1. 1 Chapter 2 Unit Vector Vector addition by components Vector addition by graphing

2. 2 Displacement (Vector)

3. 3 Speed & Velocity Average and Instantaneous Average Speed = Distance/ Elapsed time ( How fast it moves, but no direction) Units and dimensions

4. 4 a) m/s b) m/s Time to go to Florida from Atlanta Average Velocity? Average Velocity

5. 5 Instantaneous Velocity Smaller and smaller gives smaller and smaller As  0 it gives the limiting value ( In general instantaneous values are given unless noted otherwise.) What if changes ( push the accelerator pedal )

6. 6 Acceleration Rate of change of velocity is acceleration Average acceleration

7. 7 Equation of Kinematics Constant acceleration (straight line- motion ) Say X 0 and t 0 = 0

8. 8 Constant acceleration Velocity increases What is the average velocity?

9. 9 If starts at and with constant acceleration.

10. 10 If you do not know “t”

11. 11 Summary Equation Variables numberEquation xavv0v0 t (2.4) v=v 0 +at ----- (2.7) x=(1/2)(v 0 +v)t ----- (2.8) x=v 0 t+(1/2)at 2 ----- (2.9) v 2 =v 0 2 +2ax ----- Table 2.1 Equations of Kinematics for Constant Acceleration

12. 12 Application of the Equations Draw the situation Pick +ve and –ve direction and co- ordinates Write the known values (with signs) Look for hidden information Enough information, what is unknown? If two objects, are they connected ?

13. 13 Free FALL No air resistanceConstant acceleration due to gravity g= 9.80 m/s 2 or 32.2 ft/s 2 (on earth surface) Gallileo’s testLeaning tower (better on moon) Air filled tubeEvacuated tube

14. 14 Example 10 A Falling Stone After 3.0 sec what is the displacement v 0 =0 m/s t=3.0 sec Known : v 0 = 0 a = g (i.e.) t = 3.0 y = v 0 t+(1/2)at 2 v = v 0 +at v 2 = v 0 2 +2ay y = v 0 t+(1/2)at 2 = 0+(1/2)(-9.8) * 9 = (-44.1m) Which eq n to use?

15. 15 What is the velocity after 3.0 sec? v=v 0 +at v 2 =v 0 2 +2ay Can use both----why? Which is easier? v = -9.8m/s 2 * 3sec = -29.4ms -ve sign indicate

16. 16 Example 12 How high the coin reach, if it is tossed up with an initial speed of 5.00 m/s? At the highest point the velocity should be zero. v 0 =5.0 m/s, a=-9.8 m/s 2 v=0.0 m/s, y=? v 2 =v 0 2 +2ay

17. 17 How long will it take to reach the top ? v=0, v 0 =5, a= -9.8 How much time it takes to come to the same level as it was thrown ? v 0 =5.0 a=-9.8 y=0 t=? Which eq n ? v=v 0 +at 0=5-9.8t t=0.5/sec

18. 18 t=0 or i.e.t=0 or Why two values? Explain each answer.

19. 19 symmetry No air resistance—acceleration is the same for the motion in both directions Velocity vector changes continuously Acceleration does not. Time to reach the highest point = time to reach the ground At same height, speed is the same.

20. 20 Bullet reaches the ground with the same speed irrespective of which direction it was fired first.

21. 21 150 m Lets say Show the speed of the bullet when it comes down to the same height. What is the velocity if shot up with initial velocity 30 m/s under gravity when it reaches the same level? Known information: v 0 =30, y=0, Which eq n ? a=-g v 2 =v 0 2 +2ay v 2 =(30) 2 +2(-9.8) * 0 v=+/-30 What is the significance of + and – sign?

22. 22 Harder way How high it reaches before changing direction? v 0 = 30, a=-g, v=?( hidden information ) y=?( needs to find) Which eq n ? v 2 =v 0 2 +2ay 0=(30) 2 +2(-9.8) * y y=(30 * 30)/(2 * 9.8)=45.92 m

23. 23 v 0 =0 v=? y=? For the downward motion, v 2 =v 0 2 +2ay =0+2(+9.8) * [(30 * 30)/(2 * 9.8)] (why +ve ?) =30 * 30 v=30 m/s (again +/-) Find the time to reach max height and then find the speed when it reaches the same level.

