Ch. 2 1-Dimensional Motion conceptual question #6, problems # 2, 9, 15, 17, 29, 31, 37, 46, 54

Transcription

1 Ch. 2 1-Dimensional Moion concepual quesion #6, problems # 2, 9, 15, 17, 29, 31, 37, 46, 54 Quaniies Scalars disance speed Vecors displacemen velociy acceleraion

2 Defininion Disance = how far apar wo poins in space are from each oher (only has magniude) Displacemen = he relaive locaion of one poin in space o anoher poin. (involves direcion) Speed = how fas an objec is raveling Velociy = how fas and in wha direcion an objec is raveling Acceleraion = he changing of an objecs velociy wih ime (his can include he velociy s direcion) for example keeping your speed consan bu driving around a curve

3 Coordinae sysems When seing up a coordinae sysem you can pick any direcion o be posiive. y x For his coordinae sysem, x and y are posiive in he righ and up direcions respecively.

4 The disance beween A and B is 5 m. N A 5 m B W E The displacemen of B wih respec o A is 5 m o he righ of A. S You could say ha poin B is 5 m eas of poin A.

5 Speed and Velociy S how fas an objec is moving a ha insan S ave = disance raveled/oal ime V how fas and in wha direcion an objec is moving. V ave = oal displacemen/oal ime V ave = (x f x i )/( f i ) or V ave = x/ y x V ave = (18 m 3 m )/(12 s 7 s) = 3 m/s x= 3 m, =7 s x= 18 m, =12 s

6 For consan velociy: V ave = (x f x i )/( f i ) Becomes V = (x f x i )/ Rearranging he equaion gives: x = x 0 + v* for consan velociy (I renamed x f => x and x i => x o )

7 example Esimae he average speed of an Apollo spacecraf in m/s if i ook 5 days o reach he moon. The moon is 3.8 x 10 8 m from he Earh.

8 Esimae he average speed of an Apollo spacecraf in m/s if i ook 5 days o reach he moon. The moon is 3.8 x 10 8 m from he Earh. 1 day is 60 x 60 x 24 seconds or s Les round his o s. So 5 days is abou s. Average Speed = disance/ime = 3.8 x 10 8 m/5 x 10 5 s = 760 m/s Example in book says ~900 m/s. Difference is from rounding s o s.

9 Confused fooball player A fooball player fields a kickoff a his own goal-line, runs o he edge of he opposie end-zone before urning around and running back. He is ackled a he same spo where he caugh he ball. (He plays for Univ. of Tennessee) a) Wha is he players oal disance raveled? b) His displacemen? c) Average velociy?

10 Drag Race Waching a drag race from overhead you see 2 cars aking 2 differen pahs from he same saring and end poins. If hey arrive a he end poin a he same ime: a) Do hey have he same average velociy? b) Do hey have he same average speed? Pah 1 Pah 2

11 Graphical Represenaion of velociy If you have a graph of posiion versus ime. The slope of he line connecing wo endpoins is he velociy. x Pah raveled slope represen velociy

12 Graphical Represenaion of velociy x As he ime separaion is changed, he slope of he blue lines change. As he ime inerval approaches zero, he slope of he angen line represens he insananeous velociy.

13 Insananeous Velociy Deermined by he slope of a line angen o he posiion vs. ime curve, a ha insan. x No very seep, lower velociy Seepes, highes velociy

14 Graphical Represenaion of velociy If you have a graph of velociy versus ime. The are posiive area under he curve represens he ne displacemen. v 1 2

15 Acceleraion Defined as he ime rae of change of he velociy. Acceleraion is a vecor quaniy. Can include he velociy s magniude changing, he direcion of he velociy changing, or boh. a ave = (v f v i )/( f i ) or a ave = v/ v Slope represens acceleraion Again, as he ime separaion is changed, he slope of he blue line changes. As he ime inerval approaches zero, he slope of he angen line represens he insananeous acceleraion.

16 Insananeous Acceleraion v No very seep, Velociy changes slowly Seepes, velociy changes rapidly

17 Mach he velociy vs. ime graph wih he corresponding acceleraion vs.ime graph. v v v a b c a a a 1 2 3

18 Mach he velociy vs. ime graph wih he corresponding acceleraion vs.ime graph. Answers: a-2 b-1 c-3 v v v a b c a a a 1 2 3

19 1-D Moion wih Consan Acceleraion a ave = (v f v i )/( f i ) becomes a = (v f v i )/ Rewrie as: v = v 0 +a* v ave = (v 0 +v)/2 when a is consan Since x = v ave *, Then x = ½ (v 0 +v)* Combining: x = ½ (v 0 +v)* and v = v 0 +a* We ge: x = ½ (v 0 + v 0 + a*)* or x = v 0 * + ½ a* 2

