Motion in Two Dimensions Chapter 3 Motion in Two Dimensions

3.1 Position and Displacement The position of an object is described by its position vector, The displacement of the object is defined as the change in its position Fig 3.1

Average Velocity The average velocity is the ratio of the displacement to the time interval for the displacement The direction of the average velocity is the direction of the displacement vector, Fig 3.2

Average Velocity, cont The average velocity between points is independent of the path taken This is because it is dependent on the displacement, which is also independent of the path If a particle starts its motion at some point and returns to this point via any path, its average velocity is zero for this trip since its displacement is zero

Instantaneous Velocity The instantaneous velocity is the limit of the average velocity as ∆t approaches zero

Instantaneous Velocity, cont The direction of the instantaneous velocity vector at any point in a particle’s path is along a line tangent to the path at that point and in the direction of motion The magnitude of the instantaneous velocity vector is the speed

Average Acceleration The average acceleration of a particle as it moves is defined as the change in the instantaneous velocity vector divided by the time interval during which that change occurs.

Average Acceleration, cont As a particle moves, can be found in different ways The average acceleration is a vector quantity directed along Fig 3.3

Instantaneous Acceleration The instantaneous acceleration is the limit of the average acceleration as approaches zero

3.2 Producing An Acceleration Various changes in a particle’s motion may produce an acceleration The magnitude of the velocity The direction of the velocity Even if the magnitude remains constant Both change simultaneously

Kinematic Equations for Two-Dimensional Motion When the two-dimensional motion has a constant acceleration, a series of equations can be developed that describe the motion These equations will be similar to those of one-dimensional kinematics

Kinematic Equations, 2 Position vector Velocity Since acceleration is constant, we can also find an expression for the velocity as a function of time:

Kinematic Equations, 3 The velocity vector can be represented by its components is generally not along the direction of either or Fig 3.4(a)

Kinematic Equations, 4 The position vector can also be expressed as a function of time: This indicates that the position vector is the sum of three other vectors: The initial position vector The displacement resulting from

Kinematic Equations, 5 is generally not in the same direction as or as and are generally not in the same direction Fig 3.4(b)

Kinematic Equations, Components The equations for final velocity and final position are vector equations, therefore they may also be written in component form This shows that two-dimensional motion at constant acceleration is equivalent to two independent motions One motion in the x-direction and the other in the y-direction

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Solution Conceptualize Establish the mental representation and thinking about what the particle is doing. Categorize Consider that because the acceleration is only in the x direction, the moving particle can be modeled as one under constant acceleration in the x direction and one under constant velocity in the y direction.

Analyze Identify vxi = 20 m/s and ax = 4. 0 m/s2 Analyze Identify vxi = 20 m/s and ax = 4.0 m/s2. The equations of kinematics for the x direction,

Solution At t = 5.0 s, the velocity expression from part A gives

拋 射 3.3 Projectile Motion An object may move in both the x and y directions simultaneously The form of two-dimensional motion we will deal with is called projectile motion

假 設 Assumptions of Projectile Motion The free-fall acceleration is constant over the range of motion and is directed downward The effect of air friction is negligible With these assumptions, an object in projectile motion will follow a parabolic path This path is called the trajectory This is the simplification model that will be used throughout this chapter

Verifying the Parabolic Trajectory Reference frame chosen y is vertical with upward positive Acceleration components ay = -g and ax = 0 Initial velocity components vxi = vi cos and vyi = vi sin

Projectile Motion – Velocity Equations The velocity components for the projectile at any time t are: vxf = vxi = vi cosi = constant vyf = vyi – gt = vi sini – gt

Projectile Motion – Position Displacements Combining the equations gives: This is in the form of y = ax – bx2 which is the standard form of a parabola

Analyzing Projectile Motion Consider the motion as the superposition of the motions in the x- and y-directions The x-direction has constant velocity ax = 0 The y-direction is free fall ay = -g The actual position at any time is given by:

Projectile Motion Vectors Fig 3.6

Projectile Motion Diagram Fig 3.5

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Projectile Motion – Implications The y-component of the velocity is zero at the maximum height of the trajectory The accleration stays the same throughout the trajectory

Horizontal Range and Maximum Height of a Projectile When analyzing projectile motion, two characteristics are of special interest The range, R, is the horizontal distance of the projectile The maximum height the projectile reaches is h Fig 3.7

Height of a Projectile, equation The maximum height of the projectile can be found in terms of the initial velocity vector:

Range of a Projectile, equation The range of a projectile can be expressed in terms of the initial velocity vector: This is valid only for symmetric trajectory

More About the Range of a Projectile

Range of a Projectile, final The maximum range occurs at qi = 45o Complementary angles will produce the same range The maximum height will be different for the two angles The times of the flight will be different for the two angles

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Projectile Motion – Problem Solving Hints Conceptualize Establish the mental representation of the projectile moving along its trajectory Categorize Confirm that the problem involves the particle in free fall (in the y-direction) and air resistance can be neglected Select a coordinate system

Problem Solving Hints, cont. Analyze Resolve the initial velocity into x and y components Remember to treat the horizontal motion independently from the vertical motion Treat the horizontal motion using constant velocity techniques Analyze the vertical motion using constant acceleration techniques Remember that both directions share the same time Finalize Check your results

Non-Symmetric Projectile Motion Follow the general rules for projectile motion Break the y-direction into parts up and down or symmetrical back to initial height and then the rest of the height May be non-symmetric in other ways

