The distance between City C and City A is just the magnitude of the vector, namely,

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1 Pysics 11 Homework III Solutions C. 3 - Problems 2, 15, 18, 23, 24, 30, 39, 58. Problem 2 So, we fly 200km due west from City A to City B, ten 300km 30 nort of west from City B to City C. (a) We want te distance between City C and City A. If we define nort as +ŷ and east as +ˆx, ten we can write eac leg of te trip as a vector (in km). Tus, r A B = (-200, 0) and r B C = (300cos(150), 300sin(150)) = (-150 3, 150), were all angles are measured positively (counterclockwise) from te +ˆx axis. Ten, simple vector addition gives r A C = r A B + r B C, wic is simply: r A C = ( 200, 0) + ( 150 3, 150), = ( ( 150 3))ˆx + ( )ŷ, ( 460, 150). Te distance between City C and City A is just te magnitude of te vector, namely, r = rx 2 + ry 2 = ( 460) km. (b) In order to determine te relative direction, we need te vector r A C wic was found above. Ten, θ = π + arctan( r y r x ) = π + arctan( ) = or π - θ = 18.1 nort of west. Te reason for te addition of π above is to take into account te convention tat arctan is defined on te interval (- π 2, π 2 ). Problem 15 A urricane travels 60 nort of west wit a speed of 41.0 km. Tree ours later, it suddenly canges eading to due nort slowing down to 25.0 km. Suppose te urricane starts at te origin of our coordinate system, (0, 0). After tree ours, te eye of te urricane is at (3*41*cos(120), 3*41*sin(120)) = (-61.5, 106.5), in units km. Ten, te urricane canges direction and eads due nort. Tus, 4.5 from start of measurements, 1

2 we find te urricane at (-61.5, 106.5) + (0, 1.5*25) = (-61.5, 144). Te distance from te origin (Granda Baama Island) after tis interval is r = ( 61.5) = 157 km. Problem 18 (a) Using axes as defined in te text (te same used above in te previous two problems), write eac force in terms of its components. So, force F 1 can be written (in units N), (120cos(60), 120sin(60)) = (60, 60 3). Also, force F 2 = (80cos(105), 80sin(105)) = (-20.7, 77.3). A brief aside to discuss terminology, te resultant force is te single force tat represents te sum of multiple individual forces. So, te resultant force is: F r = (60, 60 3) + ( 20.7, 77.3) = (39.3, 181.2), F r = = 185.4N, θ = arctan( ) = So, te resultant force as magnitude N and points 77.8 measured counterclockwise from te +ˆx axis. x f x i v x Problem 23 We ave te following information: v x = (200 mi )(1609 m mi )( s ) = 89.4m s, i = 100m, x i = 0m, x f = 100m, a x = 0, Now, te time it takes te falcon to travel 100 m (orizontally) is t = = 1.12s. Tus, in te same time interval, te falcon free falls = 100m 89.4 m s (from rest) a distance x = 1 2 (9.8 m s 2 )(1.12s) 2 = 6.13 m. Problem 24 We ave te following information: 2

3 v x = 18.0 m s, i = 50.0m, f = 0m, a x = 0 m s, 2 a y = g, (v y ) i = 0 m s. Te stone strikes te beac at t = given by: 2 g = 3.19 s. Its final velocity is v = (18.0, g t) = (18.0, 31.3) m s, v = = 36.1 m s, θ = arctan( ) = 60.1, measured clockwise from te -ˆx axis. So, te stone strikes te beac at a speed of 36.1 m s measured from te -ˆx axis. at an angle of 60.1 Problem 30 We ave te following information: x = 36.0m, = 3.05m, v i = 20.0 m s, θ = 53, a x = 0 m s, 2 (a) We want to determine if te ball clears te crossbar, and if so, by ow muc. Traveling at a constant orizontal velocity v x = 20.0cos(53) = 12.0 m x. A ball traveling at tis speed would take t = s v x = 3.00s. At tis time, te y-position of te ball is 3

4 y(t = 3.00) = (20.0 sin(53))(3.00) ( 9.8)(3.00)2 = 3.82m. Since y(t=3.00) >, te ball clears te goal post. It clears te post by y(t=3.00) - = 0.77m. If we solve te y equation wen y = 0, we find 0 = 20.0 sin(53)t 1 2 ( 9.8)t2, = 16t 4.9t 2, = t(16 4.9t), wic as solutions t = 0, 3.27s. Tus, te football lands a distance x(t=3.27) = (20.0cos(53))(3.27) - x = 3.36m past te goal post. (b) To determine weter te ball is rising or falling, we must determine were (orizontally) te (vertical) apex of ball s trajectory occurs. Recall 20.0 sin(53) tat at tis point, v y = 0. Tis occurs at time t = = 1.63s. At tis 9.8 time, te orizontal position is given by x(t = 1.63) = (20.0 cos(53))(1.63) = 19.6m. Since x(t=1.63) < x, te ball crosses te goal post wile falling. Problem 39 A boat crosses a river mi wide wit a velocity of 3.30 mi at an angle 62.5 nort of west. Te river carries an eastward current of 1.25 mi. How far upstream is te boat wen it reaces te opposite sore? Let s define perpendicular and parallel directions in relation to te riverbank (as well as te current). Ten, te perpendicular component of velocity is given by v = 3.30sin(117.5) = 2.93 mi. Tus, it takes te boat t = = to reac te opposite side. Tus, te boat ends up 2.93 x = (3.30cos(117.5))(0.173) + (1.25)(0.173) = ( mi)( 5280ft ) = 250 ft. 1mi upstream. Problem 58 We ave te following information: 4

5 y i = 2.00m, y f = 3.05m, θ = 40.0, x = 10.0m, a x = 0, We want to determine te initial velocity of te basketball so tat it goes in te oop witout toucing te backboard. Let s call it v 0. Ten, wit tis velocity, te time it takes te ball to reac te oop is given by t = x v 0 cos(θ). Substituting tis value for t into te expression for y as a function of t yields y as a function of v 0. x y f = y i + v 0 sin(θ)( v 0 cos(θ) ) a( x v 0 cos(θ) )2, = y i + x tan(θ) + Solving te above for v 0 yields gx 2 2v 2 0 cos 2 (θ). v 0 = = gx 2 y f y i x tan(θ), ( 9.8)(10.0) tan(40), = 11.6 m s. 5