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Engineering Fundamentals Session 8 (3 hours) Motion Distance Vs Displacement • Distance距離 is a scalar quantity which refers to "how much ground an object has covered" during its motion. • Displacement s 位移 is a vector quantity which refers to "how far out of place an object is"; it is the object's change in position. Exercise Start here A teacher walks 4 meters East, 2 meters South, 4 meters West, and finally 2 meters North. Distance = ___________ Displacement = _____________ Exercise The skier moves from A to B to C to D. displacement =___________ distance traveled=____________ Answer to previous page: distance=12;displacement=[0,0] Speed and Velocity • Average speed 速率 – Average speed = total distance / total time taken – ( or rate of change of distance, or changes in distance per unit time) – a scalar • Average velocity v 速度 – v = total displacement/total time taken v = ∆s / ∆t – (or rate of change of displacement, or changes in displacement per unit time) – a vector – Unit: ms-1 Answer to previous page: displacement=140m,distance=420m Example A car moves 3 km north for 10 minutes and then 3 km east for 10 minutes. Find its average speed and velocity. 3km 3km Average speed = 6km / 20 minutes = 6 km / (1/3 hour) = 18km/hr Average velocity = √18 km/ (1/3 hour) at an angle of 45 =12.73 km/hr at 45。 。 Exercise displacement B A C time • Which part(s) is the car having a positive velocity? ________ • Which part(s) is the car having a negative velocity?_________ • What is the velocity at part B?____ Acceleration加速度 • Car speeds up velocity increases and there is an acceleration a • Car slows down velocity decreases and there is a -ve acceleration, or deceleration. • Average acceleration = changes in velocity / total Average acceleration a = ∆v / ∆t time taken • (or rate of change of velocity, or changes of velocity per unit time) • SI unit: ms-2 Answer to previous page: A, C, 0 Example A sports car can go from rest to 100 km hr-1 within 10 seconds. What is its acceleration? Answer Changes in velocity = 100 - 0 km hr-1 = 100 /3600 km s-1 = 100/3.6 m s-1 Average acceleration = (10 / 3.6) /10 ms-2 Exercise • Which part(s) of the curve shows an acceleration? ____ • Which part(s) of the curve shows a deceleration? _____ • Which part(s) of the curve shows a constant (stable) velocity?_____ velocity B A C D time Exercise • A car is originally at rest. It accelerates at 2 ms-2 for 1 second. What is its velocity afterwards? _________ • A car is originally moving at a constant velocity of 1 ms-1. It then accelerates at 2 ms-2 for 0.5 second. What is its velocity afterwards? _________ Instantaneous Velocity • Average velocity 平均速度= ∆s / ∆t (average over the time interval ∆t ) • Instantaneous velocity 瞬時速度= velocity at an instant 瞬閒of time. (∆t 0) • Instantaneous velocity at a time instant t1 = slope of tangent line at t1. instantaneous velocity (at time t) = slope of tangent at t Displacement s t1 t Instantaneous Velocity vs Average Velocity instantaneous velocity at t=2 is 1 ms-1 Instantaneous velocity at t=3 is 0 ms-1 Instantaneous velocity at t=4 is __________ Instantaneous velocity at t=8 is __________ Instantaneous velocity at t=2 is undefined since it is different at 2+ (slightly > 2) and 2- (slightly < 2). Average velocity between t=0 and t=2 is 1 ms-1 Average velocity between t=0 and 7 is _____________ Average velocity between t=7 and 9 is _____________ Average velocity between t=0 and 9 is _____________ S(m) 2 2 7 9 T(s) s Exercise 6 4 3 1.5 v 10 13 t Plot the v-t graph below: 1 0.5 t Exercise s t t1 t2 t3 t4 Time instants at which velocities are positive: _________ Time instants at which velocities are negative: ________ Compare velocity at t1 and velocity at t2:___________ Velocity at t3 = ______________ Realistic Displacement-Time s curve Discontinuous velocity (not realistic) t v Velocity gradually increases (realistic) Red