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Two-Dimensional Motion and Vectors Chapter 3 SCALARS Column 1 Money Mass Speed Temperature Volume Column 2 VECTORS Displacement Velocity Acceleration Force Position Two-Dimensional Motion and Vectors • Vectors – are quantities that have magnitude and direction. – Displacement, velocity, acceleration, force, and momentum are examples of vectors. – are shown with boldface symbols. – use arrows to show direction and size.. • A scalar is a physical quantity that has magnitude but no direction. How do we draw vectors? Arrows, scaled drawings, and directions. http://www.physicsclassroom.com/class/vectors/U3L1a.cfm How do we designate vector direction? 1. The direction of a vector is often expressed as an angle of rotation of the vector about its “tail” from east, west, north, or south. 30o N or W or 60o W of N 2. The direction of a vector is often expressed as a counterclockwise angle of rotation of the vector about its “tail” from due East. http://www.physicsclassroom.com/class/vectors/U3L1a.cfm Two-Dimensional Motion and Vectors A resultant vector represents the sum of two or more vectors. It is the result of adding vectors. Two-Dimensional Motion and Vectors How do we determine the resultant vector when the individual vectors are at angles to each other? 1) Graphically head-to-tail using a scaled diagram, or 2) Mathematically, using Pythagorean theorem and a little trigonometry. Two-Dimensional Motion and Vectors • Vectors can be added graphically. Draw the first vector to scale at the correct angle using a ruler and a protractor. Place the tail of the second vector on the tip of the first vector and then draw it to scale at the appropriate angle. Continue doing this until all vectors have been drawn. The resultant vector is the final vector (arrow) drawn from the tail of the first vector (origin) to the tip of the last vector. Measure its length and angle. Chapter 3 Properties of Vectors • Vectors can be added in any order. A + B = C and B + A = C • To subtract a vector, add its opposite. A + (-B) = D • Multiplying or dividing vectors by scalars results in vectors. • Video 1 • Video 2 Two-Dimensional Motion and Vectors • To determine the resultant vector. vectors can be moved parallel to themselves. vectors can be added in any order. Two-Dimensional Motion and Vectors Algebraic Addition of Vectors 3.2 Vector Operations 1. Determine the resultant magnitude and the resultant direction. 2. Resolve any vector into components. 3. Add multiple vectors by adding components and then determine the magnitude of the resultant using the Pythagorean theorem and direction of the resultant using SOH CAH TOA. Algebraic Addition of Vectors 3.2 Vector Operations – To add vectors that are at right angles use the Pythagorean theorem to determine the resultant magnitude and inverse tangent to determine the resultant direction. Two-Dimensional Motion and Vectors 3.2 Determining Resultant Magnitude and Direction Eric leaves the base camp and hikes 11 km, north and then hikes 11 km east. Determine Eric's resulting displacement. Two-Dimensional Motion and Vectors R = 11 km R = 50 km Two-Dimensional Motion and Vectors 3.2 Determining Resultant Direction Chief Soh Cah Toa sin q = Soh opp hyp cos q = adj hyp Cah tan q = opp adj Toa Two-Dimensional Motion and Vectors 3.2 Determining Resultant Magnitude and Direction 12 m/s E cos q 8 m/s N R = 14.4 m/s = sin q = hyp q = sin-1 = = 8 m/s 14.4 m/s 8 m/s = 56o 14.4 m/s tan q 12 m/s hyp q = cos-1 q opp adj = opp adj = 12 m/s 8 m/s 14.4 m/s q = tan-1 12 m/s 12 m/s 14.4 m/s = 56o 8m/s = 56o Two-Dimensional Motion and Vectors Determine the direction of the resultant. R = 11 km R = 50 km q = 27o W of N R = 53o S of W Two-Dimensional Motion and Vectors Practice A p. 89 Two-Dimensional Motion and Vectors All vectors can be resolved into a horizontal (x) component and a vertical (y) component. The components of a vector are the projections of the vector along the axes of a coordinate system. Resolving a vector allows you to analyze the motion in each direction. Chapter 3 Section 2 Vector Operations Resolving Vectors into Components Consider an airplane flying at 95 km/h at an angle of 20o to the ground. • The hypotenuse (vplane) is the resultant vector that describes the airplane’s total velocity. • The adjacent leg represents the x component (vx), which describes the airplane’s horizontal speed. • The opposite leg represents the y component (vy), which describes the airplane’s vertical speed. