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### Topic 5 kinematics of particle

• 1. CHAPTER 5 KINEMATICS OF PARTICLE 1 Dynamics deals with the accelerated motion of a body. Dynamics involved: a) Kinematics is a study of the geometry of the motion. b) Kinetics is a study of the forces that cause the motion. Kinematics is the study of the geometry of motion of particles, rigid bodies, etc., without considering the forces that cause them to move. It is often referred to as the "geometry of motion" There are four (4) kinematic equations, Since acceleration is uniform at u v   average v t u v   2 2 1 at ut s   as u v 2 2 2   5.2 CONCEPT OF KINEMATICS OF PARTICLES 5 5.1 INTRODUCTION OF DYNAMICS s = displacement a = acceleration t = time v = final velocity u = initial velocity
• 2. CHAPTER 5 KINEMATICS OF PARTICLE 2 Rectilinear motion is another name for straight-line motion. Position, velocity, and acceleration of a particle as it moves along a straight line. The linear motion can be of two types: uniform linear motion with constant velocity or zero acceleration; non uniform linear motion with variable velocity or non-zero acceleration. Where something is located, a position vector r starting from the origin O. It is specific location of particle P at any instant. Magnitude of s = Distance from O to P - The sense (arrowhead direction of r) is defined by algebraic sign on s. => +ve = right of origin, -ve = left of origin 5.3 RECTILINEAR MOTION 5.3.1 Position 1) Single coordinate axis, s 2) Origin, O 3) Position vector r – specific location of particle P at any instant. 4) Algebraic Scalar s in metres O P r s s Position
• 3. CHAPTER 5 KINEMATICS OF PARTICLE 3 Displacement – change in its position, vector quantity If particle moves from P to P’ is +ve if particle’s position is right of its initial position is -ve if particle’s position is left of its initial position Speed is a measure of how fast something moves. Speed refers to the distance traveled by an object during a specific amount of time. time ce dis Speed tan  Units for measuring speed: km/h, mi/h (mph), m/s 5.3.1 Speed 5.3.1 Displacement P’ O P r s s Δr Δs s' r' Displacement Scalar quantity, specified only by its magnitude.
• 4. CHAPTER 5 KINEMATICS OF PARTICLE 4 Average speed is the whole distance covered divided by the total time of travel. Average speed = total distance travelled/time taken average v t u v   Velocity is defined as a speed in a given direction; the rate of change of displacement of a particle. Velocity = speed and direction; velocity is a vector.   s m dt ds v  Constant velocity = constant speed and no change in direction Average velocity, t r vavg    Instantaneous velocity is defined as dt dr vins   5.3.2 Average Speed 5.3.3 Velocity P’ O P s Δs v Velocity
• 5. CHAPTER 5 KINEMATICS OF PARTICLE 5 Acceleration is the rate of change of velocity of a particle per unit of time. It tells you how fast (the rate) velocity changes: Acceleration = change in velocity/time interval Acceleration is a vector Define the following terms; i. kinematics of particle ii. rectilinear motion iii. position iv. displacement v. average speed vi. velocity vii. constant velocity viii. constant acceleration 5.3.4 Acceleration Example 1
• 6. CHAPTER 5 KINEMATICS OF PARTICLE 6 A car accelerates from 4 m/s to 40 m/s in 20 seconds. Determine i. the acceleration ii. the average velocity iii. distance travelled s m v 4 1  s m v 40 2  sec 20  v i. Acceleration s m t v v a 8 . 1 20 4 40 1 2      ii. Average velocity s m v v vavg 22 2 40 4 2 2 1      iii. Distance travelled m x xt v s avg 440 20 22    Example 2
