Presentation on theme: "CHAPTER 2: Kinematics of Linear Motion (5 hours)"— Presentation transcript:
1CHAPTER 2: Kinematics of Linear Motion (5 hours) SF017CHAPTER 2: Kinematics of Linear Motion (5 hours)
22.0 Kinematics of Linear motion SF0172.0 Kinematics of Linear motionis defined as the studies of motion of an objects without considering the effects that produce the motion.There are two types of motion:Linear or straight line motion (1-D)with constant (uniform) velocitywith constant (uniform) acceleration, e.g. free fall motionProjectile motion (2-D)x-component (horizontal)y-component (vertical)
3Learning Outcomes : 2.1 Linear Motion (1 hour) SF017Learning Outcomes :2.1 Linear Motion (1 hour)At the end of this chapter, students should be able to:Define and distinguish betweenDistance and displacementSpeed and velocityInstantaneous velocity, average velocity and uniform velocityInstantaneous acceleration, average acceleration and uniform acceleration,Sketch graphs of displacement-time, velocity-time and acceleration-time.Determine the distance travelled, displacement, velocity and acceleration from appropriate graphs.
42.1. Linear motion (1-D) 2.1.1. Distance, d scalar quantity. SF0172.1. Linear motion (1-D)Distance, dscalar quantity.is defined as the length of actual path between two points.For example :The length of the path from P to Q is 25 cm.PQ
52.1.2 Displacement, Example 2.1 : vector quantity. SF017Displacement,vector quantity.is defined as the distance between initial point and final point in a straight line.The S.I. unit of displacement is metre (m).Example 2.1 :An object P moves 30 m to the east after that 15 m to the southand finally moves 40 m to west. Determine the displacement of Prelative to the original position.Solution :NEWSOP30 m15 m10 m
62.1.3 Speed, v The magnitude of the displacement is given by SF017The magnitude of the displacement is given byand its direction is2.1.3 Speed, vis defined as the rate of change of distance.scalar quantity.Equation:
72.1.4 Velocity, is a vector quantity. SF017Velocity,is a vector quantity.The S.I. unit for velocity is m s-1.Average velocity, vavis defined as the rate of change of displacement.Equation:Its direction is in the same direction of the change in displacement.
8Instantaneous velocity, v SF017Instantaneous velocity, vis defined as the instantaneous rate of change of displacement.Equation:An object moves in a uniform velocity whenand the instantaneous velocity equals to the average velocity at any time.
9Gradient of s-t graph = velocity SF017Thereforests1t1The gradient of the tangent to the curve at point Q= the instantaneous velocity at time, t = t1QGradient of s-t graph = velocity
102.1.5 Acceleration, vector quantity. SF017Acceleration,vector quantity.The S.I. unit for acceleration is m s-2.Average acceleration, aavis defined as the rate of change of velocity.Equation:Its direction is in the same direction of motion.The acceleration of an object is uniform when the magnitude of velocity changes at a constant rate and along fixed direction.
11Instantaneous acceleration, a SF017Instantaneous acceleration, ais defined as the instantaneous rate of change of velocity.Equation:An object moves in a uniform acceleration whenand the instantaneous acceleration equals to the average acceleration at any time.
12Gradient of v-t graph = acceleration SF017Deceleration, ais a negative acceleration.The object is slowing down meaning the speed of the object decreases with time.ThereforevtQv1t1The gradient of the tangent to the curve at point Q= the instantaneous acceleration at time, t = t1Gradient of v-t graph = acceleration
132.1.6 Graphical methods s s t t s t SF017Graphical methodsDisplacement against time graph (s-t)ststGradient increases with timeGradient = constantst(a) Uniform velocity(b) The velocity increases with time(c)QGradient at point R is negative.RPThe direction of velocity is changing.Gradient at point Q is zero.The velocity is zero.
14Area under the v-t graph = displacement SF017Velocity versus time graph (v-t)The gradient at point A is positive – a > 0(speeding up)The gradient at point B is zero – a= 0The gradient at point C is negative – a < 0(slowing down)t1t2vt(a)t2t1vt(b)t1t2vt(c)BUniform accelerationCUniform velocityAArea under the v-t graph = displacement
15From the equation of instantaneous velocity, SF017From the equation of instantaneous velocity,ThereforeSimulation 2.1Simulation 2.2Simulation 2.3
16SF017Example 2.2 :A toy train moves slowly along a straight track according to the displacement, s against time, t graph in Figure 2.1.a. Explain qualitatively the motion of the toy train.b. Sketch a velocity (cm s-1) against time (s) graph.c. Determine the average velocity for the whole journey.d. Calculate the instantaneous velocity at t = 12 s.e. Determine the distance travelled by the toy train.2468101214t (s)s (cm)Figure 2.1
17v (cm s1) t (s) Solution : SF017Solution :a. 0 to 6 s : The train moves at a constant velocity of6 to 10 s : The train stops.10 to 14 s : The train moves in the same direction at a constant velocity ofb.2468101214t (s)0.681.50v (cm s1)
18e. The distance travelled by the toy train is 10 cm. SF017Solution :c.d.e. The distance travelled by the toy train is 10 cm.