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 3 Section 2 Vector Operations Resolving Vectors into Components, continued • The sine and cosine functions can be used to find the components of a vector. • The sine and cosine functions are defined in terms of the lengths of the sides of right triangles. opposite leg sine of angle q = hypotenuse adjacent leg cosine of angle q = hypotenuse Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Two-Dimensional Motion and Vectors 3.2 Resolving Vectors into Components Suppose you kick a ball into the air with an initial velocity of 14 m/s at a 40o angle. The ball has an upward (vertical motion) and a horizontal motion. What are these velocity components? vx v =14 m/s vy 40o vx = (14 m/s)(cos 40o) = 11 m/s v =14 m/s vy 40o vx vy = (14 m/s)(sin 40o) = 9 m/s Two-Dimensional Motion and Vectors 3.2 Resolving Vectors into Components Practice B p. 92 Two-Dimensional Motion and Vectors 3.2 Adding Multiple Vectors that are not Perpendicular 1. Resolve each vector into its x- and y- components. 2. Add all x-components to get a net x-component for the resultant vector. 3. Add all y-components to get a net y-component for the resultant vector. 4. Draw and label the x- and y- components of the resultant and sketch the resultant vector. 5. Use the Pythagorean Theorem to determine the magnitude of the resultant and tan-1q to determine the direction. Section 2 Vector Operations Chapter 3 Adding Vectors That Are Not Perpendicular • Suppose that a plane travels first 5 km at an angle of 35°, then climbs at 10° for 22 km, as shown below. How can you find the total displacement? • Because the original displacement vectors do not form a right triangle, you cannot directly apply the tangent function or the Pythagorean theorem. d2 d1 Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 3 Section 2 Vector Operations Adding Vectors That Are Not Perpendicular, • You can find the magnitude and the direction of the resultant by resolving each of the plane’s displacement vectors into its x and y components. • Then the components along each axis can be added together. As shown in the figure, these sums will be the two perpendicular components of the resultant, d. The resultant’s magnitude can then be found by using the Pythagorean theorem, and its direction can be found by using the inverse tangent function. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 3 Section 2 Vector Operations Adding Vectors That Are Not Perpendicular Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 3 Section 2 Vector Operations Sample Problem Adding Vectors Algebraically A hiker walks 27.0 km from her base camp at 35° south of east. The next day, she walks 41.0 km in a direction 65° north of east and discovers a forest ranger’s tower. Find the magnitude and direction of her resultant displacement Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 3 Section 2 Vector Operations Sample Problem, continued 1 . Select a coordinate system. Then sketch and label each vector. Given: d1 = 27.0 km d2 = 41.0 km q1 = –35° q2 = 65° Tip: q1 is negative, because clockwise movement from the positive x-axis is negative by convention. Unknown: d=? q=? Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 3 Section 2 Vector Operations Sample Problem, continued 2 . Find the x and y components of all vectors. Make a separate sketch of the displacements for each day. Use the cosine and sine functions to find the components. For day 1 : x1 = d1 cosq1 = (27.0 km)(cos –35) = 22 km y1 = d1 sin q1 = (27.0 km)(sin –35) = –15 km For day 2 : x2 = d2 cosq 2 = (41.0 km)(cos 65) = 17 km y2 = d2 sin q 2 = (41.0 km)(sin 65) = 37 km Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 3 Section 2 Vector Operations Sample Problem, continued 3 . Find the x and y components of the total displacement. xtot = x1 x2 = 22 km + 17 km = 39 km ytot = y1 y2 = –15 km + 37 km = 22 km 4 . Use the Pythagorean theorem to find the magnitude of the resultant vector. d 2 = (xtot )2 (ytot )2 d = (xtot )2 (ytot )2 = (39 km)2 (22 km)2 d = 45 km Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 3 Section 2 Vector Operations Sample Problem, continued 5 . Use a suitable trigonometric function to find the angle. y –1 22 km q = tan = tan x 39 km q = 29 north of east –1 Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Two-Dimensional Motion and Vectors 3.2 Adding Vectors that are not Perpendicular A B A By B Bx Rx = Ax + (-Bx) Ry Rx Ry = Ay + By Ay Ax Two-Dimensional Motion and Vectors 3.2 Vector Operations Practice p. 89 p. 92 p. 94 Two-Dimensional Motion and Vectors 3.3 Projectile Motion – the curved path that an object follows when thrown, launched, or otherwise projected near the surface of the earth – 2-dimensional motion – has a vertical (y) component and a horizontal (x) component – follows parabolic paths – vx constant (horizontal component is assumed constant) – ay =-g – is freefall combined with an initial horizontal velocity – punted football, thrown ball, water drop cascading down a waterfall Two-Dimensional Motion and Vectors 3.3 Projectile Motion – Projectiles can be launched horizontally or at an angle. – Projectiles launched horizontally do not have any initial vertical (y-direction) velocity. The initial velocity is only in the horizontal direction (x-direction). – Projectiles launched at an angle have an initial vertical velocity (y-direction) and an initial horizontal velocity (x-direction). – So, when analyzing projectile motion we have to look at the vertical and horizontal motions. These motions are independent of each other. – We use our motion equations to analyze projectiles. Horizontally Launched Projectiles Horizontally Launched Projectiles physicsphenomena.com Horizontally Launched Projectiles physicsclassroom.com Two-Dimensional Motion and Vectors 3.3 Motion Equations Reminder x = v= t xf - x i t vf = vi + at vf 2 = vi 2 + 2ax x = vit + ½at2 Two-Dimensional Motion and Vectors 3.3 Vertical Motion of a Projectile Launched Horizontally vy,i = 0 vy,f = vy,i + at vy,f = ayt vy,f 2 = vy,i 2 + 2ayy vy,f 2 = 2ayy y = vy,it + ½ayt2 y = ½ ayt2 Two-Dimensional Motion and Vectors 3.3 Horizontal Motion of a Projectile Remember vx is constant so x vx = t vx = vx,i = constant x = vxt x t = v x or . Two-Dimensional Motion and Vectors Suppose that a baseball leaves the hand of the pitcher, traveling horizontally with a velocity of 15 m/s, and suppose that it is 2.0 m above the ground at that instant. Where will it hit the ground? v = 15 m/s = vx 2.0 m y = ½ t = What do we want to know and what do we know? v = 15 m/s = vx x = ? y = - 2.0 m tx = ? vy,i = 0 tx = ty= ? ay = -9.81 m/s2 x vx = t ayt2 2y x = vxt x = (15 m/s) (0.64 s) x = 9.6 m ay (2)(- 2.0 m) t = - 9.81 m/s2 = 0.64 s The ball will hit the ground 9.6 m in front of the pitcher. Two-Dimensional Motion and Vectors Solving projectile problems: 1. Use one of the independent motions (horizontal or vertical) to find the time in flight, then use it to determine the position of impact. 2. If projectile impacts a vertical surface (wall) before hitting the ground, start with the horizontal motion to determine time in flight. 3. If projectile impacts a horizontal surface (ground) first start with the vertical motion to determine time in flight. 