Lecture 3Eng (2) PDF

Mechanics Dr. Yasmin Mohamed Yousef Bakier Physics Department - Faculty of Science - Assiut University - Egypt 𝟓𝒕𝒉 floor, Room no. 510 Motion in One Dimension  Displacement, Velocity and Speed Displacement: Change of the position of a particle....

Mechanics Dr. Yasmin Mohamed Yousef Bakier Physics Department - Faculty of Science - Assiut University - Egypt 𝟓𝒕𝒉 floor, Room no. 510 Motion in One Dimension  Displacement, Velocity and Speed Displacement: Change of the position of a particle. xi xf It is a vector quantity. A B    Displacement x = x f − xi  if x f  xi  x = +ve    if x f  xi  x = −ve Note: Displacement ≠ Distance travelled xi xf Distance travelled = AC + CB A B C Displaceme nt = AB  Displacement, Velocity and Speed The motion of a particle is known if its position always are known (position -time graph) A particle is moved from point A to point B:  Point A : position = xi , time = ti  Point B : position = x f , time = t f xi xf    A B Displaceme nt x = x f − xi Time interval t = t f − ti Average velocity: the ratio of its displacement ∆x and the time interval ∆t    x x f − xi vx = = Dimension vx  = L Units of m / s v  SI system t t f − ti T  ft / s Br.Eng. system  Displacement, Velocity and Speed Geometrical meaning of v: v is the slope of the straight-line joining initial A and final B points on the position-time graph.  x Slope of AB = = vx t Average Speed: Total distance travelled Average Speed = Total time Average speed is a scalar quantity (has no direction ). T( sec)  Displacement, Velocity and Speed Example: A car is moving along the x-axis from point A to point F.  Point A : xi = 30 m, ti = 0 sec  Point B : x = 52 m, t = 10 sec  Point F : x f = -53m, t f = 50 sec Find: Displacement , Average velocity, Average speed Answer:    (i) Displacement x = x f − xi = −53 − 30 = −83 m  Δx − 83 − 83 − 83 m (ii)Averag e velocity = = = = = −1.7 m / s t (t f − ti ) 50 − 0 50 s Note that: The displacement and the velocity are negative because the car is moving to left Total distance travelled (AB) + (BF) (22) + (105) iii) Average Speed = = = = 2.5m / s Total time 50 50  Displacement, Velocity and Speed Instantaneous velocity and speed Instantaneous velocity The limit of the average velocity as the time interval approaches zero.  x Particle motion from A to B v x1 = 1 = slope of AB t1 B B   x2 Particle motion from A to B v x 2 = = slope of AB t 2  x3 Particle motion from A to B v x 3 = = slope of AB t3 t3  t 2  t1  Displacement, Velocity and Speed If B approches A  t → 0 B B  Instantaneous speed Speed of a particle is defined to Slope = Slope of the tangent of the curve be equal to the magnitude of its = Instantaneous velocity instantaneous velocity.   x dx  x = limt = t →0 t dt Therefore: The instantaneous velocity → velocity   x Instantaneous velocity  equals the limiting value of the ratio t Instantaneous velocity can be positive, negative or zero  dx  dx at piont A = v = = (+) at point B = v = = (0) dt dt  dx at point C = v = = ( −) dt  Displacement, Velocity and Speed Example: A particle moves along the x-axis. Its x-coordinates varies with time according to the expression x = -4t + 2t2 where x is in meters and t is in seconds. The position–time graph for this motion is shown below. a) Determine the displacement of the particle in the time intervals (i) t = 0 to t = 1s and (ii) t = 1s to t = 3s. b) Calculate the average velocity during these two time intervals. t = 0 to t = 1s and t = 1s to t = 3s c) Find the instantaneous velocity of the particle at t = 2.5s t (s) x (m) 0 0 1 -4+2=-2 2 -8+8=0 3 -12+18=6 4 -6+32=16  Displacement, Velocity and Speed (a-1) ∆x = xf - xi = xB- xA = [-4(1)+ 