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Fundamentals of Physics 8 th Edition HALLIDAY * RESNICK Motion Along a Straight Line الحركة على طول خط مستقيم Ch-2.

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Presentation on theme: "Fundamentals of Physics 8 th Edition HALLIDAY * RESNICK Motion Along a Straight Line الحركة على طول خط مستقيم Ch-2."— Presentation transcript:

1 Fundamentals of Physics 8 th Edition HALLIDAY * RESNICK Motion Along a Straight Line الحركة على طول خط مستقيم Ch-2

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3 Kinematics Motion it self One dimensional motion Mechanics dynamics Newton laws Two dimensional motion statics three dimensional motion

4 To locate an object means to find it’s position relative to reference point origin ( or zero point ) of an axis such as the X axis. If the particle move from the position X= 3m to the position X= -3m A change from initial position to final position Unit of X is m Δ X is vector quantity

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6 Three pairs of initial and final positions along an x axis represent the location of objects at two successive times: (pair 1) –3 m, +5 m; (pair 2) –3 m, –7 m; (pair 3) 7 m, –3 m. (a) Which pairs give a negative displacement? (b) Calculate the value of the displacement in each case using vector notation.

7 Velocity and Speed Average Velocity SI Unit of Average Velocity: meter per second (m/s) SI Unit of Average Velocity: meter per second (m/s)

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9 Average Velocity SI Unit of Average Velocity: meter per second (m/s)  The ratio of displacement that occurs during a particular time interval to that interval interval to that interval  The ratio of distance that occurs during a particular time interval to that interval interval to that interval  Unit of s avg is m/s  v avg is a scalar quantity S avg = --------------- total distance ∆t

10 t=0 x=-5m t=3s x=0m t=4s x=2m A B

11 X(m) t(s) 1 2 3 4 -2 -3 -4 -5 -6 1 2 3 4 5 (1) (2) (3) 0 C

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13 station drive stop 8.4 Km 70 Km/h 2 Km 30 min 2 Km 45 min walk

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18 numerically

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23 Cheak point (2)

24 1. X = 3t – 2 V ins = dx / dt = d (3t – 2 ) = 3 m/s =================== 2. x =-4t 2 – 2 V ins = dx/ st = d(-4t 2 – 2 = -8t m/s ===========================

25 3. x = 2/t 2 X = 2t -2 V ins = dx/dt = d (2t -2 )= -4t -3 m/s =================== 4. x=-2 V ins = dx/dt = o. ==================== a) v constant 1,4 b) v negative direction 2,3

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27 Sample problem 2-3

28 X= 7.8 + 9.2t – 2.1t 3 V ins – dx/d t 9.2 – 6.3 t 2 At t = 3.5 s v = 9.2 – (6.3 ) ( 3.5 ) 2 = -68 m/s 1 -the particle is moving in the negative direction of x 2 -the velocity v depend on t so is continuously changing

29 when a particale, s velocity change, the particle is said to undergo acceleration.

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33 A constant Acceleration: is the particle move’s with constant velocity in equal time  If the acceleration is constant, the average acceleration and instantaneous acceleration are equal.

34 Ch 2-6 Acceleration  If the sign of the velocity and acceleration of a particle are the same, the speed of particle increases.  If the sign are opposite, the speed decreases.

35  A wombat moves along an x axis. What is the sign of its acceleration if it is moving:  a) in the positive direction with increasing speed,  b) in the positive direction with decreasing speed  c) in the negative direction with increasing speed,  d) in the negative direction with decreasing speed? Check Point 2-4 Ans: a) plus b) minus c) minus d) plus

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37 Equations for Motion with Constant Acceleration  v=v 0 +at  x-x 0 =v 0 t+(at 2 )/2  v 2 =v 0 2 +2a(x-x 0 )  x-x 0 =t(v+v 0 )/2  x-x 0 =vt-(at 2 )/2

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39  The following equations give the position x(t) of a particle in four situations: 1) x=3t-4 2) x=-5t 3 +4t 2 +6 3) x=2/t 2 -4/t 4) x=5t 2 -3  To which of these situations do the equations of Table 2-1 apply? Ans: Table 2-1 deals with constant acceleration case hence calculate acceleration for each equation: 1) a = d 2 x/dt 2 =0 2) a = d 2 x/dt 2 =-30t+8 3) a = d 2 x/dt 2 = 12/t 4 -8/t 2 4) a = d 2 x/dt 2 = 10 Ans: 1 and 4 ( constant acceleration case) Check Point 2-5

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42 The direction of motion are now along a vertical y axis instead of the x axis Direction Upward positive (y) downward negative (y)

43 Upward V (+) V₀ (+) Y (+) (-a ( downward V (-) V₀ (-) Y (-) (-a ( free-fall acceleration is always negative and thus downward

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46 بعض المصطلحات الموجودة في Ch2 المصطلح العلمي ( انجليزى ) المصطلح العلمي ( العربي ) Motion Along a Straight Line الحركة على طول خط مستقيم Study دراسة object جسم Motion حركة Displacement الازاحة Position موقع direction اتجاه Negative سالب Positive موجب coordinates احداثيات final position الموقع النهائي Initial Position موقع البداية final Velocity النهائية Initial Velocity البداية distance المسافة acceleration Average التسارع المتوسط Average Velocity السرعه المتوسطة Velocity السرعة Speed السرعة Instantaneous الحظية free-fall سقوط حر dropped سقط

47 بعض المصطلحات الموجودة في Ch2 المصطلح العلمي ( انجليزى ) المصطلح العلمي ( العربي ) Motion Along a Straight Line الحركة على طول خط مستقيم Study دراسة object جسم Motion حركة Displacement الازاحة Position موقع direction اتجاه Negative سالب Positive موجب coordinates احداثيات final position الموقع النهائي Initial Position موقع البداية final Velocity النهائية Initial Velocity البداية distance المسافة acceleration Average التسارع المتوسط Average Velocity السرعه المتوسطة Velocity السرعة Speed السرعة Instantaneous الحظية free-fall سقوط حر dropped سقط


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