1. PHY 151: Lecture 2B 2.5 Motion Diagrams 2.6 Particle Under Constant Acceleration 2.7 Freely Falling Objects 2.8 Kinematic Equations Derived from Calculus

2. PHY 151: Lecture 2B Motion in One Dimension 2.5 Motion Diagrams

3. A motion diagram can be formed by imagining the stroboscope photograph of a moving object Red arrows represent velocity Purple arrows represent acceleration

4. PHY 151: Lecture 2B Motion in One Dimension 2.6 Particle Under Constant Acceleration

5. Constant Velocity Images are equally spaced The car is moving with constant positive velocity (shown by red arrows maintaining the same size) Acceleration equals zero

6. Acceleration and Velocity - 1 Images become farther apart as time increases Velocity and acceleration are in the same direction Acceleration is uniform (violet arrows maintain the same length) Velocity is increasing (red arrows are getting longer) This shows positive acceleration and positive velocity

7. Acceleration and Velocity - 2 Images become closer together as time increases Acceleration and velocity are in opposite directions Acceleration is uniform (violet arrows maintain the same length) Velocity is decreasing (red arrows getting shorter) Positive velocity and negative acceleration

8. Kinematic Equations - 1 The kinematic equations can be used with any particle under uniform acceleration The kinematic equations may be used to solve any problem involving one-dimensional motion with a constant acceleration You may need to use two of the equations to solve one problem Many times there is more than one way to solve a problem

9. Kinematic Equations - 2 For constant a x, Can determine an object’s velocity at any time t when we know its initial velocity and its acceleration –Assumes t i = 0 and t f = t Does not give any information about displacement

10. Kinematic Equations - 3 For constant acceleration, The average velocity can be expressed as the arithmetic mean of the initial and final velocities –This applies only in situations where the acceleration is constant.

11. Kinematic Equations - 4 For constant acceleration, This gives you the position of the particle in terms of time and velocities Doesn’t give you the acceleration

12. Kinematic Equations - 5 For constant acceleration, Gives final position in terms of velocity, time, and acceleration Doesn’t tell you about final velocity

13. Kinematic Equations = 6 For constant a, Gives final velocity in terms of acceleration and displacement Does not give any information about the time

14. When a = 0 When the acceleration is zero, –v xf = v xi = v x –x f = x i + v t The constant acceleration model reduces to the constant velocity model

15. Kinematic Equations – summary

16. Graphical Look at Motion: Displacement – Time curve The slope of the curve is the velocity The curved line indicates the velocity is changing –Therefore, there is an acceleration

17. Graphical Look at Motion: Velocity – Time curve The slope gives the acceleration The straight line indicates a constant acceleration Section 2.6

18. Graphical Look at Motion: Acceleration – Time curve The zero slope indicates a constant acceleration Section 2.6

19. Scalar vs. Vector Magnitude Scalar can be negative or positive –For example, temperature, work, and energy can be negative or positive Magnitude of a vector is always positive –For example, displacement, velocity, and acceleration magnitude is always positive Negative sign on a magnitude is an indication that the vector points in the negative (opposite) direction

20. 1-Dimension Horizontal Motion Example 1 Basketball player starts from rest and accelerates uniformly to speed 6.0 m/s in 1.5 s What distance does the player run?  v f = v i + at  x f = x i + v i t + ½at 2  v f 2 = v i 2 + 2a(x f – x i )  x i = 0x f = ?  v i = 0v f = 6.0  a = ?t = 1.5  6=0+a(1.5)Equation has a (1 st )  x f =0+0(1.5)+½a(1.5) 2 Equation has 2 unknowns, x f, a (2 nd )  6 2 =0 2 +2a(x f –0)Equation has two unknowns, x f, a  6=0+a(1.5); a = 6/1.5 = 4 m/s 2  x f =0+0(1.5)+½(4)(1.5) 2 = 4.5 m/s

