2. KINEMATICS. By Liew Sau Poh

Transcription

1 2. KINEMATICS By Liew Sau Poh 1

2 OBJECTIVES 2.1 Linear motion 2.2 Projectiles 2.3 Free falls and air resistance 2

3 OUTCOMES Derive and use equations of motion with constant acceleration Sketch and use the graphs of displacementtime, velocity-time and acceleration-time for the motion of a body with constant acceleration Solve problems on projectile motion without air resistance Explain the effects of air resistance on the motion of bodies in air 3

4 2.1 LINEAR MOTION Derive and use equations of motion with constant acceleration Sketch and use the graphs of displacement-time, velocity-time and acceleration-time for the motion of a body with constant acceleration 4

5 2.1 LINEAR MOTION Linear Motion: The object is moving along a straight line Linear Motion with constant velocity:... at constant velocity Linear Motion with constant acceleration: at a constant change rate of velocity. 5

6 MOTION WITH CONSTANT ACCELERATION Derivation: Let u = initial velocity, v = final velocity after a time, t, then, from the definition, The uniform acceleration a = (change of velocity)/(time taken) a = (v u) / t (1) v = u + at (2) 6

7 MOTION WITH CONSTANT ACCELERATION v = u + at (2) Let s = displacement of the body in time t, then s = (average velocity) (time) s = [(u + v)/2] t s = ½ (u + v)t (3) (2) (3): s = ½ [u + (u + at)]t s = ut + ½ at 2 (4) 7

8 MOTION WITH CONSTANT ACCELERATION v = u + at (2) s = ½ (u + v)t (3) From (2): v u = at (5) v + u = 2s/t (6) (6) (5): (v + u)( v u ) = (2s/t)(at) v 2 u 2 = 2as v 2 = u 2 + 2as (7) 8

9 MOTION WITH CONSTANT ACCELERATION Reminder: a = (v u) / t v = u + at s = ½ (u + v)t s = ut + ½ at 2 v 2 = u 2 + 2as 9

10 EXAMPLE 1 1. An object which is initially at rest at the centre of coordinate 0 (x = 0) starts to move along the x-axis with constant acceleration. It moves to point P and then to point Q at velocity 15 m s 1 and 25 m s 1 respectively. The distance PQ = 100 m. Determine (a) the acceleration of the object, (b) the distance OP. 10

11 ANSWER 1 U - 0 S 1 V 1 = 15 ms -1 V 2 = 25 ms -1 O S 2 = S (a) Use v 2 = u 2 + 2as for motion between P and Q, 25 2 = a(s 2 s 1 ); 25 2 = a(100) a = 2 m s -2 (b) Use v 2 = u 2 + 2as for motion between O and P, 15 2 = 0 + 2(2) s 1 s 1 = 56.3 m P 100 m Q 11

12 EXAMPLE 2 2. An object moves along a straight line with constant acceleration. Its initial velocity is 20 m s 1. After 5.0 s, the velocity becomes 40 m s 1. Determine the distance travelled during the third second. 12

13 ANSWER 2 U = 20 m s -1 S 1 S 2 v = 40 m s -1 t 1 = 2 s t 2 = 3 s Use v = u + at 40 = 20 + a(5) a = 4 m s -2 Use s = ut + ½ at 2 Third second s 1 = 20t 1 + ½ (4)t 2 1 = 20t 1 + 2t 2 1 (1) s 2 = 20t 2 + ½ (4)t 2 2 = 20t 2 + 2t 2 2 (2) (2) (1): s 2 -s 1 = 20(t 2 t 1 ) + 2(t 2 2 t 12 ) = 30 m 13

14 EXAMPLE 3 3. An object moves along a straight line with constant acceleration. At a particular moment, the velocity of the object is 30 m s -1. After travelling through 50 m, the object has velocity 20 m s 1. How far does the object have to travel before it comes to a stop? 14

15 ANSWER 3 U = 30 ms -1 v = 20 ms -1 P 50 m Q s R Motion from P to Q: Use v 2 = u 2 + 2as 20 2 = a(50) a = -5 m s -2 Motion from Q to R: Use v 2 = u 2 + 2as 0 = (-5)s s = 40 m 15

16 EXAMPLE 4 4. A car which is initially at rest starts to move along a straight line with constant acceleration. It reaches a velocity of 60 m s 1 after travelling through a distance of 100 m. Determine (a) the acceleration, (b) the time taken to reach the velocity of 60 m s 1, (c) the velocity at the third second. 16

17 ANSWER 4 (a) Use v 2 = u 2 + 2as 60 2 = 0 + 2a(100) a = 18 m s -2 (b) Use v = u + at 60 = t t = 3.3 s (c) Use v = u + at = 0 + (18) (3) = 54 m s -1 17

