Chapter 2 Problems. s = d t up. = 40km / hr d t down. 60km / hr. d t total. + t down. = t up. = 40km / hr + d. 60km / hr + 40km / hr

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1 Chaper 2 Problems 2.2 A car ravels up a hill a a consan speed of 40km/h and reurns down he hill a a consan speed of 60 km/h. Calculae he average speed for he rip. This problem is a bi more suble han i migh seem. I is emping o simply average he wo velociies ha are given. The problem is ha he car does no spend he same amoun of ime a each velociy, so his approach will give he wrong answer. We need o use he definiion of average speed o ge he correc answer. s oaldis oalime For his problem, we ll consider he disance from he boom o he op of he hill o be d and calculae he ime up and he ime down he hill o find he oal ime. d up 40km / hr d down 60km / h d oal up + down 40km / hr + d 60km / hr 60km / hr + 40km / hr 40km / hr 60km / hr d s 2 d 2d 2 40km / hr 60km / hr oal 60km / hr + 40km / hr 40km / hr 60km / hr d 60km / hr + 40km / hr 48km / hr 2.3 During a hard sneeze, your eyes migh shu for 0.5s. If you are driving a car a 90km/h during such a sneeze, how far does he car move during ha ime s 90km 1000m h 1km 1h 3600s 25m / s d s 25m / s 0.5s 12.5m / s 2.5 The posiion of an objec moving in a sraigh line is given by x , where x is in meers and in seconds (a). Wha is he posiion of he objec a 1,2,3, and 4s? (b) Wha is he objec s displacemen beween 0 and 4s. (c) Wha is he average velociy for he ime inerval from 2 s o 4s? (d) Graph x vs for 0 4s and indicae how he answer for c can be found from he graph. (a-d) We plug in o calculae posiions.

2 x(1) 0m x(2) 2.0m x(3) 0m x(4) 12m (e) We can calculae he displacemen from he posiions. Δx x(4) x(0) 12m 0m 12m (f) We calculae he average velociy using he displacemens and ime inerval. v x(4) x(2) 12m ( 2m) 7m/ s 4s 2s 2s (g) A graph of x vs.. The average velociy can be compued by connecing x(4) and x(2) wih a sraigh line and compuing he slope. Noe: Graph done wih Mahemaica Traffic Shock wave. An abrup slowdown in concenraed raffic can ravel as a pulse, ermed a shock wave, along ahe line of cars, eiher downsreame (in he raffic direcion) or upsream, or i can be saionary. Figure 2-23 shows a uniformly spaced line of cars moving a speed v 25m / s oward a uniformly spaced line of slow cars, moving a speed 5m / s. Assume ha each faser car adds lengh L 12m (car lengh plus buffer zone) o he line of cars when i joins he line, and assume i slows abruply a he las insan. (a) For wha separaion disance d beween he faser cars does he shock wave remain saionary? If he separaion is wice ha amoun, wha are he (b) speed and (c) direcion (upsream or downsream) of he shock wave. To do his problem, is useful o consider he las slow car and he firs fas car. We ll coun he rear of he car as he place whre we measure is posiion. If he pulse is o remain saionary, he picure looks like

3 v vs fas slow L d fas slow L In a ime, he slow car goes a disance L and he fas car goes a disance d + L. We can wrie consan acceleraion equaions for boh cars. We hen solve he equaion for he slow car for and hen use ha ime in he equaion L 0 + Fas Car x if L d 0 v 25m / s x if + v 0 L d + v Slow Car x is 0 x fs L 5m / s x fs x is + v f L 0 + L 0 L d + v 0 L d + v L d L( v 1) L( v 25m / s 5m / s ) 12m ( ) 5m / s d 48m We can find he movemen of he wave by finding he posiion of he car when i reaches he wave. We do his by wriing he posiion of each car, recognizing ha he slow car is a disance L in fron of he fas car when he fas car joins he line. We use his condiion o find he ime when he cars mee. Using he final posiion of he fas car and he ime, we can find he velociy of he pulse.

4 v vs fas slow L d fas slow xff xfsxff+l Fas Car x if L d? v 25m / s x if + v L d + v Slow Car x is 0 x fs + L 5m / s x fs x is + v f + L 0 + L d + v + L L L d + v + L (1 v ) L( v 1) d ( v ) L( v ) d ( )d L v 5m / s ( ) 96m 12m 25m / s 5m / s 12m + L 12m +12m 4.8s 5m / s v pulse 12m / s m / s

5 2.14 The posiion funcion x() of a paricle moving along an x axis is x() where x() is in m and in seconds. (a) A wha ime and (b) where does he paricle (momanarily) sop? A wha (c) negaive ime and (d) posiive ime does he paricle pass hrough he origin? (e) Graph x() vs. for he range -5s o +5s. (f) To shif he curve righward on he graph should we include he erm +20 or he erm 20 in he x()? (g) Does ha incluseion increase or decrease he value of x a which he paricle momenarily sops. Le s begin by ploing he posiion as a funcion of ime beween -5 and 5 Here is a close up--ploed -1 o +1. We can see from he graph ha he paricle sops a x 4, 0. We can ge his from our equaion by seing he velociy o zero and solving for he ime.

