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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1 Chaper 2 Problems 2.1 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 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. 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) 4s 2s = 12m ( 2m) 2s = 7m/ s (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.

2 2.6 The 1992 world speed record for bicycle (human-powered vehicle) was se by Chris Huber. His ime hrough he measured 200m srech was sizzling s, a which he commened Cogio ergo zoom! (I hink, herefore I go fas!) In 2001 Sam Whiingham bea Huber s record by 19.0 km/hr Wha was whiingham s ime hrough he 200m? We begin by compuing Huber s speed in km/hr. v = d = m / s v = m 1km s 1000m 3600s = km / hr 1hr We can now compue Whiingham s speed, firs in km/hr and hen in m/s. We ll hen use ha speed o find he ime. v = km / hr + 19km / hr = km / hr v = km 1hr hr 3600s 1000m 1km = m / s = d v = 200m m / s = 5.555s 2.7 Two rains, each having a speed of 30km/h, are headed a each oher on he same sraigh rack. A bird ha can fly 60km/h flies off he fron of one rain when hey are 60 km apar and heads direcly for he oher rain. On reaching he oher rain, he bird flies direcly bac o he firs rain, and so forh. (We have no idea why a bird would behave in his way.) Wha is he oal disance he bird ravels before he rains collide? The easies way o hink abou and do his problem is o noe ha he bird flies a a consan speed for he enire ime ha he rains ravel unil colliding. The disance he bird ravels is hus

3 km d = v = 60 h We need o compue he ime ha he rains run before colliding. By symmery, we can argue ha he rains will collide in he middle, a 30 km. The ime o collision is easy o compue--i s jus he ime for eiher ime o ravel 30 km, which is 1 hour. Now ha we know how long he rains will run, we can see ha he disance he bird will ravel is 60 km You are o drive o an inerview in anoher own a a disance of 300 km on an expressway. The inerview is a 11:15 AM. You plan o drive 100 km/h, so you eave a 8:00 AM o allow some exra ime. You drive a ha speed for he firs 100 km, bu hen consrucion work forces you o slow o 40 km/h for 40 km. Wha would be he leas speed needed for he res of he rip o arrive in ime for he inerview? We need o compue how much ime has passed and how far you have gone. If you drive a 100 km/h for 100 km, you spend 1 hour driving ha segmen. The nex segmen is 40 km a 40 km/ h. This akes 1 hour as well. So you have driven 2 hours and covered 140 km. You allowed 3.25 hours for he enire 300km rip and you now have 1.25 hours lef o cover he remaining 160km. This means ha you need o drive a o complee he rip on ime. 160km km v = = h h 2.12 Traffic Shock wave. An abrup slowdown in concenraed raffic can ravel as a pulse, ermed a shock wave, along he line of cars, eiher downsream (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 where we measure is posiion. If he pulse is o remain saionary, he picure looks like

4 v vs fas slow L d fas slow L In his reamen, we ll consider he posiion of he back of each car. 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 s = L = 5m / s s = x is + 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.

5 v vs fas slow L d fas slow xff Fas Car x if = L d =? v = 25m / s = x if + v = L d + v Slow Car x is = 0 s = + L = 5m / s s = x is + + L = 0 + xfs=xff+l = L d + v + L = 0 + = + 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 4.8 = 2.5m / s

6 2.14 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 velociy and se i o zero. We can also look a he plo and read off he ime for he v = dx d = d d (16 e ) =16 e 16 e = 16(1 )e 0 = 16(1 )e = 1s x(1) =16 1 e 1 m = 5.886m posiion ime 2.15 (a) If a paricle s posiion is given by x = (where is in seconds and x is in meers), wha is is velociy a =1 s? (b) Is i moving oward increasing or decreasing x jus hen? (c) Wha is is speed jus hen? (d) Is he speed larger or smaller a laer imes? (Try answering he nex wo quesion wihou furher calculaion.) (e) Is here ever an insan when he velociy is zero? (f) Is here a ime afer =3s when he paricle is moving oward decreasing x? To proceed, we begin by aking he derivaive o find v. x = v = dx = 6 12 d a. A 1s, v(1) = 6m/ s b. Direcion is negaive--oward he lef and more negaive since x(1) = 5m.