• 7. CHAPTER 5 KINEMATICS OF PARTICLE 7 A car travelling at 110 km/h decelerates uniformly to rest at rate 4 m/s2 . Calculate the time and distance taken to stop. s m x v 56 . 30 3600 1000 110 1   s m v 0 2  2 4 s m a  i. Time taken t v v a 1 2   sec 64 . 7 4 56 . 30 0 1 2      a v v t ii. Distance as u v 2 2 2     m a v v s 74 . 116 4 2 56 . 30 0 2 2 2 2 1 2 2      Example 3
• 8. CHAPTER 5 KINEMATICS OF PARTICLE 8 A vehicle accelerates from 5 m/s to 45 m/s in 20 seconds. Determine the acceleration. Also find the average velocity and distance travelled. . m x vt d cetravelle Dis v v ocity Averagevel t v v on Accelerati s m 500 20 25 tan 25 2 5 45 2 2 20 5 45 1 2 1 2 2              A missile is fired vertically with an initial velocity of 300 m/s. It is acted on by gravity. Calculate; i. The height it reaches ii. The time taken to go up and down again. s m v 300 1  s m v 0 2  2 81 . 9 s m a   i. The height it reaches as v v 2 2 1 2 2   Example 5 Example 4
• 9. CHAPTER 5 KINEMATICS OF PARTICLE 9  s 81 . 9 2 300 0 2 2        m s 15 . 4587 81 . 9 2 300 2     ii. Time taken t v v a 1 2   sec 58 . 30 81 . 9 300 0 1 2       a v v t A graph can be established describing the relationship with any two of the variables, a, v, s, t Using the kinematics equations a = dv/dt, v = ds/dt, a ds = v dv To determine the particle’s velocity as a function of time, the v-t Graph, use v = ds/dt. Velocity as any instant is determined by measuring the slope of the s-t graph v dt ds  Slope of s-t graph = velocity In figure 5.1 are examples of the displacement-time graphs you will encounter. 5.4 GRAPHICAL METHOD OF RECTILINEAR MOTION 5.4.1 The s-t Graph
• 10. CHAPTER 5 KINEMATICS OF PARTICLE 10 Time (t) Time (t) Displacement (s) Displacement (s) An object stationary over a period of time. The gradient is zero, so the object has zero velocity. Displacement (s) Time (t) An object moving at a constant velocity. The displacement is increasing as time goes on. The gradient, however, stays constant, so the velocity is constant. Here the gradient is positive, so the object is moving in the direction we have defined as positive. An object moving at a constant acceleration. Both the displacement and the velocity (gradient of the graph) increase with time. The gradient is increasing with time, thus the velocity is increasing with time and the object is accelerating.
• 11. CHAPTER 5 KINEMATICS OF PARTICLE 11 Velocity (m/s) Time (t) (sec) Constant deceleration Constant acceleration Constant velocity The gradient of the velocity-time gradient gives a value of the changing rate in velocity, which is the acceleration of the object. The area below the velocity-time graph gives a value of the object's displacement. On a velocity-time graph a horizontal (flat) line indicates the object is travelling at a constant velocity. A straight diagonal line indicates the object's velocity is changing. In the graph on the left, the line sloping upwards shows the object is accelerating and the line sloping downwards in this case towards v = 0, shows it is decelerating. A lift is accelerated from rest to 15 m/s at a rate of 5m/s2 . It then moves at constant velocity fro 8 seconds and then decelerates to rest at 1.4 m/s2 . Draw the velocity-time graph and deduce the distance travelled during the journey. Also deduce the average velocity for the journey. 5.4.1 The v-t Graph Velocity – Time Graphs Example 6
• 12. CHAPTER 5 KINEMATICS OF PARTICLE 12 Time taken for accelerate t v v a 1 2   sec 3 5 0 15 1 2      a v v t Time taken for decelerate sec 71 . 10 4 . 1 15 0 1 2       a v v t Distance travelled (area under graph)       m x x x s 75 . 222 25 . 80 120 5 . 22 15 71 . 10 2 1 15 8 15 3 2 1        Velocity (m/s) Time (t) (sec) a=5m/s 2 a=-1.4m/s2 t2 t1 8 15
