Bachiller En Ciencias Y Letras

4
Motion in Two Dimensions
CHAPTER OUTLINE
4.1 4.2 4.3 4.4 4.5 4.6 The Position, Velocity, and Acceleration Vectors Two-Dimensional Motion with Constant Acceleration Projectile Motion Uniform Circular Motion Tangential and Radial Acceleration Relative Velocity and Relative Acceleration

ANSWERS TO QUESTIONS
*Q4.1 The car’s acceleration must have an inward component and a forward component:answer (f ). Another argument: Draw a final velocity vector of two units west. Add to it a vector of one unit south. This represents subtracting the initial velocity from the final velocity, on the way to finding the acceleration. The direction of the resultant is that of vector (f ). No, you cannot determine the instantaneous velocity. Yes, you can determine the average velocity. The points could bewidely separated. In this case, you can only determine the average velocity, which is v avg = ∆x ∆t

Q4.2

Q4.3

(a)

(b)

*Q4.4

(i) The 45° angle means that at point A the horizontal and vertical velocity components are equal. The horizontal velocity component is the same at A, B, and C. The vertical velocity component is zero at B and negative at C. The assembled answer is a = b =c = e > d = 0 > f (ii) The x-component of acceleration is everywhere zero and the y-component is everywhere – 9.8 m s2. Then we have a = c = e = 0 > b = d = f. A parabola results, because the originally forward velocity component stays constant and the rocket motor gives the spacecraft constant acceleration in a perpendicular direction. (a) yes (b) no: the escaping jet exhaust exerts an extraforce on the plane. (c) no (d) yes (e) no: the stone is only a few times more dense than water, so friction is a significant force on the stone. The answer is (a) and (d). The projectile is in free fall. Its vertical component of acceleration is the downward acceleration of gravity. Its horizontal component of acceleration is zero. (a) no (b) yes (c) yes (d) no. Answer: (b) and (c)

Q4.5

Q4.6Q4.7

Q4.8 *Q4.9

The projectile on the moon is in flight for a time interval six times larger, with the same range of vertical speeds and with the same constant horizontal speed as on Earth. Then (i) its range is (d) six times larger and (ii) its maximum altitude is (d) six times larger. Apollo astronauts performed the experiment with golf balls.
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Chapter 4

Q4.10

(a) no. Its velocity is constant in magnitude and direction. (b) yes. The particle is continuously changing the direction of its velocity vector. (a) straight ahead (b) either in a circle or straight ahead. The acceleration magnitude can be constant either with a nonzero or with a zero value.

Q4.11

*Q4.12 (i) a = v2 r becomes 32 3 = 3 timeslarger: answer (b). (ii) T = 2π r v changes by a factor of 3 3 = 1. The answer is (a). Q4.13

The skater starts at the center of the eight, goes clockwise around the left circle and then counterclockwise around the right circle. *Q4.14 With radius half as large, speed should be smaller by a factor of 1 same. The answer is (d). 2, so that a = v2 r can be the

*Q4.15 The wrench will hit (b) at thebase of the mast. If air resistance is a factor, it will hit slightly leeward of the base of the mast, displaced in the direction in which air is moving relative to the deck. If the boat is scudding before the wind, for example, the wrench’s impact point can be in front of the mast. *Q4.16 Let the positive x direction be that of the girl’s motion. The x component of the velocity of the ball relativeto the ground is +5 – 12 m s = –7 m s. The x-velocity of the ball relative to the girl is –7 – 8 m s = –15 m s. The relative speed of the ball is +15 m s, answer (d).

SOLUTIONS TO PROBLEMS
Section 4.1 P4.1 The Position, Velocity, and Acceleration Vectors y (m) −3 600 0 1 270 −2 330 m

x (m) 0 −3 000 −1 270 −4 270 m

(a) Net displacement = x 2 + y 2 at tan −1 ( y x ) R = 4.87 km at 28.6°...