Physics 2001 Problem Set 2 Solutions

Transcription

1 Physics 2001 Problem Se 2 Soluions Jeff Kissel Sepember 12, An objec moves from one poin in space o anoher. Afer i arrives a is desinaion, is displacemen is (a) greaer han or equal o he oal disance raveled. (b) always greaer han he oal disance raveled. (c) always equal o he oal disance raveled. (d) smaller han or equal o he oal disance raveled. (e) always smaller han he oal disance raveled. (f) eiher smaller or larger han he oal disance raveled. The answer is d; afer he objec moves o is desinaion, he displacemen is smaller han or equal o he oal disance raveled. Imagine his objec is you, rying o ge o your dorm afer he fooball game. The displacemen, s is defined as he disance as he crow flies beween your dorm and Tiger Sadium. If you were walking, you could ge here via his sraigh pah and hus your disance raveled, D would be equal o your displacemen (D = s). However, you could also ask your parens, who re in own for he game, o drive you home. This would be a much longer pah, and he disance you raveled, will be greaer han your displacemen (D > s). Wording i as he problem asks, your displacemen will be smaller han your disance raveled ( s < D). The displacemen could no be greaer your disance raveled, because hen you would no have made i o your dorm! 1

2 2. The following figure shows he velociy of a car as a funcion of ime. Wha can you say abou he car s velociy, acceleraion, and disance raveled in each of he areas labeled I hrough IV? One can explain his mos clearly looking a he definiions of acceleraion, a and velociy, v: v = x v, and a = From his, we see ha he slope a any given poin of a disance vs. ime graph should be he velociy. Similarly, he slope of any region of he velociy vs. ime graph is he he acceleraion. You can also see his in equaion form: Equaion of a : M ah V ersion P hysics V ersion consan value: f(x) = a a() = g (2) (1) line: g(x) = mx + b v() = g + v 0 (3) parabola: h(x) = ax 2 + bx 2 + c y() = ( 1 2 g)2 + v 0 + y 0 (4) From his, for hose ha know basic differeniaion (calculus), one can see ha v() = dx dv d ẋ() and a() = d ẍ(). So le s use his knowledge o draw wo plos, one of a vs., and he oher x vs.. Region I) In his region, here is no velociy, so hen he acceleraion and disance is he same as a he sar of he region, i.e. a consan, x 0 = 0.. In algebraic form, we see v I () = x() = 0 x I () = (x 1 x 0 ) = 0 (5) a I () = vi() (0) = a I () = 0 (6) 2

3 Alernaively, using calculus: ẋ I () = 0 ẍ I () = dẋ I() = 0 (7) d x I () = 0 d x I () = 0 (8) Region 2) In his region, he velociy has increased consanly from v 1 = 0 o some value, v 2. This line of velociy has a posiive consan acceleraion, a II (). From Eqs. 2, 3, and 4, we know ha a period wih a consan acceleraion, g is a line in v, and a parabola in x so ha s exacly wha we draw. Algebraic Proof: Finding he acceleraion is quick v II () = g + v 0 0 v II () = g a II () = v II() a II () = v II () a II () = g (9) by direc comparison; he equaion of a consan. For he disance we know ha x II () should by a parabola, so we ll use Eq. 4: x II () = ( 1 2 g) 2 + v x 0 1 = 1 2 g2 (10) 3

4 Using calculus: ẋ II () = g (Eq of a line!) ẍ II () = dẋ II() d = g (A consan!) x II () = g d = ( ) 1 2 g 2 + ( v x 0 0 ) x II () = 1 2 g2 (Eq of a parabola!) (11) Region 3) Now he velociy urns o a posiive consan value, v 3. Jus as a consan acceleraion resuled in a line of ha consan slope in he velociy graph for Region 2, he consan velociy will resul in a line of ha consan slope in he disance graph. And as in Region 1, no change in velociy means zero acceleraion. Algebraic Proof: Calculus Proof: v III () = x = v 3 a III () = v I() = (0) a I () = 0 (12) v 3 = (x III() x 2 ) x III () = v 3 + x 2 (13) ẋ III () = v 3 ẍ III () = dẋ III() d x III () = v 3 d = 0 = v 3 + x 2 (14) 4

5 Region 4) This one is ricky. We see ha he velociy is a line wih negaive slope, so we immediaely know ha he acceleraion is going o be a consan negaive. Bu wha does he disance plo look like? There s an acceleraion, so we know i should look parabolic. The velociy never goes below zero, so he disance curve won urn in he opposie direcion, and in fac he final velociy a 5 is zero so i mus go horizonal again in x. Noe ha his disance will be furher away from he origin hough, because even hough velociy is decreasing, i is sill posiive. So i mus look like a sideways parabola, connecing he sar of region 4 wih he end! So finally, all ogeher, he acceleraion, velociy, and posiion vs. ime graphs should look like his: 5