6 x() v d x d x(0) 4.0 We can find when he paricle passes hrough zero... x() ± 2 3 s ±0.8165s To shif he curve o he righ, we need o include +20. If we plo wih his erm included we can see he shif we wan. Where and when does he sop occur? We can calculae his

7 x() v d x d s 4 3 s x( 4 3 ) 20m The graph is misleading (i mislead me!) in ha he curve looks he same and Mahemaica has rescaled he verical axis--so i looks as if he x posiion is he same as before. If you look carefully a he new scale, you can see ha he sopping poin is a 20m An elecron moving along he x axis has a posiion given by x 16 e m where is in seconds. How far is he elecron from he origin when i momenarily sops. We need o find ou when he elecron sops. If we know when i sops, we can find ou where i is. To find ou when i sops, we find he insananeous v dx d d d (16 e ) 16 e 16(1 )e 0 16(1 )e 1s 16 e x(1) 16 1 e 1 m 5.886m posiion ime

8 2.18 (a) If he posiion of a paricle is given by x() where x() is in m and is in seconds, when if ever is he paricle s velociy zero? (b) When is is acceleraion a zero? (c) For wha ime range (posiive or negaive) is a negaive? (c) Posiive (e) Graph x(), v() and a( ) Firs we wrie expressions for x(), v() and a( ). (a) We can solv for when he velociy is zero: (b) The acceleraion is zero a 0s (c) Acceleraion is posiive when < 0s (d) Acceleraion is negaive when > 0s x() v() dx d a( ) dv d 30 v() ± s ±1.155s

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10 2.23 An elecron has a consan acceleraion of +3.2m / s 2. A a cerain insan is velociy is +9.6m/s. Wha is is velociy 2.5 s earlier and (b) 2.5s laer? Take he ime o be zero when he velociy is 9.6 m/s. a. A -2.5s, v 1.6m/ s b. A +2.5s, v 17.6m/ s. a 3.2m / s 2 v 0 9.6m / s v v 0 + a 2.26 On a dry road, a car wih good ires may be able o brake wih a consan deceleraion of 4.92 m / s 2 (a) How long does such a car, iniially raveling a 24.6 m/s ake o sop? (b) How far does i ravel in his ime? Graph x verses and v versus for he deceleraion. 24.6m/s This is a classic consan acceleraion problem. We begin by wriing ou wha we know and hen solving for ime. x i 0 x f? v i 24.6m / s v f 0 a 4.92m / s 2? b. Now we can find he sopping poin v f v i + a 0 v i + a v i a 24.6 m / s 4.92m / s 2 5s x f x i + v i a 2 x f 0 + (24.6m / s) 5s+ 1 2 ( 4.92m / s2 ) (5s) m

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12 2.30 A world s land speed record was se by Colonel John P. Sapp when in March 1954 he rode a rocke-propelled sled ha moved along a rack a 1020 km/h. He and he sled were brough o a sop in 1.4 s. In erms of g, wha acceleraion did he experience while sopping? v i 1020km hr v f 0 1.4s 1000m 1km 1hr 3600s 283.3m / s v f v i + a a v f v i # g's m / s 1.4s m / s2 9.8m / s m / s When a high-speed passenger rain raveling a 161 km/h rounds a bend, he engineer is shocked o see ha a locomoive has improperly enered ono he rack from he siding is a disance D676 m ahead (Fig 2-24). The locomoive is moving a 29 km/h. he engineer of he high speed rain immediaely applies he brakes (a) Wha mus be he magniude of he resuling consan deceleraion if a collision is o be jus avoided? (b) Assume he engineer is a x0 when, a 0, he firs spos he locomoive. Skech curves for he e locomoive and he high speed rain for he case in which a collision is jus avoided and is no quie avoided. We begin by wriing wha we know abou each rain High Speed Train x i h 0 x f h? v i h 161km h v f h? a h? 1000m 1km 1h 3600s 44.72m / s Low Speed Train x i l 676m x f l? v i l 29km h v f v i a l m 1km 1h 3600s 8.06m / s The correc condiion is ha he velociy of he fas rain maches he velociy of he slow rain. If he rains have no collided by he ime his happens, hey never will. Afer his occurs, he fas rain will fall furher and furher behind. We need o se he wo posiions equal o each oher o find ou when he collision will occur.

13 x i h + v i h a h 2 x i l + v i l a l v i h a h 2 x i l + v i l + 0 v i l v i h + a h v i l v i h a h v v i h i l v i h a v i l v i h h a h a h 2 v x i l + v i l i l v i h v i h (v i l v i h ) + 1 (v i l v i h ) 2 x i l + v (v i l i l v i h ) a h 2 a h a h v i h (v i l v i h ) (v v i l i h )2 a h x i l + v i l (v i l v i h ) v i h (v i l v i h ) v i l (v i l v i h ) (v v i l i h )2 a h x i l a h (v i h v i l )(v i l v i h ) (v i l v i h )2 a h x i l (v i h v i l )(v i l v i h ) + 1 a h 2 (v v i l i h )2 a h x i l

14 a 0 m / s 2 a 0.5m / s 2

15 a 0.9 m / s 2 a m / s 2 Noice ha in he las picure, he slopes of he curves mach jus where hey mee.

16 2.47 (a) Wih wha speed mus a ball be hrown verically from ground level o rise o a maximum heigh of 50m. (b) How long will i be in he air. Skech y, v, a, vs. (a). We compue he iniial velociy firs. y i 0m y f 50m v i? v f 0m / s? a g v f 2 v i 2 + 2a (y f y i ) 0 v i 2 2g (y f 0) v i 2g (50m) 31.3m / s (b) Now ha we know iniial velociy, we can find he ime o reach he highes poin. v f v i + a 0 v i g v i g 31.3m / s 9.8m / s 3.19s 2 The pah is symmeric, so he enire ime of fligh is 6.38s.

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Chapter 2 Problems. 3600s = 25m / s d = s t = 25m / s 0.5s = 12.5m. Δx = x(4) x(0) =12m 0m =12m

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