1. PHY 151: Lecture 4A 4.1 Position, Velocity, and Acceleration Vectors 4.2 Two-Dimensional Motion with Constant Acceleration 4.3 Projectile Motion

2. PHY 151: Lecture 4A Motion in Two Dimensions 4.1 Position, Velocity, and Acceleration Vectors

3. 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

4. General Motion Ideas In two- or three-dimensional kinematics, everything is the same as in one- dimensional motion except that we must now use full vector notation –Positive and negative signs are no longer sufficient to determine the direction

5. 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 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

6. Instantaneous Velocity - 1 The instantaneous velocity is the limit of the average velocity as Δt approaches zero –As the time interval becomes smaller, the direction of the displacement approaches that of the line tangent to the curve

7. Instantaneous Velocity - 2 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 –The speed is a scalar quantity

8. Average Acceleration - 1 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

9. Average Acceleration - 2 As a particle moves, the direction of the change in velocity is found by vector subtraction The average acceleration is a vector quantity directed along

10. Instantaneous Acceleration The instantaneous acceleration is the limiting value of the ratio as Δt approaches zero –The instantaneous equals the derivative of the velocity vector with respect to time

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

12. PHY 151: Lecture 4A Motion in Two Dimensions 4.2 Two-Dimensional Motion with Constant Acceleration

13. 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 Motion in two dimensions can be modeled as two independent motions in each of the two perpendicular directions associated with the x and y axes –Any influence in the y direction does not affect the motion in the x direction

14. Kinematic Equations - 2 Position vector for a particle moving in the xy plane The velocity vector can be found from the position vector –Since acceleration is constant, we can also find an expression for the velocity as a function of time:

15. Kinematic Equations 3 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 the initial velocity The displacement resulting from the acceleration

16. Kinematic Equations, Graphical Representation of Final Velocity The velocity vector can be represented by its components is generally not along the direction of either or

17. Kinematic Equations, Graphical Representation of Final Position The vector representation of the position vector is generally not along the same direction as, or and are generally not in the same direction

18. PHY 151: Lecture 4A Motion in Two Dimensions 4.3 Projectile Motion

19. Projectile Motion - 1 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

20. Assumptions of Projectile Motion Horizontal  v xf = v xi  a x = 0 m/s 2 Vertical  a y = -9.8 m/s 2  g = acceleration of gravity = 9.8 m/s 2  g is positive  Use g as a y = –g  No air resistance

21. Projectile Motion Diagram

22. Acceleration at the Highest Point The vertical velocity is zero at the top The acceleration is not zero anywhere along the trajectory –If the projectile experienced zero acceleration at the highest point, its velocity at the point would not change The projectile would move with a constant horizontal velocity from that point on

23. Analyzing Projectile Motion Consider the motion as the superposition of the motions in the x- and y-directions The actual position at any time is given by: The initial velocity can be expressed in terms of its components. –v xi = v i cos  and v yi = v i sin  The x-direction has constant velocity –a x = 0 The y-direction is free fall –a y = -g

24. Projectile Motion Vectors The final position is the vector sum of the initial position, the position resulting from the initial velocity and the position resulting from the acceleration

25. 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

26. Height of a Projectile, equation The maximum height of the projectile can be found in terms of the initial velocity vector: This equation is valid only for symmetric motion

27. Description of Range Object is fired at angle  to horizontal Object is fired at a height 0 Object returns to the height 0 This is a symmetric trajectory Trajectory is a parabola Horizontal distance from starting to final position is the Range

28. Range Formula y-motion gives time-of-flight  y = v i sin  t – (1/2)gt 2  0 = v i sin  t – (1/2)gt 2  0 = v i sin  – (1/2)gt  t =2v i sin  /g x-motion give Range, R  x = R= (v i cos  )t  R = (v i cos  )(2v i sin  /g)  R=(v i 2 /g)2sin  cos   R=(v i 2 /g)sin2 

29. Range Formula – Example 1 An M16 bullet has an initial velocity of 950 m/s The bullet is fired at an angle of 45 0 What is it’s maximum range?  R=(v i 2 /g)sin2   R=(950 2 /9.8)sin(2 x 45) = 92.1 km

30. Range Formula – Example 2 A motorcycle travelling at 50 m/s wants to jump a gap to 200 m At what angle should the motorcycle be launched?  R=(v i 2 /g)sin(2  )  200=(2500/9.8)sin(2  )  sin(2  )  = 200(9.8)/2500 = 0.785   = 25.86 0 and 64.14 0