21. 1-Dimension Horizontal Motion Example 2a Car accelerates from rest at rate of 2.0 m/s 2 for 5.0 s (a) What is the speed of the car at the end of that time?  v f = v i + at  x f = x i + v i t + ½at 2  v f 2 = v i 2 + 2a(x f – x i )  x i = 0x f = ?  v i = 0v f = ?  a = 2.0t = 5.0  v f =0+2(5)Equation has v f  x f =0+0(5)+½(2)(5) 2 Equation has x f  v f 2 =0 2 +2(2)(x f –0)Equation has two unknowns, x f, v f  v f =0+2(5)=10 m/s

22. 1-Dimension Horizontal Motion Example 2b Car accelerates from rest at rate of 2.0 m/s 2 for 5.0 s (b) How far does the car travel in this time?  v f = v i + at  x f = x i + v i t + ½at 2  v f 2 = v i 2 + 2a(x f – x i )  x i = 0x f = ?  v i = 0v f = 10  a = 2.0t = 5.0  10=0+2(5)Can’t use Equation  x f =0+0(5)+½(2)(5) 2 Equation has x f  10 2 =0 2 +2(2)(x f –0)Equation has x f  x f =0+0(5)+½(2)(5) 2 = 25 m

23. 1-Dimension Horizontal Motion Example 3a A car traveling at 15 m/s stops in 35 m (a)What is the acceleration?  v f = v i + at  x f = x i + v i t + ½at 2  v f 2 = v i 2 + 2a(x f – x i )  x i = 0x f = 35  v i = 15v f = 0  a = ?t = ?  0=15+atEquation has a, t  35=0+15t+½(a)(t) 2 Equation has a, t  0 2 =15 2 +2a(35–0)Equation has a  0 2 =15 2 +2a(35–0)  a = -225/70 = -3.2 m/s 2

24. 1-Dimension Horizontal Motion Example 3b A car traveling at 15 m/s stops in 35 m (b)What is time during this deceleration until car stops?  v f = v i + at  x f = x i + v i t + ½at 2  v f 2 = v i 2 + 2a(x f – x i )  x i = 0x f = 35  v i = 15v f = 0  a = -3.2t = ?  0=15-3.2tEquation has t  35=0+15t+½(-3.2)(t) 2 Equation has t  0 2 =15 2 +2(-3.2)(35–0)Can’t use equation  0=15-3.2t  t = -15/-3.2 = 4.67 s

25. 1-Dimension Horizontal Motion Example 4 A plane accelerates at 8 m/s 2 on a runway that is 500 m long The take off speed of the plane is 80 m/sec Can the plane takeoff?  v f = v i + at  x f = x i + v i t + ½at 2  v f 2 = v i 2 + 2a(x f – x i )  x i = 0x f = 500  v i = 0v f = ?  a = 8t = ?  v f =0+8tEquation has v f, t  500=0+0t+½(8)(t) 2 Equation has t  v f 2 =0 2 +2(8)(500–0)Equation has v f  v f 2 =0 2 +2(8)(500–0) = 500(16)  v f =sqrt(500(16) = 89.4 m/s The plane can takeoff

26. 1-Dimension Horizontal Motion Example 5 A plane accelerates at 8 m/s 2 The take off speed of the plane is 80 m/sec What is minimum length of run way for plane to reach take off speed?  v f = v i + at  x f = x i + v i t + ½at 2  v f 2 = v i 2 + 2a(x f – x i )  x i = 0x f = ?  v i = 0v f = 80  a = 8t = ?  80=0+8tEquation has t  x f =0+0t+½(8)(t) 2 Equation has x f, t  80 2 =0 2 +2(8)(x f –0)Equation has x f  80 2 =0 2 +2(8)(x f –0)  x f =6400/16 = 400 m

27. 1-Dimension Horizontal Motion Example 6 Bull Bull runs 8 m/sec A boy at rest has a head start of 12 m When he sees the bull he accelerates at 2 m/s 2 Does the bull catch the boy?  v f = v i + at  x f = x i + v i t + ½at 2  v f 2 = v i 2 + 2a(x f – x i )  x i = 0x f = ?  v i = 8v f = 8  a = 0t = ?  8=8+0tCan’t use this equation  x f =0+8t+½(0)(t) 2 Equation has x f, t  8 2 =8 2 +2(0)(x f –0)Can’t use this equation