• 13. CHAPTER 5 KINEMATICS OF PARTICLE 13 Average velocity s m vavg 26 . 10 71 . 21 75 . 222   A car start from rest and accelerates uniformly for 7 seconds and reaches a velocity of 5 m/s at the end of the acceleration. Its velocity is maintained for 12 seconds and then it decelerates and finally stop in 4 seconds. i. Draw the graph of velocity versus time ii. Calculate distance travelled for the journey iii. Calculate the average velocity iv. Calculate the acceleration over first part of the journey . i. Graph of velocity versus time Example 7 Velocity (m/s) Time (t) (sec) A 23 7 19 5 B C
• 14. CHAPTER 5 KINEMATICS OF PARTICLE 14 ii. Distance travelled Total distance = area A + area B + area C           m x x x s 5 . 87 10 60 5 . 17 5 19 23 2 1 5 7 19 7 5 2 1          iii. the average velocity s m vavg 8 . 3 23 5 . 87   iv. acceleration over first part of the journey acceleration = first slope of graph 2 71 . 0 0 7 0 5 s m a     A car moves with a uniform velocity of 20 m/s for 25 seconds and then accelerates uniformly at 4m/s2 to a velocity of 30 m/s. After that it continues with that velocity of 30 m/s before it decelerates and finally stop in 20 seconds. The total time the car takes to stop is 60 seconds i. Draw the graph of velocity versus time ii. Calculate the deceleration iii. Calculate the time taken for constant acceleration Example 8
• 15. CHAPTER 5 KINEMATICS OF PARTICLE 15 i. graph of velocity versus time ii. Decelerate 2 1 2 5 . 1 20 30 0 s m t v v a       iii. Time taken for constant acceleration t v v a 1 2   sec 2 5 20 30 1 2      a v v t Velocity (m/s) Time (t) (sec) a=5m/s2 20 sec 20 15 30 15 25 15
• 16. CHAPTER 5 KINEMATICS OF PARTICLE 16 Non uniform acceleration means that the acceleration of an object is not constant, but either increasing or decreasing over a passage of time, as shown in the diagram above. It refers to variation in the rate of change in velocity. This variation can be expressed either in terms of time (t) or position (s). Instantaneous acceleration at time t is found by taking smaller and smaller values of Δt and corresponding smaller and smaller values of Δv, dt dv dt s d a dt ds v    2 2 We obtain expression for velocity as below    dt t a v ) ( We obtain expression for position as below 1.4 NON-UNIFORM ACCELERATION 5.3.1 VELOCITY AND ACCELERATION IN EXPRESSED IN TERM OF TIME (t)
• 17. CHAPTER 5 KINEMATICS OF PARTICLE 17    dt t v s ) ( A particle moves along a straight line such that its position is define by s = (2t3 – 2t2 + 8)m. Determine the average velocity, average speed and acceleration of the particle when t = 3s. s = (2t3 – 2t2 + 8) when t=0 sec, s= 8 t=3 sec, s=2(33)-2(32) + 8 = 44 i) V average= s m t s Vaverage / 12 3 8 44       ii) Speed average when t=0 sec, s= 8 t=1 sec, s=2(13)-2(12) + 8 = 8 t=2 sec, s=2(23)-2(22) + 8 = 16 t=3 sec, s=2(33)-2(32) + 8 = 44 s m taken time travel Total Speedaverage / 12 3 36 _ _    iii) Velocity Example 8
• 18. CHAPTER 5 KINEMATICS OF PARTICLE 18 t t dt ds v 4 6 2    when t=3 sec s m v 42 ) 3 ( 4 ) 3 ( 6 2    iv) Acceleration 4 12    t dt dv a when t=3 sec 2 32 4 ) 3 ( 12 s m a    The position of particle which moves along a straight line is defined by the relation s=3t3-5t2-25t+30, where s is expressed in meter and t in seconds. Determine: a) The time at which the velocity will be zero b) The position and distance traveled by the particle at that time c) The acceleration of the particle at that time d) The distance traveled by the particle from t=4s to t=6s Example 8