18 GRAPHICAL REPRESENTATION Displacement-time graph (s-t) S / m t / s 18

19 DISPLACEMENT-TIME GRAPH (S-T) S / m t / s Note that the instantaneous velocity, v = ds/dt = gradient of the graph at time t. The instantaneous velocity at time 4s, v = (18-5)/(4-0) = 13/4 = 3.25 ms

20 DISPLACEMENT-TIME GRAPH AZ (S-T) S / m t / s At time 0s to 1s, the velocity is constant; from time 1s to 3s, the velocity is increasing and from time 3s to 5s, the velocity is decreasing. 20

21 DISPLACEMENT-TIME GRAPH (S-T) S / m Negative gradient at time 6s onward shows that the object is moving in the opposite direction t / s 21

22 VELOCITY-TIME GRAPH (V-T) v / ms t / s Note that the acceleration, a = dv/dt = gradient of graph. The acceleration at time 0.5 s is given by a = (20-0)/(2-0) = 10 ms -2 22

23 VELOCITY-TIME GRAPH (V-T) Time Acceleration 0 s to 1 s constant 3s to 5s decreasing 5s to 7s 0 ms -2 (Constant) 7s to 8s v / ms deceleration t / s 23

24 VELOCITY-TIME GRAPH (V-T) The displacement within a period can be found by finding the area under the graph. Displacement between t = 0 and t = 1, s 1 = ½ 1 20 = 10 m Displacement between t = 1 and t = 4, s2 = 1 4 vdt v / ms t / s 24

25 EXAMPLE 5 A body moves along the x-axis. Assume that a positive sign represents a direction to the right. The velocity v of the body is related to time t through the equation v =2 3t 2 where v and t are measured in m s -1 and s respectively. t = 0 when x = 0. Determine (a) the displacement, (b) the acceleration, at the instant of time t = 1 s. 25

26 ANSWER 5 (a) v = ds/dt = 2 3t 2 s = 01 (2 3t 2 )dt = [2t t 3 ] 1 0 = 1.0 m (b) a = dv/dt = -6t = -6(1) = -6 m s -2 26

27 ACCELERATION-TIME GRAPH (A-T) a / ms Constant acceleration t / s The change in velocity within a period can be found by finding the area under the graph. Change in velocity between t = 2 and t = 6, v = 15 (6 2) = 60 ms -1 OR, v = 2 6 a dt = dt = [15a]2 6 = 15 (6-2) = 60ms -1 27

28 2.2 PROJECTILES Solve problems on projectile motion without air resistance 28

29 2.2 PROJECTILE (PROJECTION) A projectile is a body (object) which is projected(or thrown) with some initial velocity, and then allowed to be acted upon by the forces of gravity and possible drag. E.g. Baseball being thrown, water fountains, fireworks displays, soccer ball being kicked & ballistics testing. 29

30 2.2 PROJECTILE (PROJECTION) h v o projectile trajectory x Definition: A projectile is an object that is projected at an angle to the horizontal and moves under the action of gravity. The path of a projectile is called its trajectory. 30

31 2.2 PROJECTILE (PROJECTION) h v o trajectory x projectile Height, h is the maximum upward distance reached by the projectil Range, x is horizontal distance travelled (or sometimes distance). 31

32 2.2 PROJECTILE (PROJECTION) h v o projectile trajectory x If a body is allowed to free-fall under gravity and is acted upon by the drag of air resistance, it reaches a maximum downward velocity known as the terminal velocity. 32

33 2.2 PROJECTILE (PROJECTION) h v o projectile The study of the motion of projectiles is called ballistics. trajectory x 33

34 2.2 PROJECTILE (PROJECTION) An object falling freely towards the ground is exerted by gravitational force, g = +9.8ms -2. For object falling freely without resistant, we have v = u + gt s = ut + ½ gt 2 v 2 = u 2 + 2gs 34

35 2.2 PROJECTILE (PROJECTION) For an object moves upwards against the gravity, g = 9.8ms -2. In this case, the displacement and velocity that pointing upwards have positive magnitude, and the magnitude is negative while pointing downwards. 35

36 2.2 PROJECTILE (PROJECTION) In general, an object falling freely to ground experience inconsistence air resistance, F a. The net force on the object, F = weight of object air resistance F = mg F a 36

37 2.2 PROJECTILE (PROJECTION) The magnitude of F a is not consistence but increases with the speed of the object. When F a = mg, the net force, F = 0 and therefore the velocity of the object becomes constant. 37

38 2.2 PROJECTILE (PROJECTION) Since the acceleration of the object is given by a = g - F a /m, the acceleration becomes zero when F a increases with time. 38