19SF017Example 2.3 :A velocity-time (v-t) graph in Figure 2.2 shows the motion of a lift.a. Describe qualitatively the motion of the lift.b. Sketch a graph of acceleration (m s2) against time (s).c. Determine the total distance travelled by the lift and itsdisplacement.d. Calculate the average acceleration between 20 s to 40 s.5101520253035t (s)-4-224v (m s1)Figure 2.2404550
2015 to 20 s : Lift moving with constant velocity of SF017Solution :a. 0 to 5 s : Lift moves upward from rest with a constant acceleration of5 to 15 s : The velocity of the lift increases from 2 m s1 to 4 m s1 but the acceleration decreasing to15 to 20 s : Lift moving with constant velocity of20 to 25 s : Lift decelerates at a constant rate of25 to 30 s : Lift at rest or stationary.30 to 35 s : Lift moves downward with a constant acceleration of35 to 40 s : Lift moving downward with constant velocityof40 to 50 s : Lift decelerates at a constant rate of and comes to rest.
21a (m s2) t (s) Solution : b. 5 10 15 20 25 30 35 40 45 50 -0.4 -0.2 SF017Solution :b.t (s)5101520253035404550-0.4-0.20.20.6a (m s2)-0.6-0.80.80.4
22v (m s1) t (s) Solution : c. i. 5 10 15 20 25 30 35 -4 -2 2 4 40 45 SF017Solution :c. i.5101520253035t (s)-4-224v (m s1)404550A2A3A1A4A5
24SF017Exercise 2.1 :Figure 2.3 shows a velocity versus time graph for an object constrained to move along a line. The positive direction is to the right.a. Describe the motion of the object in 10 s.b. Sketch a graph of acceleration (m s-2) against time (s) forthe whole journey.c. Calculate the displacement of the object in 10 s.ANS. : 6 mFigure 2.3
25SF017Exercise 2.1 :A train pulls out of a station and accelerates steadily for 20 s until its velocity reaches 8 m s1. It then travels at a constant velocity for 100 s, then it decelerates steadily to rest in a further time of 30 s.a. Sketch a velocity-time graph for the journey.b. Calculate the acceleration and the distance travelled in each part of the journey.c. Calculate the average velocity for the journey.Physics For Advanced Level, 4th edition, Jim Breithaupt, Nelson Thornes, pg.15, no. 1.11ANS. : 0.4 m s2,0 m s2, m s2, 80 m, 800 m, 120 m;6.67 m s1.
26Learning Outcome : 2.2 Uniformly accelerated motion (1 hour) SF017Learning Outcome :2.2 Uniformly accelerated motion (1 hour)At the end of this chapter, students should be able to:Derive and apply equations of motion with uniform acceleration:
272.2. Uniformly accelerated motion SF0172.2. Uniformly accelerated motionFrom the definition of average acceleration, uniform (constant) acceleration is given bywhere v : final velocityu : initial velocitya : uniform (constant) accelerationt : time(1)
28SF017From equation (1), the velocity-time graph is shown in Figure 2.4 :From the graph,The displacement after time, s = shaded area under the graph= the area of trapeziumHence,velocityvutimetFigure 2.4(2)
29By substituting eq. (1) into eq. (2) thus SF017By substituting eq. (1) into eq. (2) thusFrom eq. (1),From eq. (2),(3)multiply(4)
30Therefore the equations (2) and (3) can be written as SF017Notes:equations (1) – (4) can be used if the motion in a straight line with constant acceleration.For a body moving at constant velocity, ( a = 0) the equations (1) and (4) becomeTherefore the equations (2) and (3) can be written asconstant velocity
31SF017Example 2.4 :A plane on a runway accelerates from rest and must attain takeoff speed of 148 m s1 before reaching the end of the runway. The plane’s acceleration is uniform along the runway and of value 914 cm s2. Calculatea. the minimum length of the runway required by the plane totakeoff.b. the time taken for the plane cover the length in (a).Solution :a. Use
32b. By using the equation of linear motion, SF017Solution :b. By using the equation of linear motion,OR
33SF017Example 2.5 :A bus travelling steadily at 30 m s1 along a straight road passes a stationary car which, 5 s later, begins to move with a uniform acceleration of 2 m s2 in the same direction as the bus. Determinea. the time taken for the car to acquire the same velocity as thebus,b. the distance travelled by the car when it is level with the bus.Solution :a. GivenUse
35SF017Example 2.6 :A particle moves along horizontal line according to the equationWhere s is displacement in meters and t is time in seconds.At time, t = 3 s, determinea. the displacement of the particle,b. Its velocity, andc. Its acceleration.Solution :a. t =3 s ;
36b. Instantaneous velocity at t = 3 s, Use SF017Solution :b. Instantaneous velocity at t = 3 s,UseThus
