Motion in Two Dimensions

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1 SOLUTIONS TO PROBLEMS Motion in Two Dimensions Section 3.1 The Position, Velocity, and Acceleation Vectos P3.1 x( m) 0!3 000!1 70!4 70 m y( m)! ! 330 m (a) Net displacement x + y 4.87 km at 8.6 S of W FIG. P3.1 (b) Aveage speed ( 0.0 m s) ( 180 s) + ( 5.0 m s) ( 10 s) + ( 30.0 m s) ( 60.0 s) Aveage velocity 4.87! 103 m 360 s 180 s + 10 s s 13.5 m s along R 3.3 m s 1

2 Motion in Two Dimensions Section 3. P3.3 Two-Dimensional Motion with Constant Acceleation ( 4.00î ĵ ) m s and v ( 0.0) ( 0.0î! 5.00ĵ ) m s (a) a x! v x! t 0.0 " m s m s a y! v y! t "5.00 " m s "0.300 m s (b)! tan "1 " ( " fom + x axis At t 5.0 s x f x i + v xi t + 1 a x t ( 5.0) + 1 ( 0.800) ( 5.0) 360 m y i + v yi t + 1 a y t! ( 5.0) + 1 (!0.300)( 5.0)!7.7 m v xf v xi + a x t ( ) 4 m s v yf v yi + a y t 1! 0.3( 5)!6.5 m s v " tan!1 y (!6.50 tan!1 4.0 (!15. v x P3.5 a 3.00ĵ m s ; 5.00î m s ; i 0î + 0ĵ (a) f i + v i t + 1 a t! 5.00tî t ĵ " m + a t ( 5.00î tĵ ) m s (b) t.00 s, f 5.00(.00) î + 1 ( 3.00) (.00) ĵ ( 10.0î ĵ ) m so x f 10.0 m, 6.00 m 5.00î v f v xf + v yf ( ) m s ( ) ĵ 5.00î ĵ ( 5.00) + ( 6.00) 7.81 m s

3 Chapte 3 3 P3.6 (a) Fo the x-component of the motion we have x f x i + v xi t + 1 a x t m 0 + ( 1.80! 10 7 m s)t + 1 ( 8! 1014 m s )t ( 4! m s )t + ( 1.80! 10 7 m s)t " 10 " m 0 t "1.80! 107 m s ± ( )("10 " m) ( ) ( 1.8! 10 7 m s) " 4 4! m s 4! m s "1.8! 107 ± 1.84! 10 7 m s 8! m s We choose the + sign to epesent the physical situation Hee 4.39! 105 m s t 8! m s 5.49! 10"10 s. y i + v yi t + 1 a y t ( 1.6! 1015 m s )( 5.49! 10 "10 s).41! 10 "4 m. So, f ( 10.0 î ĵ ) mm. (b) ( )( 5.49! 10 "10 s) ( ) î + ( 4.39! 105 m s) î + ( 8.78! 105 m s) ĵ ( ) î + ( 8.78! 105 m s) ĵ + at 1.80! 10 7 m s î + 8! 1014 m s î + 1.6! m s ĵ 1.80! 10 7 m s 1.84! 10 7 m s ( 1.84! 10 7 m s) + ( 8.78! 10 5 m s) 1.85! 10 7 m s v (d)! tan "1 y ( 8.78 ) 10 5 tan" ) 10 7 (.73 v x

4 4 Motion in Two Dimensions Section 3.3 Pojectile Motion *P3.9 At the maximum height v y 0, and the time to each this height is found fom v yf y + a y t as t v yf! y a y 0! y!g y g. The vetical displacement that has occued duing this time is Thus, if!y ( ) max 1 ft " (!y) max v y,avg t v yf + y t " 0 + y " " 1 m 3.81 ft 3.66 m, then y g y g. y g (!y) max ( 9.80 m s )( 3.66 m) 8.47 m s, and if the angle of pojection is! 45, the launch speed is P3.11 Take the oigin at the mouth of the cannon. y sin! 8.47 m s 1.0 m s. sin 45 x f v xi t Theefoe, v yi t + 1 a y t : 000 m ( m s) t t.00 s 800 m ( m s)sin! i t + 1 ("9.80 m s )t ( ) ".00 s 800 m ( m s)sin! i ( m s 800 m( cos! i ) 000 m( sin! i ) " 19.6 m 19.6 m m( cos! i ) 000 m 1 " cos! i ( ) ".00 s cos! i (31360)cos! i + (640000)cos 4! i ( )cos! i " ( )cos 4! i cos 4! i " cos! i cos! i ± ( ) " 4( )(384) o ! i.4 o 89.4 (Both solutions ae valid.)

