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4-1 Position and Displacement A position vector locates a particle in space o Extends from a reference point (origin) to the particle Example o Position vector (-3m, 2m, 5m) © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-1 Position and Displacement Change in position vector is a displacement We can rewrite this as: Or express it in terms of changes in each coordinate: © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-2 Average Velocity and Instantaneous Velocity Average velocity is o A displacement divided by its time interval We can write this in component form: © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-2 Average Velocity and Instantaneous Velocity Average velocity is o A displacement divided by its time interval Example o A particle moves through displacement (12 m)i + (3.0 m)k in 2.0 s: © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-2 Average Velocity and Instantaneous Velocity Instantaneous velocity is o The velocity of a particle at a single point in time o The limit of avg. velocity as the time interval shrinks to 0 © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-2 Average Velocity and Instantaneous Velocity Instantaneous velocity is Visualize displacement and instantaneous velocity: © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-2 Average Velocity and Instantaneous Velocity Visualize displacement and instantaneous velocity: © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-2 Average Velocity and Instantaneous Velocity In unit-vector form, we write: Which can also be written: © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-2 Average Velocity and Instantaneous Velocity Note: a velocity vector does not extend from one point to another, only shows direction and magnitude © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-2 Average Velocity and Instantaneous Velocity © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-2 Average Velocity and Instantaneous Velocity Answer: (a) Quadrant I © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-2 Average Velocity and Instantaneous Velocity Answer: (a) Quadrant I (b) Quadrant III © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-3 Average Acceleration and Instantaneous Acceleration Average acceleration is o A change in velocity divided by its time interval Instantaneous acceleration is again the limit t → 0: © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-3 Average Acceleration and Instantaneous Acceleration Instantaneous acceleration is again the limit t → 0: In unit-vector form: © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-3 Average Acceleration and Instantaneous Acceleration We can rewrite as: © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-3 Average Acceleration and Instantaneous Acceleration We can rewrite as: To get the components of acceleration, we differentiate the components of velocity with respect to time © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-3 Average Acceleration and Instantaneous Acceleration Note: as with velocity, an acceleration vector does not extend from one point to another, only shows direction and magnitude © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-3 Average Acceleration and Instantaneous Acceleration Note: as with velocity, an acceleration vector does not extend from one point to another, only shows direction and magnitude Answer: (1) x:yes, y:yes, a:yes © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-3 Average Acceleration and Instantaneous Acceleration Note: as with velocity, an acceleration vector does not extend from one point to another, only shows direction and magnitude Answer: (1) x:yes, y:yes, a:yes (2) x:no, y:yes, a:no © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-3 Average Acceleration and Instantaneous Acceleration Note: as with velocity, an acceleration vector does not extend from one point to another, only shows direction and magnitude Answer: (1) x:yes, y:yes, a:yes (3) x:yes, y:yes, a:yes (2) x:no, y:yes, a:no © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-3 Average Acceleration and Instantaneous Acceleration Note: as with velocity, an acceleration vector does not extend from one point to another, only shows direction and magnitude Answer: (1) x:yes, y:yes, a:yes (3) x:yes, y:yes, a:yes (2) x:no, y:yes, a:no (4) x:no, y:yes, a:no © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-4 Projectile Motion A projectile is o A particle moving both horizontally and vertically, in a 2-D plane, o Launched with some initial velocity, at some angle from the horizontal, o Whose acceleration vertically is always free-fall acceleration (g) And… Whose acceleration horizontally is usually considered to be ZERO © 2014 John Wiley & Sons, Inc. All rights reserved.

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Projectile Motion Examples of projectile motion. Notice the effects of air resistance

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Projectile Motion A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.

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4-4 Projectile Motion Launched with an initial velocity v 0 © 2014 John Wiley & Sons, Inc. All rights reserved.

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Understood by analyzing the horizontal and vertical motions separately. Projectile Motion

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4-4 Projectile Motion Decompose two-dimensional motion into TWO linked one-dimensional problems © 2014 John Wiley & Sons, Inc. All rights reserved.

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Projectile Motion Photo shows 2 balls starting to fall at the same time. Yellow ball on right has an initial speed in the x-direction. Red ball on left has v x0 = 0 Vertical positions of the balls are identical at identical times Horizontal position of yellow ball increases linearly in time.

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Projectile Motion The speed in the x-direction is constant; in the y-direction the object moves with constant acceleration g.

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The x and y motion are separable We can analyze projectile motion as horizontal motion with constant velocity and vertical motion with constant acceleration: a x = 0 and a y = g.

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4-4 Projectile Motion © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-4 Projectile Motion Answer: Yes. The y-velocity is negative, so the ball is now falling. © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-4 Projectile Motion Horizontal motion: o No acceleration, so velocity is constant: Vertical motion: o Assume +y (up) is positive. Acceleration will always be -g © 2014 John Wiley & Sons, Inc. All rights reserved.

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Solving Problems Involving Projectile Motion Projectile motion is motion with constant acceleration in two dimensions, Usually where vertical acceleration is g and coordinate system assumes UP = + motion!

