1. PHY 101: Lecture 3 3.1 Displacement, Velocity, and Acceleration 3.2 Equations of Kinematics in Two Dimensions 3.3 Projectile Motion 3.4 Relative Velocity

2. PHY 101: Lecture 3 Kinematics in Two Dimensions 3.1 Displacement, Velocity, and Acceleration

3. Distance and Speed Distance and Speed in 2-Dimensions are the same as in 1-Dimension

4. Definition of Initial Position Type of Quantity: Vector Initial position of an object is indicted by a position vector, r i, from the origin to the position r i (x i, y i ) Magnitude of r i is distance from the origin to point (x i, y i ) Direction of r i is the angle made with the +x-axis by the line from the origin to the point (x i, y i ) Symbol: r i SI Unit: meter (m) Subscript f means final position, (x f, y f ) rfrf riri

5. Definition of Displacement Type of Quantity: Vector Vector drawn from initial position to final position  r i +  r = r f   r = r f - r i Magnitude equals shortest distance between initial and final positions Direction points from initial to final position Displacement depends on the initial and final positions and not on the path length Vector components of r i are x i, and y i Symbol:  r SI Unit: meter (m)

6. Definition of Average Velocity Type of Quantity: Vector Average Velocity = Displacement / Time v avg  r/t = (r f – r i )/t Direction is the same as the direction of displacement Vector components of v are:  V x,avg = (x f – x i )/t  V y,avg = (x f – x i )/t Symbol: v avg SI Unit: meter/second (m/s)

7. Average Velocity - Example Magnitude of velocity vector is 7.0 The x-component is 3.0 What is the y-component? Use Pythagorean Theorem  v avg 2 = v x,avg 2 + v y,avg 2  7 2 = 3 2 + v y,avg 2  V y,avg = sqrt(49 – 9) = 6.3 m/s

8. PHY 101: Lecture 3 Kinematics in Two Dimensions 3.2 Equations of Kinematics in Two Dimensions Included in 3.3

9. PHY 101: Lecture 3 Kinematics in Two Dimensions 3.3 Projectile Motion

10. 2-Dimensions x motion is independent of y motion

11. Projectile motion Horizontal  v f = v i  a = 0 m/s 2 Vertical  a = -9.8 m/s 2  g = acceleration of gravity = 9.8 m/s 2  g is positive  Use g as a = –g

12. 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

13. Description of Range Object is fired at angle  to horizontal Object is fired at a height 0 Object follows a parabola Object returns to the height 0 Horizontal distance from starting to final position is the Range

14. 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 

15. 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

16. 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

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

18. 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

19. 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

20. 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

21. 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

22. 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

23. 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

24. 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

25. 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

26. 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

27. 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

28. 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

29. PHY 101: Lecture 3 Kinematics in Two Dimensions 3.4 Relative Velocity Skipped