5 Chapte 3 5 P3.1 (a) x f v xi t 8.00cos 0.0 ( 3.00).6 m (b) Taking y positive downwads, and the final point just befoe gound impact, v yi t + 1 g t 8.00sin 0.0 ( 3.00) + 1 ( 9.80) ( 3.00) 5.3 m. Now take the final point 10 m below the window ( sin 0.0 )t + 1 ( 9.80)t 4.90t +.74t! t!.74 ± (.74 ) s 9.80 P3.13 Conside the motion fom oiginal zeo height to maximum height h: v yf v yi + a y (! y i ) gives 0 v yi! g h! 0 v yi gh ( ) o Now conside the motion fom the oiginal point to half the maximum height: v yf v yi + a y (! y i ) gives v yh gh +!g v yh gh ( ) " 1 h! 0 so At maximum height, the speed is v x 1 v x + v yh 1 v x + gh Solving, v x gh 3 Now the pojection angle is! i tan "1 v yi v x tan "1 gh gh/3 tan"

6 6 Motion in Two Dimensions P3.15 (a) We use the tajectoy equation: x f tan! i " v i cos. With x f 36.0 m,! i 0.0 m s, and! 53.0 we find ( 9.80 m s ) 36.0 m ( 36.0 m)tan 53.0! 0.0 m s ( 3.94! 3.05) m m. ( ) ( ) cos ( 53.0 ) gx f 3.94 m. The ball cleas the ba by (b) The time the ball takes to each the maximum height is t 1 sin! i g ( )( sin53.0 ) 0.0 m s 1.63 s m s The time to tavel 36.0 m hoizontally is t x f x t 36.0 m (0.0 m s )( cos 53.0 ).99 s. Since t > t 1 the ball cleas the goal on its way down. P3.16 The hoizontal component of displacement is x f v xi t ( )t. Theefoe, the time equied to each the building a distance d away is t d. At this time, the altitude of the wate is v yi t + 1 a " d y t sin! i ( g " d Theefoe the wate stikes the building at a height h above gound level of h d tan! i " gd cos! i.. Section 3.4 The Paticle in Unifom Cicula Motion ( ) P3.3 a c v 0.0 m s 1.06 m 377 m s The mass is unnecessay infomation.

7 Chapte 3 7 P m ; v t! T a v R ( )! m ( ) 60.0 s 00 ev m s 10.5 m s 19 m s inwad P3.7 The satellite is in fee fall. Its acceleation is due to gavity and is by effect a centipetal acceleation. a c g so v g. Solving fo the velocity, v g ( 6, ) ( 103 m) ( 8.1 m s ) 7.58! 10 3 m s. v! T and T! v! ( 7,000 " 10 3 m) 5.80 " 10 3 s 7.58 " 10 3 m s T 5.80 " 10 3 s 1 min 60 s ( 96.7 min. Section 3.5 Tangential and Radial Acceleation P3.8 (b) We do pat (b) fist. The tangential speed is descibed by + a t t 0.7 m s 0 + a t ( 1.75 s) so a t m s fowad (a) Now at t 1.5 s, + a t t 0 + ( 0.4 m s )1.5 s 0.5 m s so ( ) a c v 0.5 m s 0. m 1.5 m s towad the cente a a + a t 0.4 m s fowad m s inwad a fowad and inwad at! tan " ( a 1.31 m s fowad and 7.3 inwad

8 8 Motion in Two Dimensions P3.30 (a) See figue to the ight. (b) The components of the 0. and the.5 m s along the ope togethe constitute the centipetal acceleation: a c (.5 m s )cos( 90.0! 36.9 ) + ( 0. m s )cos m s Section 3.6 a c v so v a c 9.7 m s ( 1.50 m) 6.67 m s tangent to cicle v 6.67 m s at 36.9 above the hoizontal Relative Velocity FIG. P3.30 P3.33 Total time in still wate t d v ! 103 s. Total time time upsteam plus time downsteam: t up (1.0! 0.500) 1.43 " 103 s t down 588 s Theefoe, t total 1.43! ! 10 3 s, 1.0 moe than if the wate wee still.. *P3.34 v ce the velocity of the ca elative to the eath. v wc the velocity of the wate elative to the ca. v we the velocity of the wate elative to the eath. These velocities ae elated as shown in the diagam at the ight. (a) (b) Since v we is vetical, v wc sin 60.0 v ce 50.0 km h o v wc 57.7 km h at 60.0 west of vetical. Since v ce has zeo vetical component, FIG. P3.34 v we v wc cos 60.0 ( 57.7 km h)cos km h downwad.

9 Additional Poblems Chapte 3 9 P3.47 x f x t t cos 40.0 Thus, when x f 10.0 m, t 10.0 m cos At this time, should be 3.05 m!.00 m 1.05 m. Thus, 1.05 m ( sin 40.0 )10.0 m cos 40.0 Fom this, 10.7 m s. + 1 " 10.0 m (!9.80 m s ) cos 40.0.

10 10 Motion in Two Dimensions P3.50 Measue heights above the level gound. The elevation y b of the ball follows y b R + 0! 1 gt with x t so y b R! gx. (a) The elevation y of points on the ock is descibed by y + x R. We will have y b y at x 0, but fo all othe x we equie the ball to be above the ock suface as in y b > y. Then y b + x > R R! gx R " R! gx + x > R + g x x > R g x x > gx R. If this inequality is satisfied fo x appoaching zeo, it will be tue fo all x. If the ball s paabolic tajectoy has lage enough adius of cuvatue at the stat, the ball will clea the whole ock: 1 > gr > gr. (b) With gr and y b 0, we have 0 R! gx gr o x R. The distance fom the ock s base is x! R (! 1)R.

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