chap3 - CHAPTER 3 KINEMATICS IN TWO DIMENSIONS ANSWERS TO...

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CHAPTER 3 KINEMATICS IN TWO DIMENSIONS ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (a) The horizontal component v x of the projectile’s velocity remains constant throughout the motion, since the acceleration a x in the horizontal direction is zero ( a x = 0 m/s 2 ). The vertical component v y , however, changes as the projectile moves. This component is greatest at point 1, decreases to zero at point 2 at the top of the trajectory, and then increases to a magnitude less than that at point 1 as the projectile approaches point 3. 3. (c) The acceleration due to gravity is the same for both balls, despite the fact that they have different velocities. 4. (d) The acceleration of a projectile is the same at all points on the trajectory. It points downward, toward the earth, and has a magnitude of 9.80 m/s 2 . 6. (c) The time for a projectile to reach the ground depends only on the y component (or vertical component) of its variables, i.e., y , v 0 y , and a y . These variables are the same for both balls. The fact that Ball 1 is moving horizontally at the top of its trajectory does not play a role in the time it takes for it to reach the ground. 14. (c) The velocity v PC of the passenger relative to the car is given by v PC = v PB + v BC , according to the subscripting method discussed in Section 3.4. However, the last term on the right of this equation is given by v BC = v BG + v GC . So, v PC = v PB + v BG + v GC = +2 m/s + 16 m/s + ( 12 m/s) = +6 m/s. 16. The magnitude v AB of the velocity of car A relative to car B is v AB = 34.2 m/s. The angle that the velocity v AB makes with respect to due east is θ = 37.9 ° south of east.
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Chapter 3 Problems 98 CHAPTER 3 KINEMATICS IN TWO DIMENSIONS PROBLEMS 11. REASONING The component method can be used to determine the magnitude and direction of the bird watcher's displacement. Once the displacement is known, Equation 3.1 can be used to find the average velocity. SOLUTION The following table gives the components of the individual displacements of the bird watcher. The last entry gives the components of the bird watcher's resultant displacement. Due east and due north have been chosen as the positive directions. Displacement East/West Component North/South Component A 0.50 km 0 B C 0 –(2.15 km) cos 35.0° = –1.76 km –0.75 km (2.15 km) sin 35.0° = 1.23 km r = A + B + C –1.26 km 0.48 km a. From the Pythagorean theorem, we have r =+ = (–1.26 km) (0.48 km) 1.35 km 22 The angle θ is given by = F H G I K J tan 0.48 km 1.26 km 21 , north of west 1 r 0.48 km 1.26 km North East b. From Equation 3.1, the average velocity is 1.35 km 0.540 km/h, 21 north of west 2.50 h t Δ == = ° Δ r v Note that the direction of the average velocity is, by definition, the same as the direction of the displacement.
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Chapter 3 Problems 99 15. REASONING Trigonometry indicates that the x and y components of the dolphin’s velocity are related to the launch angle θ according to tan = v y / v x .
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This note was uploaded on 10/23/2011 for the course BIO 201 taught by Professor Frenkel/cervantes during the Spring '10 term at Rutgers.

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chap3 - CHAPTER 3 KINEMATICS IN TWO DIMENSIONS ANSWERS TO...

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