31. More About the Range of a Projectile

32. 2-Dimensions Projectile Motion – Curve x-motion equations  v x = v i cos   x = (v i cos  )t y-motion equations  v y = v i sin  gt  y = v i sin  t – (1/2)gt 2 Solve for t in terms of x gives This is a parabola

33. Angle for Maximum Range Maximum range when sin(2  )  = 1 2  = 90 0  = 45 0

34. Range of a Projectile, final The maximum range occurs at  i = 45 o 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

35. Projectile Motion – Problem Solving Hints Categorize –Confirm air resistance is neglected –Select a coordinate system with x in the horizontal and y in the vertical direction Analyze –If the initial velocity is given, resolve it into x and y components –Treat the horizontal and vertical motions independently –Analyze the horizontal motion with the particle-under- constant-velocity model –Analyze the vertical motion with the particle-under- constant-acceleration model –Remember that both directions share the same time

36. Non-Symmetric Projectile Motion

37. Projectile Motion – Example 1a Ball with horizontal speed of 1.5 m/s rolls off bench 2.0 m high (a)How long will it take the ball to reach floor?  Horizontal  x f = x i + vt  x i = 0x f = ?  v = 1.5 t = ?  x f = 0 + 1.5tEquation has x f, t  Vertical  v f = v i + at  y f = y i + v i t + ½at 2  v f 2 = v i 2 + 2a(y f – y i )  y i = 2y f = 0  v i = 0v f = ?  a = -9.8t = ?  v f = 0 – 9.8tEquation has v f, t  0 = 2 + 0t + ½(-9.8)t 2 Equation has t  0 2 = 40 2 +2(-9.8)(y f –100)Equation has y f  0 = 2 + 0t + ½(-9.8)t 2  t = sqrt(-4/-9.8) = 0.64 s

38. 2-Dimensions Projectile Motion – Example 1b Ball with horizontal speed of 1.5 m/s rolls off bench 2.0 m high How far from point on floor directly below edge of the bench will ball land?  Horizontal  x f = x i + vt  x i = 0x f = ?  v = 1.5 t = 0.64  x f = 0 + 1.5(0.64)Equation has x f  x f = 0 + 1.5(0.64) = 0.96 m

39. 2-Dimensions Projectile Motion – Example 2 (Find Time) Ball rolls horizontally with speed of 7.6 m/s off edge of tall platform Ball lands 8.7 m from the point on ground directly below edge of platform What is height of platform?  Horizontal  x f = x i + vt  x i = 0  x f = 8.7  v = 7.6  t = ?  8.7 = 0 + 7.6tEquation has t  t = 8.7 / 7.6 = 1.14 s

40. Projectile Motion – Example 2 (Find Height) Ball rolls horizontally with speed of 7.6 m/s off edge of tall platform Ball lands 8.7 m from point on ground directly below platform What is height of platform?  Vertical  v f = v i + at  y f = y i + v i t + ½at 2  v f 2 = v i 2 + 2a(y f – y i )  y i = ?y f = 0  v i = 0v f = ?  a = -9.8t = 1.14  v f = 0 – 9.8(1.14)Equation has v f  0 = y i + 0(1.14) + ½(-9.8)(1.14) 2 Equation has y i  v f 2 = 0 2 +2(-9.8)(y f –y i )Equation has y i, y f, v f  0 = y i + 0(1.14) + ½(-9.8)(1.14) 2  y i =-(1/2)(-9.8)(1.14) 2 = 6.37 m

41. Projectile Motion – Example 3a A rifle fires a bullet at a speed of 250 m/s at an angle of 37 0 above the horizontal. (a)What height does the bullet reach?  Horizontal  x f = x i + vt  x i = 0x f = ?  v=250cos(37)=199.7 t = ?  x f = 0 + 199.7tEquation has x f, t  Vertical  v f = v i + at  y f = y i + v i t + ½at 2  v f 2 = v i 2 + 2a(y f – y i )  y i = 0y f = ?  v i = 250sin(37)=150.5v f = 0  a = -9.8t = 1.14  0 = 150.5 – 9.8tEquation has t  y f = 0 + 150.5t + ½(-9.8)t 2 Equation has y f, t  0 2 = 150.5 2 +2(-9.8)(y f –0)Equation has y f  0 2 = 150.5 2 +2(-9.8)(y f –0)  y f =-150.5 2 /2/-9.8 = 1155 m/s