37c. Instantaneous acceleration at t = 3 s, Use SF017Solution :c. Instantaneous acceleration at t = 3 s,UseHence
38SF017Exercise 2.2 :A speedboat moving at 30.0 m s-1 approaches stationary buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of m s-2 by reducing the throttle.a. How long does it take the boat to reach the buoy?b. What is the velocity of the boat when it reaches the buoy?No. 23,pg. 51,Physics for scientists and engineers with modern physics, Serway & Jewett,6th edition.ANS. : 4.53 s; 14.1 m s1An unmarked police car travelling a constant 95 km h-1 is passed by a speeder traveling 140 km h-1. Precisely 1.00 s after the speeder passes, the policemen steps on the accelerator; if the police car’s acceleration is 2.00 m s-2, how much time passes before the police car overtakes the speeder (assumed moving at constant speed)?No. 44, pg. 41,Physics for scientists and engineers with modern physics, Douglas C. Giancoli,3rd edition.ANS. : 14.4 s
39SF017Exercise 2.2 :A car traveling 90 km h-1 is 100 m behind a truck traveling 75 km h-1. Assuming both vehicles moving at constant velocity, calculate the time taken for the car to reach the truck.No. 15, pg. 39,Physics for scientists and engineers with modern physics, Douglas C. Giancoli,3rd edition.ANS. : 24 sA car driver, travelling in his car at a constant velocity of m s-1, sees a dog walking across the road 30 m ahead. The driver’s reaction time is 0.2 s, and the brakes are capable of producing a deceleration of 1.2 m s-2. Calculate the distance from where the car stops to where the dog is crossing, assuming the driver reacts and brakes as quickly as possible.ANS. : 1.73 m
40Learning Outcome : 2.3 Freely falling bodies (1 hour) SF017Learning Outcome :2.3 Freely falling bodies (1 hour)At the end of this chapter, students should be able to:Describe and use equations for freely falling bodies.For upward and downward motion, usea = g = 9.81 m s2
41SF0172.3 Freely falling bodiesis defined as the vertical motion of a body at constant acceleration, g under gravitational field without air resistance.In the earth’s gravitational field, the constant accelerationknown as acceleration due to gravity or free-fall acceleration or gravitational acceleration.the value is g = 9.81 m s2the direction is towards the centre of the earth (downward).Note:In solving any problem involves freely falling bodies or free fall motion, the assumption made is ignore the air resistance.
42Linear motion Freely falling bodies SF017Sign convention:Table 2.1 shows the equations of linear motion and freely falling bodies.+-From the sign convention thus,Linear motionFreely falling bodiesTable 2.1
43SF017An example of freely falling body is the motion of a ball thrown vertically upwards with initial velocity, u as shown in Figure 2.5.Assuming air resistance is negligible, the acceleration of the ball, a = g when the ball moves upward and its velocity decreases to zero when the ball reaches the maximum height, H.velocity = 0HuFigure 2.5v
44s v =0 H t t1 2t1 v u t t1 2t1 u a t t1 2t1 g SF017tsHt12t1v =0The graphs in Figure 2.6 show the motion of the ball moves up and down.Derivation of equationsAt the maximum height or displacement, H where t = t1, its velocity,hencetherefore the time taken for the ball reaches H,tvuut12t1tagt12t1Simulation 2.4Figure 2.6
45To calculate the maximum height or displacement, H: use either SF017To calculate the maximum height or displacement, H:use eithermaximum height,Another form of freely falling bodies expressions areWhere s = HOR
46SF017Example 2.7 :A ball is thrown from the top of a building is given an initial velocity of 10.0 m s1 straight upward. The building is 30.0 m high and the ball just misses the edge of the roof on its way down, as shown in figure 2.7. Calculatea. the maximum height of the stone from point A.b. the time taken from point A to C.c. the time taken from point A to D.d. the velocity of the ball when it reaches point D.(Given g = 9.81 m s2)ABCDu =10.0 m s130.0 mFigure 2.7
47SF017Solution :a. At the maximum height, H, vy = 0 and u = uy = 10.0 m s1 thusb. From point A to C, the vertical displacement, sy= 0 m thusABCDu30.0 m
48SF017Solution :c. From point A to D, the vertical displacement, sy= 30.0 m thusBy usingABCDu30.0 mabcTime don’t have negative value.OR
49u Solution : d. Time taken from A to D is t = 3.69 s thus B SF017Solution :d. Time taken from A to D is t = 3.69 s thusFrom A to D, sy = 30.0 mTherefore the ball’s velocity at D isABCDu30.0 mOR
