Obtencion,

CHAPTER 4

MOTION IN TWO AND THREE DIMENSIONS

Select odd-numbered solutions, marked with a dagger (†), appear in the Student Solutions Manual, available for purchase. Answers to all solutions below are underscored.
4-1. (a) Total displacement is (3.2 + 2.6) km 45° E of N + 4.5 km 50° W of N = 7.0 km , 5° E of N (see Problem 3, Chapter 3) (b) Avg vel = ∆r/∆t = 7.0 km/1.25 h, 5° E of N = 5.6km/h, 5° E of N (c) Avg speed =

total distance 3.2 + 2.6 + 4.5 km = = 8.24 km/h time 1.25 h

4-2.

avg speed = | Avg vel | =

distance π × r π × 1.50 × 1011 m = = = 29.9 km/s time 1/ 2 yr 1/ 2 × (365.25 × 24 × 3600) s ∆r 2×r 2 × 1.50 × 1011 m = = = 19.0 km/s ∆t 1/ 2 yr 1/ 2 × 3.16 × 107 s

4-3.

171 km/h = 47.5 m/s i j If the bullet is fired at an angle θ, its position vector is givenby rbul = 366 sin θ t ˆ + 366 cos θ t ˆ For the bird, rbird = (47.5)t ˆ + 30 ˆ i j For the bird to be hit rbul = r bird. Comparing the x-components, we have 366 sin θ = 47.5 θ = sin−1(47.5/366) = 7.45° Then d = 30 tan θ = 30 tan 7.45° = 3.93 m. Thus the hunter must aim 3.93 m ahead of the bird in order to hit the bird. At t = 0, the velocity is v = 30 km/h N, or t = 10 s v = 0i + (30 km/h) j. Att = 10 s,
v = 30 km/h @ 45° W of N, or v = −(21 km/h)i + (21 km/h) j. At t = 20 s, v = 30 km/h W, or v = −(30 km/h)i + 0 j. At t = 30 s, v = 30 km/h @ 45° S of W, or v = −(21 km/h)i − (21 km/h) j. At t = 40 s, the driver is halfway around and v = 30 km/h S, or v = (30 km/h)i + 0 j.

4-4.

t = 20 s t = 30 s

t=0

t = 40 s

4-5.

r = (4 + 2t)i + (3 + 5t + 4t2)j + (2 − 2t − 3t 2)k (a) v= dr/dt = 2i + (5 + 8t ) j − (2 + 6t )k (b) a = dv/dt = 8 j − 6k magnitude, | a | =

(8) 2 + (6) 2 = 10 m/s 2

⎛6⎞ direction θ = −tan−1 ⎜ ⎟ = −37° (37° below the y -axis in the z -y plane) ⎝8⎠

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4-6.

4

(a) The components of v are vx = dx/dt and vy = dy/dt. Also, the components of a are ax = dvx/dt and ay = dvy/dt. vx = − Ab sin bt v y = Ab cos bt
ax = − Ab 2 cos bt

ay = − Ab 2sin bt
A2b 2 (sin 2 bt + cos 2 bt = Ab

(b) | v | =
|a|= †4-7.

(− Ab sin bt ) 2 + ( ab cos bt ) 2 =

2 2 2 (− Ab 2 cos bt ) 2 + (− Ab 2sin bt ) 2 = Ab2 cos bt + sin bt = Ab

At t = 2.0 s, the missile has been falling with acceleration a = 0i − (9.81 m/s 2 ) j and horizontal velocity equal to the velocity of the airplane. This means the missile is still directly below the gt 2(9.81 m/s 2 )(2.0 s) 2 j = 0i − j= airplane. Its displacement relative to the plane is r = 0i − 2 2 0i − (19.6 m) j, or 19.6 m @ 90º below the direction of travel of the airplane. At t = 3.0 s, which is 1.0 s after igniting the engine, the acceleration is a = (6.0 m/s 2 )i − (9.81 m/s 2 ) j. Its ax t 2 gt 2 i− j 2 2 (6 m/s 2 )t 2 gt 2 (6 m/s 2 )(1.0 s) 2 (9.81 m/s 2 )(1.0 s) 2 i− j= i− j = (3.0m)i − (4.9 m) j. Now = 2 2 2 2 the missile’s total displacement relative to the plane is r2 = (0 m)i − (19.6 m) j + (3.0 m)i − (4.9 m) j = (3.0 m)i − (24.5 m) j. The magnitude is −24.5 r2 = (3.0 m) 2 + (24.5 m) 2 = 24.7 m. The direction is given by θ = tan −1 = − 83.0°, or 3 24.7 m @ 83º below the direction of travel of the plane. dr r = (5t + 4t 2 )i + (3t 2 + 2t 3 ) j + 0k m. v = = (5 + 8t )i +(6t + 6t 2 ) j + 0k m/s. dt d 2 r dv a= 2 = = (8)i + (6 + 12t ) j + 0k m/s 2 . At t = 2.0 s, v = (21 m/s)i + (36 m/s) j + 0k m/s, dt dt displacement during this 1.0 s interval is r1 =

4-8.

which gives a speed of v = (21 m/s) 2 + (36 m/s) 2 = 42 m/s. †4-9. ∆r , where ∆r is the total displacement and ∆t = 1.5 h is the total elapsed time. To find ∆t the total displacement, find the displacementduring each part of the trip: ∆r1 = (300 km/h @ 30° N of E) × 0.50 h = 150 km @ 30° N of E. Taking the y direction to point N and the x direction to point E, this is ∆r1 = (150 km)( cos 30°)i + (150 km)( sin 30°) j = (130 km)i + (75 km) j. For the second part of the trip, ∆r2 = (300 km/h @ 30° W of S) × 1.0 h = 300 km @ 30° W of S. In terms of x and y, this is ∆r2 = − (300 km)( sin 30°)i − (300...