42. Projectile Motion – Example 3b A rifle fires a bullet at a speed of 250 m/s at an angle of 37 0 above the horizontal. (b)How long is the bullet in the air?  Horizontal  x f = x i + vt  x i = 0x f = ?  v=250cos(37)=199.7 t = ?  x f = 0 + 199.7tEquation has x f, t  Vertical  v f = v i + at  y f = y i + v i t + ½at 2  v f 2 = v i 2 + 2a(y f – y i )  y i = 0y f = 0  v i = 250sin(37)=150.5v f = ?  a = -9.8t = ?  v f = 150.5 – 9.8tEquation has v f, t  0 = 0 + 150.5t + ½(-9.8)t 2 Equation has t  v f 2 = 150.5 2 +2(-9.8)(0–0)Equation has v f  0 = 0 + 150.5t + ½(-9.8)t 2  t = -150.5(2)/-9.8 = 30.7 s

43. Projectile Motion – Example 3c A rifle fires a bullet at a speed of 250 m/s at an angle of 37 0 above the horizontal. (b)What is the balls range?  Horizontal  x f = x i + vt  x i = 0x f = ?  v=250cos(37)=199.7 t = 30.7  x f = 0 + 199.7(30.7)Equation has x f  Vertical  v f = v i + at  y f = y i + v i t + ½at 2  v f 2 = v i 2 + 2a(y f – y i )  y i = 0y f = 0  v i = 250sin(37)=150.5v f = ?  a = -9.8t = 30.7  v f = 150.5 – 9.8(30.7)Equation has v f  0=0+150.5(30.7)+½(-9.8)(30.7) 2 Can’t use equation  v f 2 = 150.5 2 +2(-9.8)(0–0)Equation has v f  x f = 0 + 199.7(30.7) = 6130.8 m

44. 2-Dimension Projectile Motion – Example 4a (Find Time) Batter hits a ball giving it a velocity of 50m/s At an angle of 36.9 degrees to the horizontal (a)Find height at which it hits fence at distance 180 m? Horizontal  x f = x i + vt  x i = 0  x f = 180  v = 50cos36.9 = 40  t = ?  180 = 0 + 40tEquation has t  t = 180 / 40 = 4.5 s

45. Projectile Motion – Example 4a (Find Height) Batter hits a ball giving it a velocity of 50m/s At an angle of 36.9 degrees to the horizontal (a)Find height at which it hits fence at distance 180 m?  Vertical  v f = v i + at  y f = y i + v i t + ½at 2  v f 2 = v i 2 + 2a(y f – y i )  y i = 0y f = ?  v i = 50sin36.9=30v f = ?  a = -9.8t = 4.5  v f = 30 – 9.8(4.5)Equation has v f  y f = 0 + 30(4.5) + ½(-9.8)(4.5) 2 Equation has y i  v f 2 = 30 2 +2(-9.8)(y f – 0)Equation has y f, v f  y f = 0 + 30(4.5) + ½(-9.8)(4.5) 2 = 35.8 m

46. 2-Dimensions Projectile Motion – Example 4b (Find Velocity 1) Batter hits a ball giving it a velocity of 50m/s At an angle of 36.9 degrees to the horizontal (b)What is velocity at which it hits a fence at distance of 180 m?  Vertical  v f = v i + at  y f = y i + v i t + ½at 2  v f 2 = v i 2 + 2a(y f – y i )  y i = 0y f = 35.8  v i = 50sin36.9=30v f = ?  a = -9.8t = 4.5  v f = 30 – 9.8(4.5)Equation has v f  35.8 = 0 + 30(4.5) + ½(-9.8)(4.5) 2 Can’t use equation  v f 2 = 30 2 +2(-9.8)(35.8 – 0)Equation has v f  v f = 30 – 9.8(4.5) = -14.1 m/s

47. 2-Dimensions Projectile Motion – Example 4b (Find Velocity 2) Batter hits a ball giving it a velocity of 50 m/s At an angle of 36.9 0 to the horizontal (b)What is velocity at which it hits a fence at distance of 180 m?  v y = -14.1 m/s  v x = 40  v = sqrt(40 2 + (-14.1) 2 ) = 42.4 m/s   = tan -1 (14.1/40) = 19.1 0  v y is negative, therefore  points into fourth quadrant