50SF017Example 2.8 :A book is dropped 150 m from the ground. Determinea. the time taken for the book reaches the ground.b. the velocity of the book when it reaches the ground.(Given g = 9.81 m s-2)Solution :a. The vertical displacement issy = 150 mHenceuy = 0 m s1150 m
51b. The book’s velocity is given by SF017Solution :b. The book’s velocity is given byTherefore the book’s velocity isOR
52SF017Exercise 2.3 :A ball is thrown directly downward, with an initial speed of 8.00 m s1, from a height of 30.0 m. Calculatea. the time taken for the ball to strike the ground,b. the ball’s speed when it reaches the ground.ANS. : 1.79 s; 25.6 m s1A falling stone takes 0.30 s to travel past a window 2.2 m tall as shown in Figure 2.8.From what height above the top of the windows did the stone fall?ANS. : 1.75 mFigure 2.8to travel this distance took 0.30 s
53SF017Exercise 2.3 :A ball is thrown directly downward, with an initial speed of 8.00 m s1, from a height of 30.0 m. Calculatea. the time taken for the ball to strike the ground,b. the ball’s speed when it reaches the ground.ANS. : 1.79 s; 25.6 m s1A falling stone takes 0.30 s to travel past a window 2.2 m tall as shown in Figure 2.8.From what height above the top of the windows did the stone fall?ANS. : 1.75 mFigure 2.8to travel this distance took 0.30 s
54Learning Outcomes : 2.4 Projectile motion (2 hours) SF017Learning Outcomes :2.4 Projectile motion (2 hours)At the end of this chapter, students should be able to:Describe and use equations for projectile,Calculate: time of flight, maximum height, range and maximum range, instantaneous position and velocity.
552.4. Projectile motion y v v1 v1y 1 v2x v1x 2 sy=H v2y u v2 uy x SF0172.4. Projectile motionA projectile motion consists of two components:vertical component (y-comp.)motion under constant acceleration, ay= ghorizontal component (x-comp.)motion with constant velocity thus ax= 0The path followed by a projectile is called trajectory is shown in Figure 2.9.yxFigure 2.9BAPQCv1vv1ysy=H1v2xt1v1xv22v2yt2uuySimulation 2.5uxsx= R
56but the y-component of the initial velocity is given by SF017From Figure 2.9,The x-component of velocity along AC (horizontal) at any point is constant,The y-component (vertical) of velocity varies from one point to another point along AC.but the y-component of the initial velocity is given by
57Velocity Point P Point Q SF017Table 2.2 shows the x and y-components, magnitude and direction of velocities at points P and Q.VelocityPoint PPoint Qx-comp.y-comp.magnitudedirectionTable 2.2
58SF017Maximum height, HThe ball reaches the highest point at point B at velocity, v wherex-component of the velocity,y-component of the velocity,y-component of the displacement,Use
592.4.2 Time taken to reach maximum height, t’ SF017Time taken to reach maximum height, t’At maximum height, HTime, t = t’ and vy= 0UseFlight time, t (from point A to point C)
602.4.4 Horizontal range, R and value of R maximum SF017Horizontal range, R and value of R maximumSince the x-component for velocity along AC is constant henceFrom the displacement formula with uniform velocity, thus the x-component of displacement along AC isand
61From the trigonometry identity, thus SF017From the trigonometry identity,thusThe value of R maximum when = 45 and sin 2 = 1 thereforeSimulation 2.6
622.4.5 Horizontal projectile SF017Horizontal projectileFigure 2.10 shows a ball bearing rolling off the end of a table with an initial velocity, u in the horizontal direction.Horizontal component along path AB.Vertical component along path AB.hxABFigure 2.10Simulation 2.7
63= Time taken for the ball to reach the floor (point B), t SF017Time taken for the ball to reach the floor (point B), tBy using the equation of freely falling bodies,Horizontal displacement, xUse condition below :Figure 2.11The time taken for the ball free fall to point AThe time taken for the ball to reach point B=(Refer to Figure 2.11)
64SF017Since the x-component of velocity along AB is constant, thus the horizontal displacement, xNote :In solving any calculation problem about projectile motion, the air resistance is negligible.and
65v1x Example 2.9 : u = 60.0 x v1 v1y v2x v2y v2 P O SF017Example 2.9 :Figure 2.12xOu = 60.0yRHv2yv1xv1yv2xQv1Pv2Figure 2.12 shows a ball thrown by superman with an initial speed, u = 200 m s-1 and makes an angle, = 60.0 to the horizontal. Determinea. the position of the ball, and the magnitude anddirection of its velocity, when t = 2.0 s.