4. Be careful to use the correct initial velocity component. Two-Dimensional Motion and Vectors A ball is thrown with a velocity of 30 ft/s at an angle of 37o above the horizontal. It leaves the pitcher’s hand 4.0 ft above the ground and 15 ft from a wall. (g = 32 ft/s2) (1) At what height above the ground will it hit the wall? (2) Will it still be going up just before it hits, or will it already be on its way down? (1) When the ball travels a horizontal distance of 15 ft, how high will it be above the ground? 37o Know or can calculate: g, vx, vy,i, tx, y for tx 4 ft 15 ft Horizontal Motion vx = (30 ft/s)(cos 37o) = 24 ft/s x vx = t x 15 ft t = = 24 ft/s = 0.63 s vx Vertical Motion Vy,i = (30 ft/s)(sin 37o) = 18 ft/s y = vy,i t + ½ ay t2 y = 18 ft/s(0.63s) + ½ (-32 ft/s2)(0.63s)2 y = 5 ft above starting point The ball hits the wall 9 ft above the ground. Two-Dimensional Motion and Vectors (2) Will it still be going up just before it hits, or will it already be on its way down? Use the final velocity equation and calculate vy,f at 0.63 s. 37o 4 ft vy,f = vy,i + at vy,f = vy,i + at vy,f = 18 ft/s + (-32 ft/s)(0.63s) vy,f = -2.2 ft/s The negative sign tells us that the velocity at 0.63s is in the negative direction which is DOWN! Two-Dimensional Motion and Vectors (2) Will it still be going up just before it hits, or will it already be on its way down? Another way How long will it take the ball reach max. height? Look at the vertical motion equations. 37o 4 ft vy,i = - at - vy,i t = a 18 ft/s t = = 0.56 s -32 ft/s2 0 vy,f = vy,i + at The ball will be on its way down since it is in the air 0.63 s before hitting the wall and it reaches its maximum height at 0.56 s. Two-Dimensional Motion and Vectors Vertical Motion vy,f = vy,i + at (vy,f) 2 = vy,i 2 + 2ayy y = vy,it + ½ayt2 vy,i = vi.sinq a = -g = -9.81m/s2 = -32ft/s2 Two-Dimensional Motion and Vectors Horizontal Motion vx,i = vi.cosq vx,i = vi.cosq = x/t 3.4 Relative Motion • On occasion objects move within a medium that is moving with respect to an observer. – An airplane usually encounters a wind - air that is moving with respect to an observer on the ground below. – A motorboat in a river is moving amidst a river current - water that is moving with respect to an observer on dry land. • In such instances as this, the magnitude of the velocity of the moving object (whether it be a plane or a motorboat) with respect to the observer on land will not be the same as the speedometer reading of the vehicle. 3.4 Relative Motion Frames of Reference • If you are moving at 80 km/h north and a car passes you going 90 km/h, to you the faster car seems to be moving north at 10 km/h. • Someone standing on the side of the road would measure the velocity of the faster car as 90 km/h toward the north. • This simple example demonstrates that velocity measurements depend on the frame of reference of the observer. 3.4 Relative Motion Frames of Reference (continued) Consider a stunt dummy dropped from a plane. (a) When viewed from the plane, the stunt dummy falls straight down. (b) When viewed from a stationary position on the ground, the stunt dummy follows a parabolic projectile path. 3.4 Relative Motion Relative Velocity • When solving relative velocity problems, write down the information in the form of velocities with subscripts. • Using our earlier car example, we have: • vse = +80 km/h north (se = slower car with respect to Earth) • vfe = +90 km/h north (fe = fast car with respect to Earth) unknown = vfs (fs = fast car with respect to slower car) • Write an equation for vfs in terms of the other velocities. The subscripts start with f and end with s. The other subscripts start with the letter that ended the preceding velocity: • vfs = vfe + ves • ves = –vse • Thus, this problem can be solved as follows: • vfs = vfe + ves = vfe – vse • vfs = (+90 km/h n) – (+80 km/h n) = +10 km/h n Chapter 3 Relative Velocity, continued • A general form of the relative velocity equation is: • vac = vab + vbc http://www.physicsclassroom.com/mmedia/vectors/plane.cfm Two-Dimensional Motion and Vectors 3.4 Relative Motion p. 105 #2