2(1)2]- [-4(0) + 2(0)2] (b) x = -4t + 2t2 ∆xBA = -2-0 = -2m dx d v= = (− 4t + 2t 2 )  v = − 4 + 4t (a-2) ∆x = xf - xi = xD- xB dt dt = [-4(3)+ 2(3)2]- [-4(1) + 2(1)2] at t = 2.5  v = − 4 + 4  2.5 = − 4 + 10 = 6m / s ∆xDB = 6+2= 8m (b)  x AB − 2 v AB = = = −2m / s t 1  xBD + 8 vBD = = = +4m / s t 2  Acceleration Average Acceleration: When the velocity of a particle changes with time, the particle is said to be accelerating. The average acceleration of the particle in the time interval ∆t= t f – ti is defined as the ratio     v x    v v − v , where v x = v xf − v xi  ax = x = xf xi t t t f − xtii SI units are m/s² & Dimensions are L/T2  Displacement, Velocity and Speed Instantaneous acceleration It is the limit of the average acceleration as t approaches zero    vx dvx a x = limt = t →0 t dt   dx  vx = dt  2  d dx d x  ax = ( ) = 2 dt dt dt  Displacement, Velocity and Speed Example: The velocity of a particle moving along the x-axis varies in time according to the expression vx = 40 -5t 2 m /s, where t is in seconds. a) Find the average acceleration in the time interval t = 0 to t = 1.0 s. b) Determine the instantaneous acceleration at t = 2.0 s. ti = 0 xi= 40 – 5(0)2 = 40 m/s d x  ax = dt tf = 1 fi= 40 – 5(1)2 = 35 m/s t (s) vx (m/s) d  a x = (40 − 5t 2 ) = 0 − 10t ∆x = vxf -vxi =20-40= -20 m/s dt 0 40-5(0)2 = 40 1 40-5(1)2 = 35 ax = -10 t at any time ∆t= tf- ti =2.0 - 0=2s 2 40-5(2)2 = 20 3 40-5(3)2 = -5 vx − 20 At t =2.0 s ax= -10 (2.0) = - 20 m/s ax = = = −10m / s 2 4 40-5(4)2 = - 40 t 2  One Dimensional Motion with Constant Acceleration When the velocity of an object changes at the same rate throughout the motion the acceleration is said to be constant. Thus, the velocity-time graph becomes a straight line 1) Velocity as a function of time: vxf − vxi  ax = = constant t f − ti v xf − v xi ti = 0 , t f = t  a x = t  vxf − vxi = axt  vxf = vxi + axt Velocity- time graph  One Dimensional Motion with Constant Acceleration 2) Displacement as a function of velocity and time: Since vx varies linearly with time, we have v xi + v xf vx = x = x f − xi = vx  t 2 Now, we can obtain the displacement using  ( x f − xi ) = 12 (vxi + vxf )t Position - time graph 3) Displacement as a function of time:  vxf = vxi + axt and ( x f − xi ) = 12 (vxi + vxf )t  ( x f − xi ) = 12 (vxi + vxi + a xt )t ( x f − xi ) = (2vxi + a x t )t 1 2 ( x f − xi ) = vxit + 12 axt 2  One Dimensional Motion with Constant Acceleration 4) Velocity as a function of displacement: vxf − vxi  vxf = vxi + axt t = ax  ( x f − xi ) = (vxf + vxi )t 1 2 vxf − vxi Substituting for t ( x f − xi ) = (vxf + vxi )  1 2 ax (v − vxf vxi + vxf vxi − v ) 2 2 (v xf2 − v xi2 ) ( x f − xi ) = xf xi ( x f − xi ) = 2a x 2a x or vxf2 − vxi2 = 2ax ( x f − xi )  vxf2 = vxi2 + 2ax ( x f − xi )  One Dimensional Motion with Constant Acceleration SUMMARY Kinematic Equations of motion in a straight line with constant acceleration Equations Information given vxf = vxi + axt Velocity as a function of time x f − xi = 12 (vxf + vxi )t Displacement as a function of These set of equations velocity and time can be used to solve x f − xi = vxit + 12 axt 2 Displacement as a function of time any problem in one dimensional motion vxf2 = vxi2 + 2ax ( x f − xi ) Velocity as a function of with constant displacement acceleration Equations Information given x f = xi + 12 (vxf + vxi )t Position as a function of velocity and time x f = xi + vxit + 12 axt 2 Position as a function of time  One Dimensional Motion with Constant Acceleration Example: A jet lands on an aircraft carrier at 140 mi/h (~ 63 m/s). (a) What is its acceleration (assumed constant) if it stops in 2.0 s ? (b) What is the displacement of the plane