20 On page 36, by doing some more subsiuing, we can derive one more equaion: Using: x = ½ (v 0 +v)* and = (v-v 0 )/a and doing some rearranging we ge: v 2 = v *a* x

21 3 very imporan equaions (found in able 2.4) For consan acceleraion: Velociy as funcion of ime v = v 0 +a* Displacemen as a funcion of ime x = v 0 * + ½ a* 2 Velociy as a funcion of displacemen. Also called imeless equaion. v 2 = v *a* x

22 Example 2.6 Runway lengh. A plane lands wih a speed of 160 mph and can decelerae a a rae of 10 mi/hr/s. If he plane moves wih consan speed of 160 mph for 1.0 s afer landing before applying he brakes, wha is he oal runway lengh needed o come o res?

23 A plane lands wih a speed of 160 mph and can decelerae a a rae of 10 mi/hr/s. If he plane moves wih consan speed of 160 mph for 1.0 s afer landing before applying he brakes, wha is he oal runway lengh needed o come o res? Firs conver all quaniies o SI unis. v 0 = 160 mph (1mph = m/s) => 71.5 m/s a = -10 mi/hr/s = m/s 2 (negaive because plane slows down in direcion of moion)

24 Ex 2.6 con. For he 1.0 s he planes ravels before applying brakes: x coasing = v 0 * + ½ a* 2 = (71.5 m/s)(1.0 s)+ ½ (0)(1.0 s) 2 = 71.5 m While braking o a sop: Use v 2 = v *a* x x braking = (v 2 v 02 )/(2a) = (0 m/s (71.5 m/s)) 2 /(2*-4.47 m/s 2 ) x braking = 572 m Toal lengh of runway is 71.5 m m = 644 m

25 Free Falling Bodies For free falling objecs he acceleraion is due o graviy; a = g = 9.8 m/s 2 in he downward direcion. (if he coordinae sysem saes ha up is posiive, he graviy is negaive) For close esimaes you can use g = 10 m/s 2 Suppose you dropped a ball from a high cliff. Afer each second inerval, he magniude of he ball s velociy increases by abou 10 m/s.

26 Time (s) Velociy (m/s)

27 Noice ha in each 1 second ime inerval, he disance raveled increases Time (s) Velociy (m/s) Remember he equaion: x = v 0 * + ½ a* 2 The displacemen as a funcion of ime is quadraic.

28 posiion If acceleraion is consan, a graph of posiion vs. ime will look like his: ime

29 Ballisic Rocke (Ex. 2.10) A rocke moves sraigh upward, saring from res wih an acceleraion of m/s 2. I runs ou of fuel afer 4.0 s and coninues upward o a maximum heigh before falling back o Earh. a) Find he rocke s velociy and posiion afer 4.0 s. b) Find he maximum heigh he rocke reaches. c) Find he velociy he insan he rocke his he ground

30 Ballisic Rocke con. Wha we know: Velociy a launch is zero. (Saring from res) Acceleraes a m/s 2 (Upward) Time of acceleraion is 4.0 While acceleraing o find he velociy and posiion afer 4.0 s, use: v = v 0 +a* and x = v 0 * + ½ a* 2

31 Ballisic Rocke con. Afer 4.0 seconds: v = v 0 + a* = 0m/s +(29.4 m/s 2 )(4.0 s) = 118 m/s x = v 0 * + ½ a* 2 = (0m/s)(4.0 s) + ½ (29.4 m/s 2 )(4.0 s) 2 x = 235 m Nex he rocke slows down unil i reaches he maximum heigh. Now: v 0 = +118 m/s, a = g = -9.8 m/s 2 We also know ha a he op, he velociy is zero o v f = 0m/s. Use : v 2 = v *a* x o find x. x = (v 2 v 02 )/(2a) = (0 m/s (118 m/s) 2 /(2*-9.8 m/s 2 )

32 x = (v 2 v 02 )/(2a) = (0 m/s (118 m/s) 2 /(2*-9.8 m/s 2 ) x = 710 m, his is he disance he rocke coninues o rise afer Running ou of fuel. To ge he oal heigh add he 235 m from Par a) for a oal of 945 m. Noice ha he book used a differen Mehod o solve par b). They boh work. c) Find he rocke s velociy jus as i his he ground. Rocke sars free falling from heigh of 945 m. a = g = -9.8 m/s 2 (downward) The ne displacemen while falling is -945 m (downward) I ll use he equaion: v 2 = v *a* x v 2 = (0 m/s) 2 + 2(-9.8 m/s 2 )(-945 m) = (m/s) 2 Taking he square roo gives us v = 136 m/s. Wha abou he sign?? The velociy when i his he ground is downward. All he oher downward quaniies in his problem were negaive. The final velociy will be negaive. (-136 m/s).