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Ch.3 Kinemtics In 2-D

v = limit. Ch.3 Kinemtics In 2-D. Displacement vector is in the x-y plane (not in x or y direction). Displacement. a = limit. Velocity Components In 2-D. Tangent to the path (instantaneous velocity). Average acceleration. Instantaneous Acceleration. Motion Only In X Direction.

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Ch.3 Kinemtics In 2-D

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  1. v = limit Ch.3 Kinemtics In 2-D Displacement vector is in the x-y plane (not in x or y direction) Displacement

  2. a = limit Velocity Components In 2-D Tangent to the path (instantaneous velocity) Average acceleration Instantaneous Acceleration

  3. Motion Only In X Direction x – motion equations

  4. Motion In Y Direction y component

  5. 2 D Motion

  6. x-Direction Data y-Direction Data Example 1

  7. Projectile Motion • 2-D motion under gravity • What is the acceleration in X direction ? • What is the acceleration in y direction ? • 2-D motion under gravity • What is the acceleration in X direction ? • What is the acceleration in y direction ? ax= 0 ay = - 9.8 m/s2 Garden Horse – watering a plant a Plane dropping a bomb

  8. Example 2: Falling Package The time to hit the ground ? y = v0y t +(1/2) ay t2 = ½ (-9.8) t2 i.e., -1050 = ½ (-9.8) t2

  9. vx θ vy R Velocity Of The Package When It Hits The Ground ? vx = v0x + ax t = v0x = 115 m/s vy = v0y + ay t = (-9.8 m/s2) 14.6 s = -143 m/s (Why Negative ?) v2 = (115)2+ (-143)2 v = 184 m/s

  10. Shoot A Bullet Straight Up In A Moving Car

  11. Example 5: Height Of A Kickoff v2 = v02 + 2ay 0 = (14)2 = 2*(-9.8)H v0x = v0 cosθ = 22 m/s cos 40.0o = 17m/s v0y = v0 sinθ = 22 m/s sin 40.0o =14m/s For y direction H = 14*14 /(2*9.8)= 10 m

  12. Time Of Flight ( Time In The Air) Time until it hits the ground y = v0yt + (½) ay t2 0 = 14 t + (½) (-9.8) t2 i.e., t(14 -4.9t) = 0 sec t = 0 sec, or t= 14/4.9 = 2.9 sec (why two answers?)

  13. vx θ vy R Time To Reach The Maximum Height vy = v0y +ay t 0 = 14 -9.8t t = 14/9.8 = 1.45 sec 2t = 2.9 (time of flight) sec Symmetry: time of flight = twice the time to reach the top --- Why? RANGE R = v0x t + (½ )ax t2 = 17 *2.9 = 49 m

  14. Same Height Same Speed

  15. Relative Velocity vPG = vPT + vTG = 2.0 m/s + 9.0 m/s = 11.0 m/s vPG = Velocity of Passenger relative to Ground (vPG = - vGP)

  16. Check your understanding 4 • Three cars A, B, C, are moving along a straight line, Relative velocities are given. • vAB vAC vCB • ? 40 m/s 30 m/s • ? 50 m/s -20 m/s • 60 m/s 20 m/s ? • -50 m/s ? 10 m/s • Three cars A, B, C, are moving along a straight line, Relative velocities are given. • vAB vAC vCB • 70 m/s 40 m/s 30 m/s • ? 50 m/s -20 m/s • 60 m/s 20 m/s ? • -50 m/s ? 10 m/s • Three cars A, B, C, are moving along a straight line, Relative velocities are given. • vAB vAC vCB • 70 m/s 40 m/s 30 m/s • 30 m/s 50 m/s -20 m/s • 60 m/s 20 m/s ? • -50 m/s ? 10 m/s • Three cars A, B, C, are moving along a straight line, Relative velocities are given. • vAB vAC vCB • 70 m/s 40 m/s 30 m/s • 30 m/s 50 m/s -20 m/s • 60 m/s 20 m/s 40 m/s • -50 m/s ? 10 m/s • Three cars A, B, C, are moving along a straight line, Relative velocities are given. • vAB vAC vCB • 70 m/s 40 m/s 30 m/s • 30 m/s 50 m/s -20 m/s • 60 m/s 20 m/s 40 m/s • -50 m/s -60 m/s 10 m/s • Three cars A, B, C, are moving along a straight line, Relative velocities are given. • vAB vAC vCB • ? 40 m/s 30 m/s • ? 50 m/s -20 m/s • 60 m/s 20 m/s ? • -50 m/s ? 10 m/s