28. 1-Dimension Horizontal Motion Example 6 Boy Bull runs 8 m/sec A boy at rest has a head start of 12 m When he sees the bull he accelerates at 2 m/s 2 Does the bull catch the boy?  v f = v i + at  x f = x i + v i t + ½at 2  v f 2 = v i 2 + 2a(x f – x i )  x i = 12x f = ?  v i = 0v f = ?  a = 2t = ?  v f =0+2tEquation has v f, t  x f =12+0t+½(2)(t) 2 Equation has x f, t  v f 2 =0 2 +2(2)(x f –12)Equation has x f, v f

29. 1-Dimension Horizontal Motion Example 6 Bull / Boy Bull runs 8 m/sec A boy at rest has a head start of 12m When he sees the bull he accelerates at 2 m/s 2 Does the bull catch the boy?  x f =0+8t+½(0)(t) 2 (from Bull equations)  x f =12+0t+½(2)(t) 2 (from Boy equations)  Substitute the first equation into the second  8t=12+t 2  t 2 -8t+12=0  Factor  (t-2)(t-6) = 0 Bull catches boy at 2 s and 6 s 2 s is bull catching up to the boy 6 s is the accelerating boy catching up to the bull

30. 1-Dimension Horizontal Motion Example 7 Car Car travels at 45.0 m/s passes a trooper One second later the trooper sets out to catch the car accelerating at 3.00 m/s 2 How long does it take the trooper to overtake the car? Let t = 0 be 1 second after the car passes the trooper The cars position is then at x i = 45 m  v f = v i + at  x f = x i + v i t + ½at 2  v f 2 = v i 2 + 2a(x f – x i )  x i = 45x f = ?  v i = 45v f = 45  a = 0t = ?  45=45+0tCan’t use this equation  x f =45+45t+½(0)(t) 2 Equation has x f, t  45 2 =45 2 +2(0)(x f –0)Can’t use this equation

31. 1-Dimension Horizontal Motion Example 7 Trooper Car travels at 45.0 m/s passes a trooper One second later the trooper sets out to catch the car accelerating at 3.00 m/s 2 How long does it take the trooper to overtake the car?  v f = v i + at  x f = x i + v i t + ½at 2  v f 2 = v i 2 + 2a(x f – x i )  x i = 0x f = ?  v i = 0v f = ?  a = 3t = ?  v f =0+3tEquation has v f, t  x f =0+0t+½(3)(t) 2 Equation has x f, t  v f 2 =0 2 +2(3)(x f –0)Equation has x f, v f

32. 1-Dimension Horizontal Motion Example 7 Car / Trooper Car travels at 45.0 m/s passes a trooper One second later the trooper sets out to catch the car accelerating at 3.00 m/s 2 How long does it take the trooper to overtake the car?  x f =45+45t+½(0)(t) 2 (from Car equations)  x f =0+0t+½(3)(t) 2 ( from Boy equations)  Substitute the first equation into the second  45+45t = 1.5t  t 2 -30t-30=0  Solve using quadratic equation  t = 31.0 s

33. PHY 151: Lecture 2B Motion in One Dimension 2.7 Freely Falling Objects

34. Galileo Galilei 1564 – 1642 Italian physicist and astronomer Formulated laws of motion for objects in free fall Supported heliocentric universe

35. Freely Falling Objects A freely falling object is any object moving freely under the influence of gravity alone It does not depend upon the initial motion of the object –Dropped – released from rest –Thrown downward –Thrown upward

36. Acceleration of Freely Falling Object - 1 The acceleration of an object in free fall is directed downward, regardless of the initial motion The magnitude of free fall acceleration is g = 9.80 m/s 2 –g decreases with increasing altitude –g varies with latitude –9.80 m/s 2 is the average at the Earth’s surface –The italicized g will be used for the acceleration due to gravity Not to be confused with g for grams

37. Acceleration of Free Fall - 2 We will neglect air resistance Free fall motion is constantly accelerated motion in one dimension –Use model of a particle under constant acceleration Let upward be positive Use the kinematic equations –With a y = -g = -9.80 m/s 2 –Note displacement is in the vertical direction