A stone is thrown from the top of a building at an angle of 30 A stone is thrown from the top of a building at an angle of 30.0° to the horizontal and with an initial speed of 20.0 m/s, as in Figure 3.9. Fig 3.9

To find t, we use the vertical motion, in which we model the stone as a particle under constant acceleration. We use Equation 3.13 with yf = 45.0 m and vyi = 10.0 m/s (we have chosen the top of the building as the origin): t = 4.22 s

Solution The y component of the velocity just before the stone strikes the ground can be obtained using Equation 3.11, with t = 4.22 s:

An Alaskan rescue plane drops a package of emergency rations to a stranded party of explorers, as shown in the pictorial representation in Figure 3.10. If the plane is traveling horizontally at 40.0 m/s at a height of 100 m above the ground, where does the package strike the ground relative to the point at which it is released? Fig 3.10

Solution Ignore air resistance Model this problem as a particle in two-dimensional free-fall, a particle under constant velocity in the x direction and a particle under constant acceleration in the y direction Define the initial position xi = 0 right under the plane at the instant the package is released.

If we know t, the time at which the package strikes the ground, we can determine xf. To find t, we turn to the equations for the vertical motion of the package. At the instant the package hits the ground, its coordinate is 100 m. The initial component of velocity vyi of the package is zero. From Equation 3.13, xf = (40.0 m/s)(4.52 s) = 181 m

An athlete throws a javelin a distance of 80 An athlete throws a javelin a distance of 80.0 m at the Olympics held at the equator, where g = 9.78 m/s2. Four years later the Olympics are held at the North Pole, where g = 9.83 m/s2. Assuming that the thrower provides the javelin with exactly the same initial velocity as she did at the equator, how far does the javelin travel at the North Pole?

= 79.6 m

3.4 Uniform Circular Motion Uniform circular motion occurs when an object moves in a circular path with a constant speed An acceleration exists since the direction of the motion is changing This change in velocity is related to an acceleration The velocity vector is always tangent to the path of the object

Fig 3.11

Centripetal Acceleration The acceleration is always perpendicular to the path of the motion The acceleration always points toward the center of the circle of motion This acceleration is called the centripetal acceleration Centripetal means center-seeking

Changing Velocity in Uniform Circular Motion The change in the velocity vector is due to the change in direction The vector diagram shows Fig 3.11

Centripetal Acceleration, cont The magnitude of the centripetal acceleration vector is given by The direction of the centripetal acceleration vector is always changing, to stay directed toward the center of the circle of motion

Period The period, T, is the time interval required for one complete revolution The speed of the particle would be the circumference of the circle of motion divided by the period Therefore, the period is

What is the centripetal acceleration of the Earth as it moves in its orbit around the Sun? Solution We model the Earth as a particle and approximate the Earth’s orbit as circular Although we don’t know the orbital speed of the Earth, with the help of Equation 3.18 we can recast Equation 3.17 in terms of the period of the Earth’s orbit, which we know is one year:

3.5 Tangential Acceleration The magnitude of the velocity could also be changing As well as the direction In this case, there would be a tangential acceleration

Total Acceleration Fig 3.12 The tangential acceleration causes the change in the speed of the particle The radial acceleration comes from a change in the direction of the velocity vector

Total Acceleration, equations The tangential acceleration: The radial acceleration: The total acceleration: Magnitude The direction of the acceleration is the same or opposite that of the velocity

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3.6 Relative Velocity Two observers moving relative to each other generally do not agree on the outcome of an experiment For example, the observer on the side of the road observes a different speed for the red car than does the observer in the blue car Fig 3.13

Relative Velocity, generalized Reference frame S is stationary Reference frame S’ is moving Define time t = 0 as that time when the origins coincide Fig 3.14

Relative Velocity, equations The positions as seen from the two reference frames are related through the velocity The derivative of the position equation will give the velocity equation This can also be expressed in terms of the observer O’

A boat heading due north crosses a wide river with a speed of 10 A boat heading due north crosses a wide river with a speed of 10.0 km/h relative to the water. The river has a current such that the water moves with uniform speed of 5.00 km/h due east relative to the ground.

3.7 Acceleration of Autos The lateral acceleration 橫向加速度is the maximum possible centripetal acceleration the car can exhibit without rolling over in a turn The lateral acceleration depends on the height of the center of mass of the vehicle and the side-to-side distance between the wheels

Beamon於1968在Mexico創下8. 90 m的跳遠紀錄，由照片得知，他起跳時重心高度約為1. 0 m，在最高點處重心高度為1 Beamon於1968在Mexico創下8.90 m的跳遠紀錄，由照片得知，他起跳時重心高度約為1.0 m，在最高點處重心高度為1.90 m，落地時重心高度0.15 m。由這些數據，求(a)他在空中停留時間。(b)起跳時速度之水平分量及垂直分量。(c)起跳角度。

An enemy ship is on the east side of a mountain island, as shown in Figure P3.61. The enemy ship has maneuvered to within 2500 m of the 1800-m-high mountain peak and can shoot projectiles with an initial speed of 250 m/s. If the western shoreline is horizontally 300 m from the peak, what are the distances from the western shore at which a ship can be safe from the bombardment of the enemy ship?

Problem 1, 3, 10, 13, 15, 19, 20, 27, 31, 36, 38, 40, 41, 48, 51, 55, 60