• 19. CHAPTER 5 KINEMATICS OF PARTICLE 19 a) The time at which the velocity will be zero 30 25 5 3 2 3     t t t s 25 10 9 2     t t dt dx v 2 . 1 sec 31 . 2 0 25 10 9 2        t t t t b) The position and distance traveled by the particle at that time When t = 2.31 sec 30 25 5 3 2 3     t t t s       m s 45 . 17 30 31 . 2 25 31 . 2 5 31 . 2 3 2 3       c) The acceleration of the particle at that time 10 18    t dt dv a   2 58 . 31 10 31 . 2 18 s m a    d) The distance traveled by the particle from t=4s to t=6s. When t = 4 sec       m s 42 30 4 25 4 5 4 3 2 3            m s 348 30 6 25 6 5 6 3 2 3      Distance traveled= 348 – 42 =306 m
• 20. CHAPTER 5 KINEMATICS OF PARTICLE 20 Integrate v dv , assuming that initially v= vo at s= so    s s c v v ds a vdv 0 0   0 2 2 2 s s a v v c o    Integrate ac=dv/dt., assuming that initially v= vo when at t=0    t c v v dt a dv 0 0 at v v o   5.3.2 Velocity as a Function of Position 5.3.3 Velocity as a Function of Position
• 21. CHAPTER 5 KINEMATICS OF PARTICLE 21 The car moves in a straight line such that for a short time its velocity is defined by v = (9t2 + 6t) m/s where t is in sec. Determine it position and acceleration when t = 5s. When t = 0, s = 0. i. Position. Noting that s = 0 when t = 0, we have   t t dt ds v 6 9 2        2 3 0 2 3 0 0 2 0 3 3 3 3 6 9 t t s t t s dt t t ds t s t s         When t = 5s,     m s 450 75 375 5 3 5 3 2 3      ii. acceleration   6 18 6 9 2      t t t dt d dt dv a When t = 5s,   2 96 6 5 18 6 18 s m t a      Example 8
• 22. CHAPTER 5 KINEMATICS OF PARTICLE 22 Rotational Displacement,  Consider a disk that rotates from A to B: Angular displacement o 1 rev = 3600 = 2 rad Measured in revolutions, degrees, or radians. θ must be expressed in the units of radians for this expression to be valid  r s   r v  t    5.4 RELATIONSHIP BETWEEN LINEAR AND ANGULAR MOTION v r – radius θ – angle rotated in radian v – linear velocity ω - angular velocity a - linear acceleration α – angular acceleration N - revolution per second s - length of arc AB - length of arc AB - linear velocity - angular velocity
• 23. CHAPTER 5 KINEMATICS OF PARTICLE 23 N   2  t 1 2      A rope is wrapped many times around a wheel of radius 0.6 m. How many revolutions of the wheel are required to raise a bucket to a height of 25 m? rad r s 7 . 41 6 . 0 25     rad rev  2 1  rev rev 64 . 6 2 1 7 . 41           Example 1 h = 25 m r - angular acceleration
• 24. CHAPTER 5 KINEMATICS OF PARTICLE 24 A motorcycle tire has a radius of 35 cm. If the wheel makes 500 revolutions, how far will the motorcycle have traveled? rad rev rev 6 . 3141 1 2 500             m rad r s 35 . 0 6 . 3141    m s 6 . 1099  Angular velocity ( ω ), is the rate of change in angular displacement. sec rad t    Angular velocity can also be given as the frequency of revolution, f (rev/sec or rpm): ω = 2 f . Angular frequency f (rev/sec). Example 2 5.4.1 Angular Velocity
• 25. CHAPTER 5 KINEMATICS OF PARTICLE 25 A wheel rotates 280º in 6 seconds. Calculate the following. i) The angle turned in radians? ii) The angular velocity in rad/s. Solution: rad rev o  2 360 1   rad rad x o 89 . 4 360 2 280 280    A rope is wrapped many times around a drum with diameter 60 cm. Calculate the angular velocity of the drum are required to lift the bucket to 18 m in 10 second? Example 4 Example 4 h = 18 m r
• 26. CHAPTER 5 KINEMATICS OF PARTICLE 26 Solution: rad r s 60 3 . 0 18     sec 6 10 60 rad t      Angular acceleration ( ) is the rate of change in angular velocity. 2 1 2 sec rad t      The angular acceleration can also be found from the change in frequency, as follows:   2 sec 2 rad t f     since   f   2  The bucket is lifted from rest until the angular velocity of the wheel is 30 rad/s after a time of 8 sec. Calculate the average angular acceleration? Solution: 8 0 30 1 2     t    2 sec 75 . 3 rad   5.4.2 Angular Acceleration Example 4