24. 24 What is the velocity of the bullet when it reaches the ground if we start from the mountain? v=? v 0 = 30, a=-9.8, y=-150 (why –ve ?) v 2 =v 0 2 +2ay v 2 =(30) 2 +2(-9.8)(-150) =(30) 2 +2.94 * 10 3 =3.84 * 10 3 v=61.97 m/s (which sign?)

25. 25 If you drop down with 30 m/s from the mountain, what is the speed? v 0 = -30, g=-9.8, y=-150 v 2 =v 0 2 +2ay =(30) 2 +2(-9.8)(-150) =(30) 2 +2.94 * 10 3 =3.84 * 10 3 v=61.97 m/s

26. 26 Example t=0 v 0 t=t 2v02v0 v=v 0 +at 0=v 0 +at v=2v 0 +at 1 0=2v 0 +at 1

27. 27 Answer-- 2t, 4y

28. 28 Graphical analysis x=0 and t=0 Slope =

29. 29 Example 16 Bicycle Trip

30. 30 1 st segment 2 nd segment 3 rd segment

31. 31 If velocity changes (i.e. acceleration) (assume v 0 =0 for simplicity) Slope of the tangent Instantaneous velocity v=v 0 +at

32. 32 Conceptual questions 2 REASONING AND SOLUTION The buses do not have equal velocities. Velocity is a vector, with both magnitude and direction. In order for two vectors to be equal, they must have the same magnitude and the same direction. The direction of the velocity of each bus points in the direction of motion of the bus. Thus, the directions of the velocities of the buses are different. Therefore, the velocities are not equal, even though the speeds are the same.

33. 33 Conceptual question 4 REASONING AND SOLUTION Consider the four traffic lights 1, 2, 3 and 4 shown below. Let the distance between lights 1 and 2 be x 12, the distance between lights 2 and 3 be x 23, and the distance between lights 3 and 4 be x 34.

34. 34 The lights can be timed so that if a car travels with a constant speed v, red lights can be avoided in the following way. Suppose that at time t = 0 s, light 1 turns green while the rest are red. Light 2 must then turn green in a time t 12, where t 12 = x 12 /v. Light 3 must turn green in a time t 23 after light 2 turns green, where t 23 = x 23 /v. Likewise, light 4 must turn green in a time t 34 after light 3 turns green, where t 34 = x 34 /v. Note that the timing of traffic lights is more complicated than indicated here when groups of cars are stopped at light 1. Then the acceleration of the cars, the reaction time of the drivers, and other factors must be considered.

35. 35 Conceptual question 5 REASONING AND SOLUTION The velocity of the car is a vector quantity with both magnitude and direction The speed of the car is a scalar quantity and has nothing to do with direction. It is possible for a car to drive around a track at constant speed. As the car drives around the track, however, the car must change direction. Therefore, the direction of the velocity changes, and the velocity cannot be constant. The incorrect statement is (a).

36. 36 Conceptual question 14 REASONING AND SOLUTION The magnitude of the muzzle velocity of the bullet can be found (to a very good approximation) by solving Equation 2.9, With v 0 = 0 m/s; that is where a is the acceleration of the bullet and x is the distance traveled by the bullet before it leaves the barrel of the gun (i.e., the length of the barrel).

37. 37 Since the muzzle velocity of the rifle with the shorter barrel is greater than the muzzle velocity of the rifle with the longer barrel, the product ax must be greater for the bullet in the rifle with the shorter barrel. But x is smaller for the rifle with the shorter barrel, thus the acceleration of the bullet must be larger in the rifle with the shorter barrel.

38. 38 Problem 4 REASONING The distance traveled by the Space Shuttle is equal to its speed multiplied by the time. The number of football fields is equal to this distance divided by the length L of one football field. SOLUTION The number of football fields is

39. 39 Problem 8 REASONING AND SOLUTION Let west be the positive direction. The average velocity of the backpacker is.

