment
x component of acceleration | ax | +24 m/s2 |
x component of initial velocity | v0x | +22 m/s |
y component of acceleration | ay | +12 m/s2 |
y component of initial velocity | v0y | +14 m/s |
Time | t | 7.0 s | Same time for x and y directions
Unknown Variables
Magnitude of final velocity | v | ? |
Direction of final velocity |  | ? |

Modeling the Problem

| Step 1 Final Velocity
| In Figure 3-7 the final velocity vector and its components vx and vy form a right triangle. Applying the Pythagorean theorem to this right triangle shows that the magnitude v of the final velocity is given in terms of the components by Equation 1a at the right. From the right triangle in Figure 3-7 it also follows that the directional angle is given by Equation 1b at the right.

| (1a)

| (1b)

| Step 2 The Components of the Final Velocity
| Values are given for the kinematic variables ax, v0x, and t in the x direction and for the corresponding variables in the y direction (see the table of knowns and unknowns). For each direction, then, these values allow us to calculate the final velocity components vx and vy by using Equation 3.3a and 3.3b from the equations of kinematics.

| (3.3a)

| (3.3b)

These expressions can be substituted into Equations 1a and 1b for the magnitude and direction of the final velocity, as shown at the right.

Solution Algebraically combining the results of each step, we find that

With the data given for the kinematic variables in the x and y directions, we find that the magnitude and direction of the final velocity of the spacecraft are

After 7.0 s, the spacecraft, at the position determined in Example 1, has a velocity of 210 m/s in a direction of 27° above the positive x axis.
Related Homework: Problem 64

Check Your Understanding 2

A power boat, starting from rest, maintains a constant acceleration. After a certain time t, its displacement and velocity are and . At time 2t, what would be its displacement and velocity, assuming the acceleration remains the same? (The answer is at the end of the book.)
(a) 2 and 2 (b) 2 and 4 (c) 4 and 2 (d) 4 and 4
Background: When an object accelerates, its displacement and velocity depend on time. If the acceleration is constant, the equations of kinematics in Table 3-1 apply.
For similar questions (including calculational counterparts), consult Self-Assessment Test 3.1. This test is described at the end of Section 3.3.
The following Reasoning Strategy gives an overview of how the equations of kinematics are applied to describe motion in two dimensions, as in Example 1 and Example 2.

| Reasoning Strategy

Applying the Equations of Kinematics in Two Dimensions
1. | Make a drawing to represent the situation being studied.
2. | Decide which directions are to be called positive (+) and negative (−) relative to a conveniently chosen coordinate origin. Do not change your decision during the course of a calculation.
3. | Remember that the time variable t has the same value for the part of the motion along the x axis and the part along the y axis.
4. | In an organized way, write down the values (with appropriate + and − signs) that are given for any of the five kinematic variables associated with the x direction and the y direction. Be on the alert for “implied data,” such as the phrase “starts from rest,” which means that the values of the initial velocity components are zero: and . The data summary boxes and tables of knowns and unknowns that are used in the examples are a good way of keeping track of this information. In addition, identify the variables that you are being asked to determine.
5. | Before attempting to solve a problem, verify that the given information contains values for at least three of the kinematic variables. Do this for the x and the y direction of the motion. Once the three known variables are identified along with the desired unknown variable, the appropriate relations from Table 3-1 can be selected.
6. | When the motion is divided into “segments,” remember that the final velocity for one segment is the initial velocity for the next segment.
7. | Keep in mind that a kinematics problem may have two possible answers. Try to visualize the different physical situations to which the answers correspond.

Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.