6 3. Assume he brakes of your car creae a consan deceleraion of 4.2m/s regardless of how fas you are driving. If you double your driving speed, does he ime required o sop increase by a facor of 2 or 4? Verify your answer by calculaing he sopping disances for iniial speeds of 16 m/s and 32 m/s From he problem saemen, we know ha he soluion will involve 4 hings: he acceleraion (a) he iniial and final velociies (v 0 and v f, respecively) and ime (). So we ll pick he equaion ha we know involves hese and only hese quaniies: he definiion of acceleraion, a = v = v f v 0 (15) A subley ha migh be missed in he firs read of he problem is ha he second senence asks for he ime required o sop which means ha we auomaically know for boh cases v f = 0. The firs senence les us know ha he acceleraion will be he same for boh cases, a = a. We ve go our knowns, unknowns, and a bi of exra info; from here i s jus algebra. For he firs case, le s jus solve for ime wih ou plugging in any variables: a = 0 v f v 0 = v 0 a (16) Now in he nex case, our iniial velociy is double ha of he firs, or v 0 = 2v 0, so a = 0 v f v 0 = v 0 a a a v 0 2v 0 = 2v 0 a (17) now we can suggesively group he answer, and use Eq. 16 o ge our answer: = 2 ( ) v 0 a = 2 (18) Apparenly, in his scenario wih a consan acceleraion, if you double your driving speed, he ime o sop also increases by a facor of 2. 6

7 We can verify his using he numerical speeds given (noing ha a deceleraion is an acceleraion wih a negaive sign): = v 0 a (16 m/s) = ( 4.2 m/s 2 ) = m/s = v 0 = a (32 m/s) ( 4.2 m/s 2 ) = m/s = 2 (19) 4. A he 18 h green of he U.S. Open, Tiger Woods needs o make a 20.5 f (6.25 m) birdie pu o win he ournamen. When he his he balls, giving i an iniial velociy of 1.57 m/s, i sops 6.0 f (1.83 m) shor of he hole. Assume he deceleraion cause by he grass is consan, and calculae wha iniial speed Tiger should have hi his ball so ha i jus reached he hole. Firs hing, as usual, le s draw a picure. Now, le s figure ou wha we ve go o work wih. For Tiger s pu, we re explicily given v 0, he disance o he hole, x f and he disance lef o he whole, d = x f x f. Implied by he problem is ha he acceleraion is a consan, a grass and he ball comes o a sop, or v f = 0. Finally, we ve se up our coordinae sysem so ha x = x f x 0 0 = x f d. Knowing all his allows us o calculae a grass, so ha we can use ha o find a beer iniial velociy, v 0 which will acually sink he ball. To find a grass we ll use he disance equaion we have ha uses everyhing we know, x = v2 f v2 0 2 a grass (20) 0 v a grass = 2 f v2 0 2 (x f x 0 0 ) 7

8 a grass = v (x f d) However, we need o ack on a negaive sign, because as we ve se up he problem, a grass is in he negaive x direcion. So, a grass = v (x f d) (21) Armed wih his knowledge, we can ou-pu Tiger! To sink he pu, we wan our x = x f, and since we now know a grass we can use Eq. 20 wih primed v s insead unprimed o find our desired v 0 : 0 v x = f 2 v a grass ( x f = v (x f d) ) 2 v0 2 +v 2 0 x f x f d = +v 0 2 v 0 = v 2 0 x f x f d (22) and here s our answer! Plugging in v 0 = 1.57 m/s, x f = 6.25 m, and d = 1.83 m, we ge v 0 (1.57 m/s) = 2 (6.25 m) (6.25 m) (1.83 m) Phew! Wha a long problem se! Ono he Quiz! = 1.86 m/s (23) 8

9 5. Quiz A car is raveling due norh a 18.1 m/s. Find he velociy of he car afer 7.50 s if is acceleraion is (a) 1.30 m/s 2 due norh, or (b) 1.15 m/s 2 due souh. How abou a picure! Since we know he iniial velociy v 0, he ime, and he acceleraions for each case, a a and a b, we ll use he definiion of acceleraion, and simply solve for he final velociy, v f. a v a = v f v 0 a = v f v 0 v f = v 0 + a (24) From here i s jus plug and chug! Noe ha because a a is norh (i.e. in he direcion of he car s velociy) i will be posiive. Conversely, a b is souh (i.e. opposie he car s direcion), so i will be negaive. v (a) f = v 0 + a a = (18.1 m/s) + (1.30 m/s 2 )(7.50 s) = m/s (25) v (b) f = v 0 + a b = (18.1 m/s) + ( 1.15 m/s 2 )(7.50 s) = m/s (26) 9