while it is stopping ? Answer: m vxi = 63 m / s vxf = 0 m.s (a) vxi = 63 and vxf = 0 s  vxf = vxi + axt  0 = 63 + a x (2) − 63 ax = = −31.5 m / s 2 (b)  ( x f − xi ) = 12 (vxi + vxf )t  ( x f − xi ) = 12 (63 + 0)  (2) You may also use ( x f − xi ) = 63 m ( x f − xi ) = vxit + 12 axt 2  Freely Falling Objects A freely falling object is any object moving freely under the influence of gravity alone, regardless of its initial motion. Acceleration: Directed downward, regardless of the initial motion of the object. Kinematic equations are valid x ⎯⎯ ⎯ ⎯→ y ax → a y = − g = −9.8m / s 2 General Case Freely Falling Object vxf = vxi + axt v yf = v yi − gt ( x f − xi ) = 12 (vxf + vxi )t ( y f − yi ) = 12 (v yf + v yi )t ( x f − xi ) = vxit + 12 axt 2 ( y f − yi ) = v yit − 12 gt 2 vxf2 = vxi2 + 2ax ( x f − xi ) v yf2 = v yi2 − 2 g ( y f − yi )  Freely Falling Objects Example: A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. The building is 50.0 m high, and the stone just misses the edge of the roof on its way down. Using 𝑡𝐴 = 0, find (a) The time at which the stone reaches its maximum height? (b) The maximum height. (c) The time at which the stone returns to the height from which it was thrown? (d) The velocity of the stone at this instant? (e) The velocity and position of the stone at t = 5 s.? Answer: (a) At max height v yB = 0  v yB = v yA + a y t = v yA − 9.8 t 20 0 = 20 − 9.8 t t = tB = 2.04 s 9.8  Freely Falling Objects (b) ymax = yB =  yAt + 12 a y t 2 (d) v yC = v yA + a yt v yA = 20 m m , a y = −9.8 2 , tC = 4.08 S a y = −9.8 m / s 2 , v yA = 20m / s , t = 2.04 s s s v yC = 20 + (−9.8 )  (4.08)  ymax = (20)  (2.04) + 12 (−9.8)  (2.04) 2 m v yC = 20 − 40 = −20 s ymax = 20.4 m Note v yC = −v yA : The motion is symmertic (c) yC − y A = v yA t + 12 a y t 2 (e) v yD = v yA + a y t m yC − y A = 0 , v yA = 20m / s , a y = −9.8 m / s 2 v yD = 20 + (−9.8 )  (5) = −29.0 Exercise: s 0 = 20t + 12 (−9.8)(t ) 2 Find the velocity of the yD = v yA t + 12 a y t 2 stone just before it hits 0 = 20t − 4.9t 2 ground and the total 0 = t (20 − 4.9t ) yD = (20)  (5) + 12 (−9.8)  (5) 2 time the stone in air ? t = 0 or (20 − 4.9t ) = 0  t = 4.08s y D = −22.5 m Note that tC = 2t B Why ?  Quick Quizzes 1 1 2 3 3  Quick Quizzes 4 5  Answer to Quick Quizzes 1 3 4 5 2  Problems 1. A particle moves according to the equation x = 10 𝑡 2 where x is in meters and t is in seconds. (a)Find the average velocity for the time interval from 2.00 s to 3.00 s. (b)Find the average velocity for the time interval from 2.00 to 2.10 s.  Problems 2. A particle starts from rest and accelerates as shown in Figure P2.12. Determine (a) the particle’s speed at t =10.0 s and at t =20.0 s, and (b) the distance traveled in the first 20.0 s.  Problems 3. An object moving with uniform acceleration has a velocity of 12.0 cm/s in the positive x direction when its x coordinate is 3.00 cm. If its x coordinate 2.00 s later is –5.00 cm, what is its acceleration?  Problems 4. A particle moves along the x axis. Its position is given by the equation x = 2 + 3t − 4t2 with x in meters and t in seconds. Determine (a) its position when it changes direction and (b) its velocity when it returns to the position it had at t = 0.  Problems 5. A golf ball is released from rest from the top of a very tall building. Neglecting air resistance, calculate the position and velocity of the ball after 1.00, 2.00, and 3.00 s.  Problems 6. A ball is dropped from rest from a height h above the ground. Another ball is thrown vertically upwards from the ground at the instant the first ball is released. Determine the speed of the second ball if the two balls are to meet at a height h/2 above the ground.