Kinematics in Two Dimensions Chapter 3 Kinematics in Two Dimensions Vectors AP Physics Motion in two dimensions Giancoli, PHYSICS,6/E © 2004. Electronically reproduced by permission of Pearson Education, Inc., Upper Saddle River, New Jersey 1

Module 5 Vectors Giancoli, Sec 3-1, 2, 3, 4 The following is an excellent lecture on this material. Giancoli, PHYSICS,6/E © 2004. Electronically reproduced by permission of Pearson Education, Inc., Upper Saddle River, New Jersey 2

Vectors A vector has magnitude as well as direction. Examples: displacement, velocity, acceleration, force, momentum A scalar has only magnitude Examples: time, mass, temperature Module 5 - 1

Vector Addition – One Dimension A person walks 8 km East and then 6 km East. Displacement = 14 km East A person walks 8 km East and then 6 km West. Displacement = 2 km Module 5 - 2

Vector Addition Example 1: A person walks 10 km East and 5.0 km North Order doesn’t matter Module 5 - 3

Graphical Method of Vector Addition Tail to Tip Method Module 5 - 4

Graphical Method of Vector Addition Tail to Tip Method Module 5 - 5

Parallelogram Method Module 5 - 6

Subtraction of Vectors Negative of vector has same magnitude but points in the opposite direction. For subtraction, we add the negative vector. Module 5 - 7

Multiplication by a Scalar A vector V can be multiplied by a scalar c; the result is a vector cV that has the same direction but a magnitude cV. If c is negative, the resultant vector points in the opposite direction. Module 5 - 8

Adding Vectors by Components Any vector can be expressed as the sum of two other vectors, which are called its components. Usually the other vectors are chosen so that they are perpendicular to each other. Module 5 - 9

Trigonometry Review Hypotenuse Opposite Adjacent Module 5 - 10

Adding Vectors by Components If the components are perpendicular, they can be found using trigonometric functions. Module 5 - 11

Adding Vectors by Components The components are effectively one-dimensional, so they can be added arithmetically: Module 5 - 12

Signs of Components Module 5 - 13

3-4 Adding Vectors by Components Draw a diagram; add the vectors graphically. Choose x and y axes. Resolve each vector into x and y components. Calculate each component using sines and cosines. Add the components in each direction. To find the length and direction of the vector, use: Module 5 - 14

Vector Problems and Relative Velocity Module 6 Vector Problems and Relative Velocity Giancoli, Sec 3- 4, 8 The following is an excellent lecture on this material. Giancoli, PHYSICS,6/E © 2004. Electronically reproduced by permission of Pearson Education, Inc., Upper Saddle River, New Jersey 17

Example 2 A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 150 cm and makes a angle of 1200 with the positive x-axis. The resultant displacement has a magnitude of 140 cm and is directed at an angle of 35.00 to the positive x axis. Find the magnitude and direction of the second displacement. Module 6 - 1

Example 2 Alternative Solution. In the solution below, the angles for vector A are measured from the negative x axis. In this case, we have to assign the signs for the components. The answer is the same. Module 6 - 2

Example 2 Continued A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 150 cm and makes a angle of 1200 with the positive x-axis. The resultant displacement has a magnitude of 140 cm and is directed at an angle of 35.00 to the positive x axis. Find the magnitude and direction of the second displacement. Module 6 - 3

Relative Velocity Will consider how observations made in different reference frames are related to each other. A person walks toward the front of a train at 5 km / h (VPT). The train is moving 80 km / h with respect to the ground (VTG). What is the person’s velocity with respect to the ground (VPG)? Module 6 - 4

Relative Velocity Boat is aimed upstream so that it will move directly across. Boat is aimed directly across, so it will land at a point downstream. Can expect similar problems with airplanes. Module 6 - 5

Example 6 An airplane is capable of flying at 400 mi/h in still air Example 6 An airplane is capable of flying at 400 mi/h in still air. At what angle should the pilot point the plane in order for it to travel due east, if there is a wind of speed 50.0 mi/h directed due south? What is the speed relative to the ground? North of East Module 6 - 6

Vectors Vector Addition Applet Module 6 - 7

Module 7 Projectile Motion Giancoli, Sec 3-5, 6, 7 Giancoli, PHYSICS,6/E © 2004. Electronically reproduced by permission of Pearson Education, Inc., Upper Saddle River, New Jersey 25

3-5 Projectile Motion A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola. Module 7 - 1

Projectile Motion Neglect air resistance Consider motion only after release and before it hits Analyze the vertical and horizontal components separately (Galileo) No acceleration in the horizontal, so velocity is constant Acceleration in the vertical is – 9.8 m/s2 due to gravity and thus velocity is not constant. Object projected horizontally will reach the ground at the same time as one dropped vertically Video 1933 (ball) & 1730 Module 7 - 2 27

Equations for Projectile Motion Horizontal Vertical ax=0 ay = - g vx= constant Module 7 - 3

Initial Velocity If the ball returns to the y = 0 point, then the velocity at that point will equal the initial velocity. At the highest point, v0 y = 0 and v = vx0 Module 7 - 4

Example 3A A football is kicked at an angle of 50.00 above the horizontal with a velocity of 18.0 m / s. Calculate the maximum height. Assume that the ball was kicked at ground level and lands at ground level. at top: Module 7 - 5

Level Horizontal Range Range is determined by time it takes for ball to return to ground level or perhaps some other vertical value. If ball hits something a fixed distance away, then time is determined by x motion If the motion is on a level field, when it hits: y = 0 Solving we find We can substitute this in the x equation to find the range R Module 7 - 6

Level Horizontal Range We can use a trig identity Greatest range:  = 450  = 300 and 600 have same range. Caution– the range formula has limited usefulness. It is only valid when the projectile returns to the same vertical position. Module 7 - 7

Assume time down = time up Example 3B A football is kicked at an angle of 50.00 above the horizontal with a velocity of 18.0 m / s. Calculate the range. Assume that the ball was kicked at ground level and lands at ground level. Assume time down = time up For Range: Could also use range formula Module 7 - 8

Example 4A A football is kicked at an angle of 50.00 above the horizontal with a velocity of 18.0 m / s. The football hits a window in a house that is 25.0 m from where it was kicked. How high was the window above the ground. Time to hit the window: Module 7 - 9

Example 4 B What is the final velocity and angle of the football that hit the window in Example 4 A. below x axis Module 7 - 10

Coordinate system is 235 m below plane Example 5. (35) A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. If the plane is traveling horizontally with a speed of 250 km /h (69.4 m / s) how far in advance of the recipients (horizontal distance) must the goods be dropped (Fig. 3–37a)? . Coordinate system is 235 m below plane Module 7 - 12

Sec. 3-7 Projectile Motion Is Parabolic In order to demonstrate that projectile motion is parabolic, the book derives y as a function of x. When we do, we find that it has the form: This is the equation for a parabola. Module 7 - 13