38. Free Fall – An Object Dropped Initial velocity is zero Let up be positive Use the kinematic equations –Generally use y instead of x since vertical Acceleration is  a y = -g = -9.80 m/s 2 v o = 0 a = -g

39. Free Fall – An Object Thrown Downward a y = -g = -9.80 m/s 2 Initial velocity  0 –With upward being positive, initial velocity will be negative v o ≠ 0 a = -g

40. Free Fall–Object Thrown Upward - 1 Initial velocity is upward, so positive The instantaneous velocity at the maximum height is zero a y = -g = -9.80 m/s 2 everywhere in the motion v = 0 v o ≠ 0 a = -g

41. Thrown upward - 2 The motion may be symmetrical –Then t up = t down –Then v = -v o The motion may not be symmetrical –Break the motion into various parts Generally up and down

42. Free Fall Example Initial velocity at A is upward (+) and acceleration is -g (-9.8 m/s 2 ) At B, the velocity is 0 and the acceleration is -g (-9.8 m/s 2 ) At C, the velocity has the same magnitude as at A, but is in the opposite direction The displacement is –50.0 m (it ends up 50.0 m below its starting point)

43. 1-Dimension Free Fall - Example 1 A bomb is dropped from 6000m With what speed does it hit the ground?  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 = 6000y f = 0  v i = 0v f = ?  a = -9.8t = ?  v f = 0 – 9.8tEquation has v f, t  0 = 6000 + 0t + ½ (-9.8)t 2 Equation has t  v f 2 = 0 2 + 2(-9.8) (0 – 6000)Equation has v f  v f 2 = 0 2 + 2(-9.8)(-6000)  v f = sqrt(117600) = -343 m/s

44. 1-Dimension Free Fall – Example 2a A student drops a ball from the top of a tall building It takes 2.8 s for the ball to reach the ground (a)What is the height of the building?  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 = 2.8  v f = 0 – 9.8(2.8)Equation has v f  0 = y i + 0(2.8) + ½(-9.8)(2.8) 2 Equation has y i  v f 2 = 0 2 + 2(-9.8) (0 – y i )Equation has y i, v f  0 = y i + (1/2)(-9.8)(2.8) 2  y i = 38.4 m

45. 1-Dimension Free Fall – Example 2b A student drops a ball from the top of a tall building It takes 2.8 s for the ball to reach the ground (b)What was the ball’s velocity just before hitting the ground?  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 = 38.4y f = 0  v i = 0v f = ?  a = -9.8t = 2.8  v f = 0 – 9.8(2.8)Equation has v f  0 = 38.4 + 0(2.8) + ½(-9.8)(2.8) 2 Can’t use equation  v f 2 = 0 2 + 2(-9.8) (0 – 38.4)Equation has v f  v f = 0 – 9.8(2.8) = -27.4 m/s(- means downward)

46. 1-Dimension Free Fall – Example 3 Boy throws stone straight upward with an initial velocity of 15 m/s What maximum height will stone reach before falling back down? At the maximum height v y = 0  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 = 15v f = 0  a = -9.8t = ?  0 = 15 – 9.8tEquation has t  y f = 0 + 15t + ½(-9.8)t 2 Equation has y f, t  0 2 = 15 2 + 2(-9.8) (y f – 0)Equation has y f  0 2 = 15 2 + 2(-9.8) (y f – 0)  y f = -15 2 /2/-9.8 = 11.48 m

47. 1-Dimension Free Fall – Example 4 Ball thrown upwards at 40m/s. Calculate time to reach 35m  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  v i = 40v f = ?  a = -9.8t = ?  v f = 40 – 9.8tEquation has v f, t (2 nd )  35 = 0 + 40t + ½(-9.8)t 2 Equation has t  v f 2 = 40 2 + 2(-9.8) (35 – 0)Equation has v f (1st)  v f 2 = 40 2 + 2(-9.8) (35 – 0) = 914  v f = +30.2 or -30.2  30.2 = 40 – 9.8t  t = (30.2 – 40) / (-9.8) = 1.0 s(time on the way up)