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Solving Problems Involving Projectile Motion 1. Read the problem carefully, and choose the object(s) you are going to analyze. 2. Draw a diagram. 3. Choose an origin and a coordinate system. 4. Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g. 5. Examine the x and y motions separately.

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Solving Problems Involving Projectile Motion 6. List known and unknown quantities. Remember that v x never changes, and that v y = 0 at the highest point. 7. Plan how you will proceed. Use the appropriate equations; you may have to combine some of them, or find time from one set and use in the other

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Solving Problems Involving Projectile Motion Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.

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Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. We WANT: v x (m/s) Solving Problems Involving Projectile Motion

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Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. v y0 = 0 m/s! We KNOW:

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Solving Problems Involving Projectile Motion Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. y = -50 m! We KNOW:

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Solving Problems Involving Projectile Motion Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. x = +90 m! We KNOW:

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Solving Problems Involving Projectile Motion Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. a x = 0! a y = -9.8! We KNOW:

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Solving Problems Involving Projectile Motion Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. For Horizontal: x = +90 m = v x t v x = 90/t How to get t ??

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Solving Problems Involving Projectile Motion Example Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance. Get t from Vertical: y = -50 m! y = v 0y t – ½ gt = - ½ gt2

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Projectile Motion If object is launched at initial angle θ 0 with the horizontal, analysis is similar except that the initial velocity has a vertical component.

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Solving Problems Involving Projectile Motion Example: A kicked football. A football is kicked at an angle θ 0 = 37.0° with a velocity of 20.0 m/s, as shown. It travels through the air (assume no resistance) and lands at the ground.

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Example: A kicked football. Calculate a)the maximum height, b)the time of travel before the football hits the ground, c)how far away it hits the ground. d)the velocity vector at the maximum height, e)the acceleration vector at maximum height.

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Solving Problems Involving Projectile Motion STEP 1: Coordinates! (and signs!) +x horizontally to the right +y vertically upwards acceleration of gravity = -9.8 m/s/s

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Solving Problems Involving Projectile Motion STEP 2: COMPONENTS! (and signs!) +v x0 horizontally to the right = v 0 cos +v y0 vertically upwards = v 0 sin a y acceleration of gravity = -9.8 m/s/s only in y

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Solving Problems Involving Projectile Motion STEP 3: LOOK FOR 3 THINGS! v f = v i + at x = ½ (v i + v f )*t x = v i *t + ½ at 2 x = v f *t – ½ at 2 V f 2 = V i 2 +2a x

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Solving Problems Involving Projectile Motion Deal with x and y SEPARATELY in +x direction, a x = 0! v fx = v ix x = v ix *t = v cos * t

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Solving Problems Involving Projectile Motion In y direction, a y = -9.8! v fy = v iy – 9.8t= v 0 sin - gt y = ½ (v iy + v fy )*t y = v iy *t – 4.9t 2= ( v 0 sin t - ½ gt 2 x = v fy *t + 4.9t 2 v fy 2 = v iy 2 – 19.6 y= ( v 0 sin 2 - 2g y

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Solving Problems Involving Projectile Motion

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Solve it! v x = 16 m/s v y = 12 m/s Max height = 7.3 m Max distance in x (“range”) = 39 m Time = 2.5 seconds

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Example: Level horizontal range. The horizontal range is defined as the horizontal distance the projectile travels before returning to its original height (which is typically the ground); that is, y (final) = y 0. Derive a formula for the horizontal range R of a projectile in terms of its initial speed v 0 and angle θ 0.

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4-4 Projectile Motion The horizontal range is: o The distance the projectile travels in x by the time it returns to its initial height © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-4 Projectile Motion The projectile's trajectory is o Its path through space (traces a parabola) The horizontal range is: o The distance the projectile travels in x by the time it returns to its initial height © 2014 John Wiley & Sons, Inc. All rights reserved.

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Solving Problems Involving Projectile Motion Example: Level horizontal range. (b) Suppose one of Napoleon’s cannons had a muzzle speed, v 0, of 60.0 m/s. At what angle should it have been aimed (ignore air resistance) to strike a target 320 m away?

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Rescue helicopter drops supplies. A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helicopter is traveling horizontally with a speed of 70 m/s (250 km/h), (a) how far in advance of the recipients (horizontal distance) must the package be dropped?

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Rescue helicopter drops supplies. (b) Suppose, instead, that the helicopter releases the package a horizontal distance of 400 m in advance of the mountain climbers. What vertical velocity should the package be given (up or down) so that it arrives precisely at the climbers’ position? (c) With what speed does the package land in the latter case?