• 27. CHAPTER 5 KINEMATICS OF PARTICLE 27 From the definition of angular displacement,  r s                 t r t r t s v   Since t      Thus, sec m r v   From the velocity relationship we have,                      t r t r t v a  . Since t      Thus, 2 sec m r a   5.4.3 Angular and Linear Velocity Linear speed = angular velocity x radius 5.4.4 Angular and Linear Acceleration Linear acceleration = angular acceleration x radius
• 28. CHAPTER 5 KINEMATICS OF PARTICLE 28 t v v s         2 2 1 t           2 2 1    at v v   1 2 t      1 2 2 1 2 1 at t v s   2 1 2 1 t t      as v v 2 2 1 2 2      2 2 1 2 2   Define the following terms; i. kinematics of particle ii. rectilinear motion iii. position iv. displacement v. average speed vi. velocity vii. constant velocity viii. constant acceleration REVIEW QUESTIONS 5.4.5 Comparison: Linear and Angular Parameter EXERCISE 2.1
• 29. CHAPTER 5 KINEMATICS OF PARTICLE 29 a) Define the following terms:- i. Constant velocity ii. Constant acceleration b) An object being thrown up from the top of 100 meters building from the ground with a velocity 20 m/s. Determine:- i. The velocity of an object exactly when it hits the ground ii. Time taken from it starts until it reaches the ground A missile was shot with an upward velocity of 90 m/s from the top of a 50 meters tower. Determine the maximum height of the missile from the ground. a) A truck travelling along a straight road at 20 km/h and the speed increases its speed 120 km/h in 15 s. Calculate the distance travelled. (The answer must be in SI unit) a) A ball in Figure 3(b) is thrown vertically upward with a speed of 15 m/s. Determine the time of flight when it returns to its original position. EXERCISE 2.1 EXERCISE 2.1 EXERCISE 2.1 EXERCISE 2.1
• 30. CHAPTER 5 KINEMATICS OF PARTICLE 30 A rocket travel upward at 75m/s. When it is 40m from the ground, the engine fails. Determine max height sB reached by the rocket and its speed just before it hits the ground . Maximum Height. Rocket traveling upward, EXERCISE 2.1
• 31. CHAPTER 5 KINEMATICS OF PARTICLE 31 vA = +75m/s when t = 0. s = sB when vB = 0 at max ht. For entire motion, acceleration aC = -9.81m/s2 (negative since it act opposite sense to positive velocity or positive displacement) ) ( 2 2 2 A B C A B s s a v v    sB = 327 m        s m s m v s s a v v C B C C B C / 1 / 80 / 1 . 80 ) ( 2 2 2 2 The negative root was chosen since the rocket is moving downward a) A train starts from rest at a station with constant acceleration of 2.5 m/s2 until it achieves velocity of 70 km/h. Then, the train decelerate until it stops in 10 s. Calculate:- i. Distance travelled by the train ii. Deceleration of the train b) A car starts from rest with a constant acceleration of 2.3 m/s2 for 16 seconds. Then, the car decelerates uniformly at 3 m/s2 and finally stops. i. Sketch a graph of velocity(v) against time(t) for the journey ii. Determine the velocity for the car iii. Calculate the time taken when it uniformly decelerates EXERCISE 2.1 EXERCISE 2.1
• 32. CHAPTER 5 KINEMATICS OF PARTICLE 32 A car is travelling along a straight road. The car takes 120 sec to travel between two sets of traffic light which are 2145 m apart. The car starts from rest at the first set of traffic lights and moves with constant acceleration for 30 sec until its speed is 22m/s. The car maintains this speed for T seconds. The car then moves constant deceleration, coming to rest at the second set of traffic lights. i. Draw, velocity-time graph ii. Calculate the value of T a) A car starts from rest and accelerates uniformly for 70 seconds and reaches a velocity of 80 m/s at the end of the acceleration. Its velocity is maintained for a while and then it stops within 65 seconds with constant deceleration. The total distance travelled by the car is 12.2 km. i. Draw a Velocity-Time graph ii. Determine the acceleration of the car iii. Calculate the time taken for the journey iv. Determine the deceleration of the car EXERCISE 2.1 EXERCISE 2.1