curves are unrealistic since the velocities are discontinuous (implies infinite acceleration) t Constant Acceleration v t a Motions Equations for Constant Acceleration • 5 variables : – – – – – t u v a s time initial velocity final velocity acceleration (constant) displacement Motion Equations (constant acceleration) velocity v v = u + at u t time v2 = u2 + 2as 1 2 s ut at 2 displacement s t time Example An object moving in a straight line with constant acceleration takes 10 s from rest to cover a distance of 100 m. Determine the acceleration of the object. By using the equation 1 2 s ut at 2 • u=0 • t = 10 sec • s = 100 m (Ans) a = 2 m/s2 Example A particle with u = 80 m/s and zero acceleration for first 5 sec. The particle is then slowed down with acceleration of -15 m/s2. How far it has travelled after 5 sec more? Find its velocity at that time. 1 2 s ut at 2 • In last 5 sec, • s = 80(5)+0.5(-15)(5)2 = 212.5 m • But it has travelled 80x5 = 400 m in the first 5 sec Example (continued) • Total distance travelled = 400 + 212.5 =612.5 m • Velocity at that time • v = u + at • v = 80 + (-15)5 • = 5 m/s Motion under the action of gravity • The acceleration due to gravity引力 g is the acceleration of a freely falling object as a result of a gravitational force. For most practical purpose is taken as being 9.81 m/s2 at the surface of the earth. Why is there Gravity? (this slide will not be tested) Newton hit by an apple. Why does the apple fall downwards? Law of Gravity萬有引力 m1 r There is gravitational force between 2 masses m2 Example An object is thrown vertically upwards with a velocity of 8 m/s. What will be the maximum height it reaches and the time taken for it to Positive reach that height ? • u = 8 m/s • a = g = -9.81 m/s2 • v = 0 at the maximum height direction g Example (Continued) Apply v2 = u2 + 2as 0 = 82 + 2(-9.81) s s = 3.26 m Also ( The maximum height ) v = u + at 0 = 8 + (-9.81) t t = 0.82 sec. Force • A force cannot be seen, only the effect of a force on a body may be seen. • Force Units: S.I. Unit ,Newton, (N) or (kN) • Force is a vector quantity. It has both magnitude and direction. Newton •Born 1643 •Newton’s Laws •Gravitational Force •Calculus Newton’s First Law (Law of Inertia) • First Law – Every body will remain at rest or continue in uniform motion in a straight line until acted upon by an external force. • Inertia慣性: tendency for a body maintains its state of rest or move at constant speed • The greater the mass質量, the larger is the resistance to change. Life Examples Once the dummy is in motion, it will be in motion Astronauts in Space Shuttle. Observe instances of Law of Inertia in the following video clip. http://spaceflight.nasa.gov/gallery/video/living/net56/fun_destiny_56.asf Newton’s Second Law • When an external force is applied to a body of constant mass. It produces an acceleration which is directly proportional to the force • The large the mass, the more force it takes to accelerate it. • The large the force, the larger the acceleration. • Force (F)= mass (m) x acceleration (a) Video Example • When a given tension (force) is given to a slider without friction (with air track). View 1-2 video of acceleration from http://www.doane.edu/Dept_Pages/PHY/P hysicsVideoLibrary/videolibrary.html (Use the flash version if you do not have quicktime.) Constant force provided by falling object Example A net force of 200 N accelerates an object with a mass of 100 kg. What is the acceleration? F = 200 N m = 100 kg F=ma a=F/m = 2 m/s2 mass meter Newton’s Third Law • Every action produces an equal and opposite reaction. • Action and Reaction Action Force = Reaction Force Life Example Life Example (continued) Concept map Gravity=9.81 Newton's third law m/s2 Uniformly accelerated motion v = u + at 1 s ut at 2 2 v2 = u2 + 2as Action = Reaction Force Motion Linear Motion Scalars and Vectors Newton's law Newton'first law Law of Inertia Newton's second law F=ma