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4-4 Projectile Motion Assume air resistance is negligible In many situations this is a poor assumption: © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-4 Projectile Motion © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-4 Projectile Motion Answer: (a) is unchanged © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-4 Projectile Motion Answer: (a) is unchanged (b) decreases (becomes negative) © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-4 Projectile Motion Answer: (a) is unchanged (b) decreases (becomes negative) (c) 0 at all times © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-4 Projectile Motion Answer: (a) is unchanged (b) decreases (becomes negative) (c) 0 at all times (d) -g (-9.8 m/s 2 ) at all times © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-5 Uniform Circular Motion A particle is in uniform circular motion if o It travels around a circle or circular arc o At a constant speed Since the velocity changes, the particle is accelerating! Velocity and acceleration have: o Constant magnitude o Changing direction Figure 4-16 © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-5 Uniform Circular Motion Acceleration is called centripetal acceleration o Means “center seeking” o Directed radially inward The period of revolution is: o The time it takes for the particle go around the closed path exactly once Eq. (4-34) Eq. (4-35) © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-5 Uniform Circular Motion © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-5 Uniform Circular Motion Answer: (a) -(4 m/s)i (b) -(8 m/s 2 )j © 2014 John Wiley & Sons, Inc. All rights reserved.

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Instantaneous velocity – heading towards circular motion! Instantaneous velocity is instantaneous rate of change of position vector with respect to time. Components of instantaneous velocity are v x = dx/dt, v y = dy/dt, and v z = dz/dt. Instantaneous velocity of a particle is always tangent to its path.

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Average acceleration The average acceleration during a time interval t is defined as the velocity change during t divided by t.

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Average acceleration The average acceleration during a time interval t is defined as the velocity change during t divided by t.

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Instantaneous acceleration Instantaneous acceleration is the instantaneous rate of change of the velocity with respect to time Any particle following a curved path is accelerating, even if it has constant speed.

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Direction of the acceleration vector Direction of acceleration vector depends on whether speed is constant, increasing, or decreasing,

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Direction of the acceleration vector Direction of acceleration vector depends on whether speed is constant, increasing, or decreasing,

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Uniform circular motion For uniform circular motion, the speed is constant and the acceleration is perpendicular to the velocity.

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Acceleration for uniform circular motion For uniform circular motion, the instantaneous acceleration always points toward the center of the circle and is called the centripetal acceleration. The magnitude of the acceleration is a rad = v 2 /R.

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Acceleration for uniform circular motion Centripetal acceleration Magnitude a rad = v 2 /R Period T is time for one revolution, and a rad = 4π 2 R/T 2.

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Nonuniform circular motion If the speed varies, the motion is nonuniform circular motion. Radial acceleration component is a rad = v 2 /R, There is also a tangential acceleration component a tan that is parallel to the instantaneous velocity.

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Acceleration of a skier Skier moving on a ski- jump ramp. Figure shows the direction of the skier’s acceleration at various points.

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Relative velocity The velocity of a moving body seen by a particular observer is called the velocity relative to that observer, or simply the relative velocity. A frame of reference is a coordinate system plus a time scale.

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Relative velocity in one dimension If point P is moving relative to reference frame A, we denote the velocity of P relative to frame A as v P/A. If P is moving relative to frame B and frame B is moving relative to frame A, then the x- velocity of P relative to frame A is v P/A-x = v P/B-x + v B/A-x.

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Relative velocity on a straight road Motion along a straight road is a case of one-dimensional motion.

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Relative velocity in two or three dimensions We extend relative velocity to two or three dimensions by using vector addition to combine velocities. A passenger’s motion is viewed in the frame of the train and the cyclist.

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Flying in a crosswind A crosswind affects the motion of an airplane.

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4-6 Relative Motion in One Dimension Measures of position and velocity depend on the reference frame of the measurer o How is the observer moving? o Our usual reference frame is that of the ground Read subscripts “PA”, “PB”, and “BA” as “P as measured by A”, “P as measured by B”, and “B as measured by A" Frames A and B are each watching the movement of object P © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-6 Relative Motion in One Dimension Positions in different frames are related by: Taking the derivative, we see velocities are related by: But accelerations (for non-accelerating reference frames, a BA = 0) are related by © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-6 Relative Motion in One Dimension Example Frame A: x = 2 m, v = 4 m/s Frame B: x = 3 m, v = -2 m/s P as measured by A: x PA = 5 m, v PA = 2 m/s, a = 1 m/s 2 So P as measured by B: o x PB = x PA + x AB = 5 m + (2m – 3m) = 4 m o v PB = v PA + v AB = 2 m/s + (4 m/s – -2m/s) = 8 m/s o a = 1 m/s 2 © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-7 Relative Motion in Two Dimensions The same as in one dimension, but now with vectors: Positions in different frames are related by: Velocities: Accelerations (for non-accelerating reference frames): Again, observers in different frames will see the same acceleration © 2014 John Wiley & Sons, Inc. All rights reserved.

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4-7 Relative Motion in Two Dimensions Figure 4-19 Frames A and B are both observing the motion of P © 2014 John Wiley & Sons, Inc. All rights reserved.

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Position Vector Locates a particle in 3-space Displacement Change in position vector Average and Instantaneous Accel. Average and Instantaneous Velocity 4 Summary © 2014 John Wiley & Sons, Inc. All rights reserved.

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Projectile Motion Flight of particle subject only to free-fall acceleration (g) Trajectory is parabolic path Horizontal range: Uniform Circular Motion Magnitude of acceleration: Time to complete a circle: Relative Motion For non-accelerating reference frames 4 Summary © 2014 John Wiley & Sons, Inc. All rights reserved.

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