• 33. CHAPTER 5 KINEMATICS OF PARTICLE 33 DISEMBER 2016 b) A train starts from rest at a station with constant acceleration of 2.5 m/s2 until it achieves velocity of 70 km/h. Then, the train decelerate until it stops in 10 s. Determine:- i. Distance travelled by the train ii. Deceleration of the train JUN 2016 b) A car starts from rest and accelerates uniformly for 70 seconds and reaches a velocity of 80 m/s at the end of the acceleration. Its velocity is maintained for a while and then it stops within 65 seconds with constant deceleration. The total distance travelled by the car is 12.2 km. i. Draw a Velocity-Time graph ii. Determine the acceleration of the car iii. Calculate the time taken for the journey c) The motion of a particle is defined by the relation x = 6t4 – 2t3 – 12t2 + 3t + 3, where x and t are expressed in meters and seconds respectively. Determine the time, the position and the velocity when a = 0 EXERCISE 2.1 EXERCISE 2.1 EXERCISE 2.5
• 34. CHAPTER 5 KINEMATICS OF PARTICLE 34 A particle moves along a straight line such that its position is defined by s=2t3+3t2-12t-10 m. Determine the velocity, average velocity and average speed of the particle when t = 3s. c) A vehicle moves in a straight line such that for a short time its velocity is defined by v = (0.9t2 + 0.6t) m/s where t is in second. When t = 3s, determine:- i. Displacement(s) ii. Acceleration(s) b) The motion of a particle is defined by the relation x = 1.5t4 – 30t2 + 5t + 10, where x and t are expressed in meters and seconds, respectively. When t = 4s, determine:- i. the position ii. the velocity iii. the acceleration of the particle EXERCISE 2.1 EXERCISE 2.1 EXERCISE 2.1
• 35. CHAPTER 5 KINEMATICS OF PARTICLE 35 JUN 2017 c) A car starts from rest with a constant acceleration of 1.25 m/s2 until it achieves 50 km/h. The velocity maintained as far as 1.2 km. When the brake is applied it stops with 10 seconds with constant deceleration. i. Calculate the total time taken during the journey ii. Determine the total distance during the journey The motion of particle is define by the relation x =t3-9t2+24t-6, where x is expressed in meter and seconds. Determine the position, velocity and acceleration when t=5s. A particle moves along a horizontal path with a velocity of v = (3t2 – 6t) m/s. if it is initially located at the origin O, determine the distance traveled in 3.5s and the particle’s average velocity and speed during the time interval. EXERCISE 2.1 EXERCISE 2.1 EXERCISE 2.1
• 36. CHAPTER 5 KINEMATICS OF PARTICLE 36 Coordinate System. Assuming positive motion to the right, measured from the origin, O Distance traveled. Since v = f(t), the position as a function of time may be found integrating v = ds/dt with t = 0, s = 0.    m t t s tdt dt t ds dt t t vdt ds s t t 2 3 0 0 0 2 2 3 6 3 6 3              m t t s tdt dt t ds dt t t vdt ds s t t 2 3 0 0 0 2 2 3 6 3 6 3           0 ≤ t < 2 s -> -ve velocity -> the particle is moving to the left, t > 2a -> +ve velocity - > the particle is moving to the right 0 0   t s m s s t 0 . 4 2   
• 37. CHAPTER 5 KINEMATICS OF PARTICLE 37 m s s t 125 . 6 5 . 3   The distance traveled in 3.5s is sT = 4.0 + 4.0 + 6.125 = 14.125m Velocity. The displacement from t = 0 to t = 3.5s is Δs = 6.125 – 0 = 6.125m And so the average velocity is        s m t s aavg / 75 . 1 0 5 . 3 125 . 6 Average speed,   s m t s v T avg sp / 04 . 4 0 5 . 3 125 . 14     
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