  17. vWS vBW vBS R Crossing A River vBS = vBW + vWS R2 = (vBW)2 + (vWS)2

  18. 10. Crossing A River Width of the river = 1800 m vBS = vBW + vWS = 4.0 m/s + 2.0 m/s = 4.5 m/s θ = tan-1(2) = 63º

  19. 11. Approaching Intersection vAG = 25.0 m/s vBG = 15.8 m/s vAB = vAG + vGB vAB = 25.0 m/s + 15.8 m/s tanθ = 15.8/25 Θ = tan-1(15.8/25) = 32.4º

  20. Raindrops On The Car vRC = vRG + vGC

  21. Concepts: Circus Clowns No air resistance vox = 4.6 m/s v0y = 10.0 m/s Θ = tan-1(v0y/v0x) = tan-1(10/4.6) = 65º

  22. Conceptual Question 2 REASONING AND SOLUTION An object thrown upward at an angle q will follow the trajectory shown below. Its acceleration is that due to gravity, and, therefore, always points downward. The acceleration is denoted by ay in the figure. In general, the velocity of the object has two components, vx and vy. Since ax = 0, vx always equals its initial value. The y component of the velocity, vy, decreases as the object rises, drops to zero when the object is at its highest point, and then increases in magnitude as the object falls downward.

  23. a.) Since vy = 0 when the object is at its highest point, the velocity of the object points only in the x direction. As suggested in the figure below, the acceleration will be perpendicular to the velocity when the object is at its highest point and vy = 0. b.) In order for the velocity and acceleration to be parallel, the x component of the velocity would have to drop to zero. However, vx always remains equal to its initial value; therefore, the velocity and the acceleration can never be parallel.

  24. Conceptual Question 4 REASONING AND SOLUTION If a baseball were pitched on the moon, it would still fall downwards as it travels toward the batter. However the acceleration due to gravity on the moon is roughly 6 times less than that on earth. Thus, in the time it takes to reach the batter, the ball will not fall as far vertically on the moon as it does on earth. Therefore, the pitcher's mound on the moon would be at a lower height than it is on earth.

  25. Conceptual Questions 13 REASONING AND SOLUTION Since the plastic bottle moves with the current, the passenger is estimating the velocity of the boat relative to the water. Therefore, the passenger cannot conclude that the boat is moving at 5 m/s with respect to the shore.

  26. Conceptual Questions 16 REASONING AND SOLUTION The time required for any given swimmer to cross the river is equal to the width of the river divided by the magnitude of the component of the velocity that is parallel to the width of the river. All three swimmers can swim equally fast relative to the water; however, all three swim at different angles relative to the current. Since swimmer A heads straight across the width of the river, swimmer A will have the largest velocity component parallel to the width of the river; therefore, swimmer A crosses the river in the least time.

  27. Problem 4 REASONING AND SOLUTION The increase in altitude represents vy = 6.80 m/s. The movement of the shadow represents vx = 15.5 m/s. The magnitude of the glider's velocity is therefore

  28. Problem 7 REASONING Trigonometry indicates that the x and y components of the dolphin’s velocity are related to the launch angle  according to tan  = vy/vx. SOLUTION Using trigonometry, we find that the y component of the dolphin’s velocity is

  29. 55o 15 m/s vx=? Problem 14 REASONING The vertical component of the ball’s velocity v0 changes as the ball approaches the opposing player. It changes due to the acceleration of gravity. However, the horizontal component does not change, assuming that air resistance can be neglected.