40. 40 Combining these equations and solving for x e (suppressing the units) gives The distance traveled is the magnitude of x e, or

41. 41 Problem 8 solution (2) N S EW y x O N S EW y x O 6.44 km A Time for 1 st segment O A is 6.44 km with an velocity of 2.68 m/s =say t 1 N S EW y x O 6.44 km A x B (X in meters) A B is x m with an velocity of.447m/s =say t 2 Time for 2 nd segment Time for displacement(O B) with an average velocity of 1.34m/s =say t

42. 42 t 1 +t 2 =t

43. 43 Problem 10 Reasoning The definition of average velocity is given by Equation 2.2 as Average velocity = Displacement/(Elapsed time). The displacement in this expression is the total displacement, which is the sum of the displacements for each part of the trip. Displacement is a vector quantity, and we must be careful to account for the fact that the displacement in the first part of the trip is north, while the displacement in the second part is south.

44. 44 Solution According to Equation 2.2, the displacement for each part of the trip is the average velocity for that part times the corresponding elapsed time. Designating north as the positive direction, we find for the total displacement that +27 m/s for t n xnxn -17 m/s xsxs Displacement = x n + x s

45. 45 Where t North and t South denote, respectively, the times for each part of the trip. Note that the minus sign indicates a direction due south. Noting that the total elapsed time is t North + t South we can use Equation 2.2 to find the average velocity for the entire trip as follows:

46. 46 But Therefore, we have that The plus sign indicates that the average velocity for the entire trip points north.

47. 47 Problem 19 REASONING AND SOLUTION Uniform acceleration, starts from rest. x=v 0 t+(1/2)at 2

48. 48 Problem 23 REASONING We know the initial and final velocities of the blood, as well as its displacement. Therefore, Equation 2.9 can be used to find the acceleration of the blood. The time it takes for the blood to reach it final velocity can be found by using Equation 2.7

49. 49 SOLUTION a.) The acceleration of the blood is b.) The time it takes for the blood, starting from 0 cm/s, to reach a final velocity of +26 cm/s is

50. 50 Problem 28 REASONING AND SOLUTION The speed of the car at the end of the first (402 m) phase can be obtained as follows: v 1 2 = v o 2 + 2a 1 x 1 The speed after the second phase (3.50  10 2 m) can be obtained in a similar fashion. v 2 2 = v 02 2 + 2a 2 x 2 v 2 =96.9 m/s

51. 51 Problem 39 REASONING The initial velocity and the elapsed time are given in the problem. Since the rock returns to the same place from which it was thrown, its displacement is zero (y = 0 m). Using this information, we can employ Equation 2.8 y = v 0 t+(1/2)at 2 to determine the acceleration a due to gravity. SOLUTION Solving Equation 2.8 for the acceleration yields

52. 52 Problem 44 REASONING AND SOLUTION a. The minus is chosen, since the diver is now moving down. Hence, v=-7.9 m/s.

53. 53 b. The diver's velocity is zero at his highest point. The position of the diver relative to the board is The position above the water is 3.0 m + 0.17 m = 3.2 m/s

54. 54 Problem 48 t t 0 =0v 0 =25.0 m/s t t 0 =1.2 secv 0 =? v=v 0 +at 0=25-9.8t t=25/9.8=2.55 sec v=v 0 +at 0=v 0 -9.8(t-1.2) v 0 = 9.8(t-1.2)= 9.8(2.55-1.2) =9.8 * 1.35=13.23 m/s

55. 55 Problem 51 v 0 =0 75 m 5.0 m/s x=? What is the time for stone to hit the water(log)? y = v 0 t+(1/2)at 2 (-75)=(1/2) * (-9.8)t 2 t 2 =2 * 75/9.8=15.3 sec If the log is moving with constant velocity during this time, the log should have moved. x = vt=5 * 3.9=19.5 m

56. 56 Problem 55 REASONING AND SOLUTION The balls pass at a time t when both are at a position y above the ground. Applying Equation 2.8 to the ball that is dropped from rest, we have (1) Note that we have taken into account the fact that y = 24 m when t = 0 s in Equation (1). For the second ball that is thrown straight upward, Equating Equations (1) and (2) for y yields

57. 57 Thus, the two balls pass at a time t, where The initial speed v 02 of the second ball is exactly the same as that with which the first ball hits the ground. To find the speed with which the first ball hits the ground, we take upward as the positive direction and use Equation 2.9 v 2 =v 0 2 +2ay. Since the first ball is dropped from rest, we find that

58. 58 At a time t = 1.11 s, the position of the first ball according to Equation (1) is which is 6.0m below the top of the cliff. Thus, the balls pass after a time