Chapter 2. One Dimensional Motion

Transcription

1 Chapter One Dimensional Motion

2 Motion in One Dimension Displacement, Velocity and Speed Acceleration Motion with Constant Acceleration MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011

3 Introduction Kinematics - Concepts needed to describe motion - displacement, velocity & acceleration. Dynamics - Deals with the effect of forces on motion. Mechanics - Kinematics + Dynamics MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 3

4 Goals of Chapter Develop an understanding of kinematics that comprehends the interrelationships among physical intuition equations graphical representations When we finish this chapter you should be able to move easily among these different aspects. MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 4

5 Kinematic Quantities Overview The words speed and velocity are used interchangably in everyday conversation but they have distinct meanings in the physics world. MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 5

6 Displacement Position along a 1-dimensional coordinate axis is denoted by the coordinate value x represents a difference in a quantity. The inital value is always subtracted from the final value MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 6

7 Displacement The displacement is a vector quantity. It s specification requires two quantities: a magnitude and a direction. The magnitude is just the size of x and the direction is represented by the sign of x MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 7

8 Displacement Example s0 = s01 + s1 = = 35ft x0 = x x0 = 5 0 = 5ft S is the variable designating distance, a scalar quantity. It is like the odometer reading on your car. x is the variable designating displacement, a vector quantity The displacement only depends on the two points MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 8

9 Displacement Example Alternatively - the displacements are additive The individual displacements are given by x01 = x1 x0 = 0 0 = 0ft x1 = x x1 = 5 0 = 15ft Adding the displacements together we see a cancellation x + x = ( x x ) + ( x x ) = x x This is equivalent to the following x0 = x01 + x1 = 0 15 = 5ft MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 9

10 Average Velocity & Speed The speed is a scalar quantity and it is always positive Average Speed Total Distance s = = Total Time t MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 10

11 Average Velocity The average velocity for the interval between t=t 1 and t=t is the slope of the straight line connecting the point P 1 (t 1,x 1 ) and and P (t,x ) on an x versus t graph. MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 11

12 Average Velocity V Avg, x x xf xi = = t t t f i Or, looking at it another way x = V t Avg, x MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 1

13 Average Velocity V Avg, x x xf xi = = t t t f i Many different motions (curves) between P 1 and P will produce the same average velocity. We only know the end points of the curve we don t know the points in between. MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 13

14 The Message of Missing Variables V Avg, x x xf xi = = t t t f i A benefit of using algebraic descriptions is that we can see what variables DON T appear in the equation. If the variable is not in the equation then the result of the calculation will not depend on the values of that variable. The average velocity doesn t depend on the x-value of the points between P 1 and P MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 14

15 Instantaneous Velocity MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 15

16 Instantaneous Velocity x vx ( t ) = lim = t 0 t dx dt MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 16

17 Instantaneous Velocity Instantaneous velocity at t = 1.8s? Greatest velocity? Zero velocity? Negative velocity? MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 17

18 Average Acceleration a Avg, x v vf vi = = t t t f i v = a t Avg, x MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 18

19 Instantaneous Acceleration v ax ( t ) = lim = t 0 t dv dt ax ( t ) = lim = = = t 0 t dt dt dt dt v dv d dx d x MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 19

20 Constant Acceleration A constant acceleration acting on a body changes the value of its velocity by a constant amount every second. Case (a) shows a falling object with an initial downward velocity v 0. Case (b) shows a rising object with an initial upward velocity v 0. MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 0

21 Is He Correct? It goes from zero to 60 in about 3 seconds. ( Sydney Harris.) MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 1

22 Some Useful Conversions 10 = hr hr km 3600s km 1 m m cm = = = 7.8 hr s s s 3 km km m hr km km m ft mi = hr hr km 1 m 580ft km 381 mi mi = = hr 580 hr hr MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011

23 Some Useful Conversions ft ft mi 3600s 3 = 3 s s 580ft hr ft 3600 mi mi mi 3 = 3 = i i i s hr s hr s hr s A constant acceleration of 3 ft/s changes the velocity of an object by about mi/hr every second. MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 3

24 He Was Correct! An object gaining mi/hr in speed every second will be going 66 mi/hr after 3 seconds. It goes from zero to 60 in about 3 seconds. ( Sydney Harris.) MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 4

25 Free Fall Assumption: acceleration due to gravity is g g = 9.8 m/s 10 m/s MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 5

26 The Most Important Graph- V vs T The values of the curve gives the instantaneous VELOCITY. The slope of the curve gives the ACCELERATION. Negative areas are possible. Area under the curve gives DISTANCE. MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 6

27 Non-Segmented V vs T Graphs For non-segmented velocity curves we would rely on taking the integral to find the area under the curve - if we know its functional form. MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 7

28 1-D Projectile Problem Given: v0y = 14.7m s v = v gt y 1 0y v = 0 = v gt t y v = g 0y 0y 1 Total Time = T = t = 1 v 0y v + 0 v Height = v t = ( ) t = t g oy oy avg, y MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 8