66b. the time taken for the ball reaches the maximum height, H and SF017b. the time taken for the ball reaches the maximum height, H andcalculate the value of H.c. the horizontal range, Rd. the magnitude and direction of its velocity when the ball reaches the ground (point P).e. the position of the ball, and the magnitude and direction of its velocity at point Q if the ball was hit from a flat-topped hill with the time at point Q is 45.0 s.(Given g = 9.81 m s-2)Solution :The component of Initial velocity :
67a. i. position of the ball when t = 2.0 s , Horizontal component : SF017Solution :a. i. position of the ball when t = 2.0 s ,Horizontal component :Vertical component :therefore the position of the ball is (200 m, 326 m)
68a. ii. magnitude and direction of ball’s velocity at t = 2.0 s , SF017Solution :a. ii. magnitude and direction of ball’s velocity at t = 2.0 s ,Horizontal component :Vertical component :Magnitude,Direction,from positive x-axis anticlockwise
69b. i. At the maximum height, H : SF017Solution :b. i. At the maximum height, H :Thus the time taken to reach maximum height is given byii. Apply
70c. Flight time = 2(the time taken to reach the maximum height) SF017Solution :c. Flight time = 2(the time taken to reach the maximum height)Hence the horizontal range, R isd. When the ball reaches point P thusThe velocity of the ball at point P,Horizontal component:Vertical component:
71therefore the direction of ball’s velocity is SF017Solution :Magnitude,Direction,therefore the direction of ball’s velocity ise. The time taken from point O to Q is 45.0 s.i. position of the ball when t = 45.0 s,Horizontal component :from positive x-axis anticlockwise
72therefore the position of the ball is (4500 m, 2148 m) SF017Solution :Vertical component :therefore the position of the ball is (4500 m, 2148 m)e. ii. magnitude and direction of ball’s velocity at t = 45.0 s ,Horizontal component :
73therefore the direction of ball’s velocity is SF017Solution :Magnitude,Direction,therefore the direction of ball’s velocity isfrom positive x-axis anticlockwise
74SF017Example 2.10 :A transport plane travelling at a constant velocity of 50 m s1 at an altitude of 300 m releases a parcel when directly above a point X on level ground. Calculatea. the flight time of the parcel,b. the velocity of impact of the parcel,c. the distance from X to the point of impact.(Given g = 9.81 m s-2)Solution :300 mdX
75The parcel’s velocity = plane’s velocity thus SF017Solution :The parcel’s velocity = plane’s velocitythusa. The vertical displacement is given byThus the flight time of the parcel isand
76b. The components of velocity of impact of the parcel : SF017Solution :b. The components of velocity of impact of the parcel :Horizontal component :Vertical component :Magnitude,Direction,therefore the direction of parcel’s velocity isfrom positive x-axis anticlockwise
77c. Let the distance from X to the point of impact is d. SF017Solution :c. Let the distance from X to the point of impact is d.Thus the distance, d is given by
78Exercise 2.4 : Use gravitational acceleration, g = 9.81 m s2 SF017Exercise 2.4 :Use gravitational acceleration, g = 9.81 m s2A basketball player who is 2.00 m tall is standing on the floor 10.0 m from the basket, as in Figure If he shoots the ball at a 40.0 angle above the horizontal, at what initial speed must he throw so that it goes through the hoop without striking the backboard? The basket height is 3.05 m.ANS. : 10.7 m s1Figure 2.13
79SF017Exercise 2.4 :An apple is thrown at an angle of 30 above the horizontal from the top of a building 20 m high. Its initial speed is m s1. Calculatea. the time taken for the apple to strikes the ground,b. the distance from the foot of the building will it strikes the ground,c. the maximum height reached by the apple from the ground.ANS. : 4.90 s; 170 m; 40.4 mA stone is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m s1 at 40 above the horizontal. How far above or below its original level will the stone strike the opposite wall?ANS. : 10.3 m below the original level.
80SF017THE END…Next Chapter…CHAPTER 3 :Momentum and Impulse