  30. Hence, the horizontal component of the ball’s velocity when the opposing player fields the ball is the same as it was initially. SOLUTION Using trigonometry, we find that the horizontal component is

  31. REASONING AND SOLUTION Using vy = 0 and voy = vo sin q = (11 m/s) sin 65 = 1.0  101 m/s and vy2 = voy2 + 2ayy, we have Problem 16 v0y=0 B v0=11m/s h=? 65O A

  32. Problem 27

  33. REASONING AND SOLUTION The time of flight of the motorcycle is given by y=0, v0y=33.5sin18, t=? y=v0t+(1/2)at2 t=0, or The horizontal distance traveled by the motorcycle is then x = vo cos qot = (33.5 m/s)(cos18.0°)(2.11 s) = 67.2 m The daredevil can jump over (67.2 m)/(2.74 m/bus) = 24.5 buses. In even numbers, this means

  34. Problem 40

  35. REASONING Using the data given in the problem, we can find the maximum flight time t of the ball using Equation 3.5b Once the flight time is known, we can use the definition of average velocity to find the minimum speed required to cover the distance x in that time. SOLUTION Equation 3.5b is quadratic in t and can be solved for t using the quadratic formula. According to Equation 3.5b, the maximum flight time is (with upward taken as the positive direction)

  36. where the first root corresponds to the time required for the ball to reach a vertical displacement of as it travels upward, and the second root corresponds to the time required for the ball to have a vertical displacement of as the ball travels upward and then downward. The desired flight time t is 2.145 s.

  37. During the 2.145 s, the horizontal distance traveled by the ball is Thus, the opponent must move in . The opponent must, therefore, move with a minimum average speed of

  38. Problem 41

  39. REASONING AND SOLUTION In the absence of air resistance, the bullet exhibits projectile motion. The x component of the motion has zero acceleration while the y component of the motion is subject to the acceleration due to gravity. The horizontal distance traveled by the bullet is given by Equation 3.5a (with ): with t equal to the time required for the bullet to reach the target. The time t can be found by considering the vertical motion. From Equation 3.3b

  40. When the bullet reaches the target, . Assuming that up and to the right are the positive directions, we have Using the fact that , we have Thus, we find that

  41. and Therefore

  42. Problem 48 REASONING Since car A is moving faster, it will eventually catch up with car B. Each car is traveling at a constant velocity, so the time t it takes for A to catch up with B is equal to the displacement between the two cars (x = +186 m) divided by the velocity vAB of A relative to B. (If the relative velocity were zero, A would never catch up with B). We can find the velocity of A relative to B by using the subscripting technique developed in Section 3.4 of the text. 24.4 m/s 18.6 m/s B A 186 m

  43. vAB = velocity of car A relative to car B vAG = velocity of car A relative to the Ground = +24.4 m/s vBG = velocity of car B relative to the Ground = +18.6 m/s We have chosen the positive direction for the displacement and velocities to be the direction in which the cars are moving. The velocities are related by vAB = vAG + vGB

  44. SOLUTION The velocity of car A relative to car B is vAB = vAG + vGB = +24.4 m/s + (18.6 m/s) = +5.8 m/s, where we have used the fact that vGB = vBG = 18.6 m/s. The time it takes for car A to catch car B is

  45. Problem 53 REASONING Let represent the velocity of the hawk relative to the balloon and represent the velocity of the balloon relative to the ground. Then, as indicated by Equation 3.7, the velocity of the hawk relative to the ground is . Since the vectors and are at right angles to each other, the vector addition can be carried out using the Pythagorean theorem.(B: Balloon, H: Hawk, G: Ground) SOLUTION Using the drawing at the right, we have from the Pythagorean theorem

  46. B: Balloon H: Hawk G: Ground =2.0 =6.0 The angle q is

  47. Problem 56

  48. REASONING AND SOLUTION The velocity of the raindrops relative to the train is given by vRT = vRG + vGT where vRGis the velocity of the raindrops relative to the ground and vGTis the velocity of the ground relative to the train Since the train moves horizontally, and the rain falls vertically, the velocity vectors are related as shown in the figure at the right. Then vGT = vRG tan q = (5.0 m/s) (tan 25°) = 2.3 m/s The train is moving at a speed of

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