29 The linear nature of the v versus t graph indicates the constant value of the acceleration. 1-D Projectile Problem Total area under the velocity curve is zero showing that the displacement was zero. The cap wound up back at its starting point MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 9

30 Car Chase Problem A speeder traveling at a constant 5 m/s passes a stationary police car at t = 0. At that instant the police car starts accelerating from rest at a p = +5 m/s Initial conditions - all motion is in one dimension, traveling to the right, which we take to be the positive x-axis. v s = v so = 5 m/s = constant v po = 0 at t = 0 ; a p = +5 m/s Questions: (a.) When does the police car catch the speeder? (b.) What is v p when he catches up with the speeder? MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 30

31 Car Chase Problem The catch up condition is when both cars have reached the same position - they have then both traveled the same distance. The condition is x p = x s MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 31

32 Problem Solving Tips Make every attempt to minimize the amount of unnecessary notation, especially subscripts. If you are working on a 1-D problem you don t need x as a subscript - it s the only coordinate in the problem. When you finally substitute numbers into the equations don t include the units in the equation itself - they are an unnecessary distraction. MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 3

33 Car Chase Problem The catch up condition x s (t c ) = v p (t c ) x = v t ; x = a t 1 s s p p v t = a t s c 1 p c Position equations Sub into catch up condition (v - a t )t = 0 There are two solutions 1 s p c c t = 0 c t = v /a = (5)/5 = 10s c s p They are together at the beginning The catch up MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 33

34 Car Chase Problem How fast is the police car going when it catches up with the speeder? v = v p po p c p + a t v = 0 + a t = 5(10) = 50 m s p c This is double the velocity of the speeder. Is this a coincidence? If two cars cover the same distance in the same time then they must have the same average velocity. MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 34

35 Car Chase Problem New question - At what time are the two cars separated by a distance D? MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 35

36 Car Chase Problem New question - At what time are the two cars separated by a distance D? At t = t x - x = D 1 1 s 1 p 1 p 1 s 1 1 s p v t - a t = D a t - v t + D = 0 a t - v t + D = 0 p 1 s 1 t = 1 v ± (v ) - 4a (D) s s p a p MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 36

37 Car Chase Problem t = 1 v ± (v ) - 4a (D) s s p a p t = 1 t = 1 1 (5) ± (i5) - 4(5)50 i5 50 ± (50) ± 1500 = t = 5 ± 15 t = 8.87s + 1 t = 1.13s - 1 The plus solution is when the police car is catching up to the speeder. The negative solution is when the speeder is moving away from the still accelerating police car. MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 37

38 Car Chase Problem MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 38

39 Chase Problem - Max Separation t = 1 s s p s p max s p max s p max max v ± (v ) - 4a (D) s 5 D max = = 6.5m i5 p a (v ) - 4a (D ) = 0 4v - 8a D = 0 v - 4a D = 0 D v = a p The D max value occurs when the radicand is zero. MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 39

40 Chase Problem - Max Separation via Calculus D = v t - s 1 1 D = 5i5-5(5 ) p 1 = (vst) - ( a pt ) = 0 s t = = = 5.0s p a t dd d d dt dt dt dd = v s - a p t = 0 dt v 5 a 5 D = 6.5m MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 40

41 The Falling Screw Problem We are describing the motion of both the elevator and the screw from the at rest coordinate system. t = t 1 should be t = t f If the screw doesn t drop then its distance from the floor remains constant MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 41

42 The Falling Screw Problem y y = v t + a t 1 F F 0 F 0y Fy y 0 = v t + a t 1 F 0y Fy Floor position y y = v t + a t 1 S S 0 S 0y Sy y h = v t + g t S 1 0 y ( ) Screw position Equate expressions for y S and y F at t = t f h + v t gt = v t + a t 1 1 0y f f 0y f Fy f h gt = a t 1 1 f Fy f t f = h a + g F MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 4

43 The Falling Screw Problem t f = h a + g F The result can be understood more simply from the rest frame of the screw. That is, the coordinate frame of reference that moves with the screw. If the elevator wasn t accelerating (a F = 0) then the result would be the same as if the elevator was sitting still. The accelerating elevator effectively changed the acceleration, as perceived by the screw, from g to g + a F MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 43

44 Kinematic Eqns via Differentiation x t x v t a t ( ) = x + x dx vx v0x axt dt = = + d x dt dv dt x = = a x MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 44

45 Kinematic Equations v = v + a t x 0 v = ( v + v ) v 1 av, x 0x x av, x x = t x = x + v t + a t x 0 x 1 0 0x x v = v + a x x MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 45

46 Kinematic Equations v av, x 1 t = t t 1 v dt x t v = x t 1 a dt x MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 46

47 Summary Motion in One Dimension Displacement, Velocity and Speed Acceleration Motion with Constant Acceleration MFMcGraw-PHY 45 Chap_0b One Dim Motion-Revised 1/16/011 47