48. 1-Dimension Free Fall Example 5 Ball 1 A ball is dropped from rest Four seconds later, a second ball is thrown down at 50m/s Calculate when and where the two balls meet Let t = 0 at 4 seconds. Ball height y=½(-g)t 2 =½(-9.8)(16) = -78.4 m. Speed v = (-g)t = -9.8(4) = -39.2 m/s  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 = -78.4y f = ?  v i = -39.2v f = ?  a = -9.8t = ?  v f = -39.2 – 9.8tEquation has v f, t  y f = -78.4 – 39.2t + ½(-9.8)t 2 Equation has y f, t  v f 2 = (-29.2) 2 +2(-9.8) (y f – (-78.4))Equation has y f, v f

49. 1-Dimension Free Fall Example 5 Ball 2 A ball is dropped from rest Four seconds later, a second ball is thrown down at 50m/s Calculate when and where the two balls meet  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 = -50v f = ?  a = -9.8t = ?  v f = -50–9.8tEquation has v f, t  y f = 0-50t+½(-9.8)t 2 Equation has y f, t  v f 2 = 50 2 +2(-9.8) (y f – 0)Equation has y f, v f

50. 1-Dimension Free Fall Example 5 Balls 1 and 2 A ball is dropped from rest Four seconds later, a second ball is thrown down at 50m/s Calculate when and where two balls meet  y f = -78.4 – 39.2t + ½(-9.8)t 2 (from ball 1)  y f = 0-50t+½(-9.8)t 2 (from ball 2)  Substitute equation 1 into equation 2  -78.4-39.2t+ ½ (-9.8)t 2 = -50t + ½(-9.8)t 2  0 = 10.8t -78.4  t = -78.4 / -10.8 = 7.3 s

51. 1-Dimension Free Fall – Example 6a From tower 100 m high, ball is thrown up with speed of 40 m/s (a)How high does it rise?  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 = 100y f = ?  v i = 40v f = 0  a = -9.8t = ?  0 = 40 – 9.8tEquation has t  y f = 100 + 40t + ½(-9.8)t 2 Equation has y f, t  0 2 = 40 2 +2(-9.8)(y f –100)Equation has y f  0 2 = 40 2 +2(-9.8)(y f –100)  y f = (-1600/2/-9.8) + 100 = 181.6 m

52. 1-Dimension Free Fall – Example 6b From tower 100 m high, ball is thrown up with speed of 40 m/s (b)When does it hit ground?  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 = 100y f = 0  v i = 40v f = ?  a = -9.8t = ?  v f = 40 – 9.8tEquation has v f, t  0 = 100 + 40t + ½(-9.8)t 2 Equation has t  v f 2 = 40 2 +2(-9.8)(0–100)Equation has v f  0 = 100 + 40t + ½(-9.8)t 2  -4.9t 2 + 40t + 100 = 0(Use quadratic equation)  t = 10.2 s

53. 1-Dimension Free Fall – Example 6c From tower 100 m high, ball is thrown up with speed of 40 m/s (c)How fast is it moving at the ground ?  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 = 100y f = 0  v i = 40v f = ?  a = -9.8t = 10.2  v f = 40 – 9.8(10.2)Equation has v f  0 = 100+40(10.2)+½(-9.8)(10.2) 2 Can’t use equation  v f 2 = 40 2 +2(-9.8)(0–100)Equation has v f  v f 2 = 40 2 +2(-9.8)(0–100) = 1600 + 1960 = 3560  v f = -59.7 m/s

54. PHY 151: Lecture 2 Motion in One Dimension 2.8 Kinematic Equations Derived from Calculus

55. Kinematic Equations from Calculus Displacement equals the area under the velocity – time curve The limit of the sum is a definite integral

56. Kinematic Equations – General Calculus Form

57. Example of Solution - 1 Definition of instantaneous velocity a x = dv x /dt a x dt = dv x Integrate, integrals from t = 0 to t  a x dt =  dv x Since a x is constant a x t = v xf - v xi

58. Example of Solution - 2 Multiply by dt and integrate again from t = 0 to t  a x t dt =  (v xf – v xi ) dt =  (dx/dt – v xi )dt ½a x t 2 = x f – x i - v xi t

59. Kinematic Equations – From Integration The integration form of v f – v i gives The integration form of x f – x i gives