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Graphical Analysis of motion in _________________ I.Distance and displacement. What is the total distance moved? What is the resultant displacement? 20 + 10 = 30 m +10 m d (m) t (s) 20 10 5 15 one direction Find the average speed in the 0-5 s interval. Repeat for 5-10 s and 10-15 s. 20 m 5 s 0 m, 5 s 10 m 5 s

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II. Uniform motion – ____________ is constant d t slope =_____ A B = ________ slope =______________ speed = _______________ Slower less slope A. Graph of d vs. t What would the graph of a slower object look like? ________d in each _____ How much slower is B? 12 velocity d/t speed v constant samet half the speed of A

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v t A B B. Graph of speed __________ for uniform motion slope =______ a slope =________________ a =_________ B is slower lower What does the area shown represent? area = L x W = _____________ =_____________ units: ______ x ______ = _________ What about B? v vs. t v/t slope =______ constant = 0 0 speed x time distance [m/s] [s] [m]

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a t C. Graph of a vs. t for uniform motion a = 0 In review: for __________________ Motion: d t A B t v A B a t a = 0 Same. a = 0 How would you graph B? Uniform

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paper tape constant v timer marks the tape at constant ________ intervals As car moves, describe pattern of marks on tape. Ex: tape timers _____________ spaced b/c car moves the __________ d between each mark. timer cart pulls tape ________________ time through timer How would tape look if car was twice as fast? Equally same

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III. Non-uniform motion: ___________________ acceleration d t slope of tangent is the ______________________ The slope ___________ b/c speed v ___________ A. Graph of d vs. t for object beginning at rest 0 1 2 Object covers ________ d in each________________ dashed lines are__________ constant (non-0) t interval more increases instantaneous v tangents ________speed v i = ___ __________________ initial 0 slope = 0 at t = 0

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v t B. Graph of speed v vs. t for __________________ acceleration a beginning _______________. slope =________ = __________ slope = _______________ a = ______________ What does the area shown represent? area = (1/2) bh = (½) ______________=__________ units: ______ x ______ = ______ constant (non-0) at rest v/t accel. a constant ≠ 0 constant time x speed distance [s] [m/s] [m]

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a t C. Graph of a vs. t for constant a a = constant In review: For _________________________acceleration d t v t a t constant, non-zero

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cart paper tape constant a As car moves, describe pattern of marks on tape. Ex: _______________ spaced b/c car moves _________ distance d between each mark. Timer tape for ___________________ Un-equally How would tape look if car had more acceleration? more constant a timer

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Compare ____________ to ________________motion: d t t v a t d t v t a t Uniform Accel. slope = vslope = a area = d area = d uniform accelerated

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Ex. Answer the questions based on the graph at right. d (m) 4 2 4 6 t (s) -8 What is the total distance traveled? What is the resultant displacement? 8 + 8 + 4 = 20 m +4 m Find the average speed in the first 2 s. Find the average speed over the entire 6 s. Find the average velocity over the entire 6 s. 0 8 m 2 s 20 m 6 s +4 m 6 s

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Ex: The graph below describes a UFO moving in a straight line. AB C Find v avg, d, and a in regions A, B and C. t (s) v (m/s) 4.0 8.0 12 20. 40.

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In A: v = v i + v f 2 = 30. m/s v = d/t d = v t = (30. m/s)(4.0 s) = 120 m = area in A a = Δv/t = (v f – v i )/t = (20. m/s + 40. m/s) / 2 = (40. m/s – 20. m/s) / 4.0 s = 5.0 m/s/ s = 5.0 m/s 2

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In B: v = v i + v f 2 = 40. m/s v = d/t d = v t = (40. m/s)(6.0 s) = 240 m = area in B a = Δv/t = (v f – v i )/t = (40. + 40.) / 2 = (40. m/s – 40. m/s) / 6.0 s = 0

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In C: v = = 20. m/s d = v t = (20. m/s)(2.0 s) = 40. m = area in ΔC: ½bh a = Δv/t = (v f – v i )/t (40. + 0) / 2 = (0 m/s – 40. m/s) / 2.0 s = -20. m/s 2

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= v 240 m t (s) 4.0 8.0 12 v (m/s) 20. 40. AB C area = d a = slope Which area, A, B or C, is uniform motion? Which area, A, B or C, is the acceleration constant, but not zero? 120 m 40 m Where is the acceleration NOT constant?

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