7 c. Speed is 6m/ s d. Speed ges smaller (zero a =2) and hen larger. e. Speed is zero a =2s. f. Afer =3s, he velociy is always posiive. See plo of posiion below The posiion of a paricle moving along an x axis is given by x = where x is in meers and is in seconds. Deermine (a) he posiion, (b) he velociy, and (c) he acceleraion of he paricle a = 3s. (d)wha is he maximum posiive coordinae reached by he paricle and (e) a wha ime is i reached? (f) Wha is he maximum posiive velociy reached by he paricle and (g) a wha ime is i reached? (h) Wha is he acceleraion o f he paricle a he insan he paricle is no moving (oher han =0)? (i) Deermine he average velociy of he paricle beween = 0 and =3 s. We begin by wriing he posiion, velociy and acceleraion. We can now evaluae for (a), (b), and (c)... x() = dx v() = = d dv a() = = d

8 x(3) = 54m v(3) = 18m/s a(3) =-12m/s 2 The maximum possible posiion occurs when he velociy reaches zero. We can find when his occurs mos easily and hen find he posiions a hese imes. v() = 0 = = (24-6 ) = 0s, 4s x(0) = 0 x(4) = 64m We can see ha he answer we are ineresed in is he x(4) = 64m. The maximum possible velociy occurs when he acceleraion is zero. As in he previous case, i s easies o firs find when his occurs. a() = 0 = = 2s v(2) = 24m/s We can compue he acceleraion when he velociy is zero a 4 s. Finally, we can compue he average velociy. a(4) =-24m/s 2 v avg = x(3) - x(0) = 18m/s 3s - 0s 2.19 A a cerain ime a paricle had a speed of 18 m/s in he posiive x direcion, and 2.4 s laer is speed was 30 m/s in he opposie direcion. Wha is he average acceleraion of he paricle during his 2.4s inerval. The average acceleraion is jus he change in velociy over he change in ime. We do need o be careful abou he signs of he velociies, however. v i = 18m / s = 30m / s a = v i Δ = 30m / s 18m / s 2.4s = 20m / s An elecron wih an iniial velociy v 0 = m / s eners a region of lengh L = 1.00cm where i is elecrically acceleraed (Fig. 2-23). I emerges wih v = m / s. Wha is acceleraion, assumed consan?

9 We begin by wriing wha we know and hen selec a consan acceleraion equaion of moion. x i = 0m = m v i = m / s = m / s 2 = v i 2 + 2a( x i ) a = v 2 2 f v i 2( x i ) = m / s A muon (an elemenary paricle) eners an elecric field wih a speed of m / s, whereupon he field slows i a he rae of m/s 2. (a) How far does he muon ake o sop? (b) Graph x vs. and v vs for he muon. v 0 = m/ s a) We solve for he ime o sop a = m/s 2 v = v 0 + a v = 0 v = v 0 + a 0 = v 0 + a = v 0 a = m / s m/s = s x = x 0 + v a 2 x = m / s s m/s 2 ( s) 2 = 0.1m Posiion vs. ime graph

10 Velociy vs. ime 2.27 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

11 2.31 Suppose a rocke ship in deep space moves wih a consan acceleraion equal o 9.8m / s 2, which gives he illusion of normal graviy during he fligh. (a) If i sars from res, how long will i ake o acquire a speed one-enh ha of he speed of ligh, which ravels a m / s. (b) How fare will i ravel in so doing. (a) How long o reach one-enh he speed of ligh? x i = 0 =? v i = 0m / s = m / s a = 9.8 m / s 2 =? (b) How far does i go in ha ime? = v i + a = 0 + a = a = m / s 9.8 m / s 2 = s = x i + v i a2 = (9.8m / s2 ) ( s) 2 = m 2.32 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 = 0 = 1.4s 1000m 1km 1hr 3600s = 283.3m / s = v i + a a = v i # g's = = m / s 1.4s m / s2 9.8m / s 2 = = m / s A car raveling a 56km/h is 24 m from a barrier when he driver slams on he brakes. The car his he barrier 2s laer. (a) Wha is he magniude of he car s consan acceleraion before impac? (b) How fas is he car raveling a impac?

12 56 km/h x i = 0 = 24m v i = 56km h =15.56m / s =? a =? = 2s 1000m 1km 1h 3600s 24m = x i + v i a 2 = 0 + v i a 2 a = v i 2 = 3.56m / s = v i + a ( ) 2s =15.56m / s 3.56m / s 2 = 8.44 m / s