39 2.2 PROJECTILE (PROJECTION) Projectile motion is a combination of uniform motion along x and uniformly accelerated motion (free fall) along y. = Sum of 2 independent motions Ignoring air resistance, horizontal motion, x has constant velocity, while vertical motion, y is accelerated by gravitational force. 39

40 2.2 PROJECTILE (PROJECTION) v y0 v y v v x v x v x0 v y 40

41 2.2 PROJECTILE (PROJECTION) Vertical component, v y Net velocity, v Horizontal component, v x Along x, the projectile travels with constant velocity, where v x = v xo and x = v xo t 41

42 2.2 PROJECTILE (PROJECTION) Vertical component, v y Net velocity, v Horizontal component, v x Along y, the projectile travels in free-fall fashion, where v y = v yo gt, y = v yo t (1/2) gt 2 and g = 9.8 m/s 2 42

43 SUMMARY OF PROJECTILE MOTION Motion Horizontal Vertical Initial Velocity u cos u sin Acceleration 0 -g Time taken t t Distance moved x = ut cos y = ut sin - gt 2 /2 43

44 SUMMARY OF PROJECTILE MOTION v y = u sin v = v x + v y v x = u cos Greatest height, h = (u sin ) 2 / 2g at t = (u sin )/g Time of flight, T = (2 u sin ) / g 44

45 SUMMARY OF PROJECTILE MOTION v y = u cos v = v x + v y v x = u sin Trajectory, y = x tan - gx 2 /(2u 2 cos 2 ) Or y 2 = 2x(ucos ) 2 /g Range, R = (u 2 sin 2 ) / g Maximum horizontal range, R max = u 2 / g When =

46 2.3 FREE FALLS AND AIR RESISTANCE Explain the effects of air resistance on the motion of bodies in air 46

47 FALLING OBJECTS What will happen as these two objects fall? 47

48 AIR RESISTANCE In air A stone falls faster than a feather Air resistance affects stone less In a vacuum A stone and a feather will fall at the same speed 48

49 AIR RESISTANCE Effects of air resistance A person in free fall reaches a terminal velocity of around 54 m/s With a parachute, terminal velocity is only 6.3 m/s Allows a safe landing 49

50 GALILEO GALILEI Applied scientific method formulated the laws that govern the motion of objects in free fall Also looked at: Inclined planes Relative motion Thermometers Pendulum 50

51 FREE FALL All objects moving under the influence of gravity only are said to be in free fall Free fall does not depend on the object s original motion All objects falling near the earth s surface fall with a constant acceleration due to gravity Acceleration due to gravity is independent of mass 51

52 ACCELERATION DUE TO GRAVITY Symbolized by g g = 9.80 m/s² When estimating, use g» 10 m/s 2 g is always directed downward Toward the center of the earth Ignoring air resistance and assuming g doesn t vary with altitude over short vertical distances, free fall is constantly accelerated motion 52

53 FREE FALL Definition: motion when gravity is the only force acting on an object This is hard to achieve in the classroom, laboratory, and even outdoors because of Air Resistance. For our working definition, assume little to no Air Resistance. 53

54 FREE FALL AN OBJECT DROPPED Initial velocity is zero Up is generally taken to be positive In the kinematic equations, generally use y instead of x v o = 0 a = g 54

55 FREE FALL AN OBJECT THROWN DOWNWARD Free upon release With up being positive, initial velocity will be negative v o = 0 a = g 55

56 FREE FALL - OBJECT THROWN UPWARD Initial velocity is upward, so positive The instantaneous velocity at the maximum height is zero a = g = m/s 2 everywhere in the motion v = 0 56

57 THROWN UPWARD, CONT. 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 57

58 NON-SYMMETRICAL FREE FALL Need to divide the motion into segments Possibilities include Upward and downward portions The symmetrical portion back to the release point and then the nonsymmetrical portion 58

59 COMBINATION MOTIONS Example: Launching of rocket Phase 1: a = 29.4ms -2 (fuel burns out) Phase 2: a = ms -2 (rocket falls freely after reaches maximum height) 59

60 EXAMPLE 6: FREE FALL A ball is dropped from rest from the top of a building. Find: a) The instantaneous velocity of the ball after 6 sec. b) How far the ball fell. c) The average velocity up to that point. Answers: -60m/s, 180m, -30m/s 60

61 EXAMPLE 7: FREE FALL ON THE MOON A hammer is dropped on the moon. It reaches the ground 1s later. If the distance it fell was 0.83m: a) Calculate the acceleration due to gravity on the surface of the moon. b) Calculate the velocity with which the hammer reached the ground and compare to the velocity it would have, if it was dropped on the earth s surface. Answer: -1.66m/s 2, -1.66m/s, -9.8m/s 61