## SAT Physics Kinematics in Two Dimensions - Independence Of Motion

INDEPENDENCE OF MOTION
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 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

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, Consider if someone throws the same ball horizontally near the surface of Earth. If the ball  Figure 4.2. Projectile velocity vectors 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 and y-variables, the individual x- and y-equations do share one very important variable—time, The mathemati­cally independent and y-motions take place simultaneously and share the same time,
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 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. RELATIVE VELOCITY
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 Table 4.3 Relative Velocity Equations 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.

Determinig 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.
Determining True Displacement A boat capable of moving at 6 meters per second attempts to cross a 60-meter­wide 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. 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. Now solve the second equation. The boat will drift 30 m downstream from point G.

PROJECTILE MOTION
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 and velocities from Table 4.4 Kinematic Equations with Gravity  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.

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 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. 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. 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 dining 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: • 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
• • For a projectile landing at its initial launch height, the time to maximum height is half the total time of flight. • 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.

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