Answers to selected problems from Essential Physics, Chapter 4

Transcription

1 Answers to selected problems rom Essential Phsics, Chapter 4 1. (a) Point our kaak perpendicular to the riverbank. To cross the river in the shortest time, ou need to maximize the component o our velocit that is directed across the river. The current in the river aects onl the component o our velocit that is directed parallel to the river, so the current can neither add to nor subtract rom our velocit directed across the river. To maximize our velocit across the river, thereore, ou must direct as much o our velocit relative to the water across the river, which means aiming our kaak perpendicular to the river. (b) To land directl across rom our starting point, ou must aim the kaak upstream, so that the component o our velocit relative to the water that is directed parallel to the river exactl cancels the velocit o the water relative to the riverbank (the current, in other words). 3. An origin can be used to answer the question. Assuming that ou both use a traditional coordinate sstem with coordinate axes directed horizontall and verticall, ou would agree on the value o the x-coordinate o the initial position, and the x- coordinate o the inal position. You would disagree on the value o the -coordinate o the initial position, and the -coordinate o the inal position, but the ke thing is that ou would agree on the magnitude o the displacement in the -direction. You would also agree on the answer to the time it takes or the object to reach the water, assuming that ou both did the calculations correctl. 5. (a) A=B=C (b) A=B=C (c) C>B>A (d) C>B>A 7. (a) G>F>E (b) G>F>E (c) E>F>G (d) G>F>E (a) 4H (b) T (c) 4R 13. (a) 0 km/h north (b) 0 km/h south (c) 10 km/h north (d) 190 km/h south The answers sta the same even ater the vehicles change positions. Finding relative velocities simpl involves adding or subtracting two velocit vectors, and the Essential Phsics, Answers to selected Chapter 4 Problems Page 1

2 velocit vectors are the same no matter what the relative positions o the car, truck, and motorccle are. 15. (a) 70 s (b) 79 s 17. (a) 3 m/s east (b) 1 m/s east (c) We can t sa. All we have is inormation about the velocities o these people relative to one another. We have no inormation about the velocit o an o them with respect to the lamppost. 19. (a) 1.5 hours (b) 150 km/h west (c) You have to l due east or 4.5 more hours. 1. (a).0 s (b).0 s (c).0 m (east o the base o the mast) 3. (a) 1.8 m (b) 6.0 m/s, directed straight up (c) zero 5. x-direction -direction i = 0 at max. height max = m inal position x = m = 0 initial velocit v =? v i =? a = 9.81 m/s () v = m/s ; time o light is s; v = m/s. i This gives an initial velocit o 18.6 m/s at an angle o 69.6 above the horizontal. 7. Essential Phsics, Answers to selected Chapter 4 Problems Page

3 x-direction -direction i = 0 inal position x = +.0 m = m initial velocit v =? v i =? inal velocit v = 0 a = 9.8 m/s () v = m/s ; time o light is 1.11 s; v = m/s. i This gives an initial velocit o 11.0 m/s at an angle o 80.5 above the horizontal. 9. x-direction -direction i = 0 inal position x = m = 0 initial velocit v =? v i =? a = 9.81 m/s time 4.56 s 4.56 s () v =+.37 m/s ; v = m/s. i This gives an initial velocit o 4.9 m/s at an angle o 63.9 above the horizontal. 31. x-direction -direction i = 0 max. height max = m inal position x = + m = +.0 m initial velocit v =? v i =? a = 9.8 m/s Essential Phsics, Answers to selected Chapter 4 Problems Page 3

4 () v =+ 8.8 m/s ; t = 1.40 s; v = m/s. i (g) (h) 33. x-direction -direction i = m inal position x =? = 0 initial velocit v =+ (1.0 m/s) cos(34.0 ) vi = + (1.0 m/s)sin(34.0 ) v = m/s v = m/s a = 9.81 m/s i () t = 4.10 s (g) 40.8 m Essential Phsics, Answers to selected Chapter 4 Problems Page 4

5 35. x-direction -direction i = 0 m inal position x =? =? initial velocit v = 0 v i = m/s inal velocit vx = axt =+ 0 m/s v = m/s (acceleration phase) acceleration a =+ 4.0 m/s a = 0 x () 56 m (g) 160 m 37. (a) 19.6 m/s, up (b) 1.5 m/s, directed downield (c) 3. m/s, at an angle o 57.5 above the horizontal m 43. (a) 15 km/h at an angle o 8.6 west o north (b) 138 km/h at an angle o 4.8 west o north 45. Powell must have been running at about 9.4 m/s when he took o. World-class sprinters run at about 10 m/s, so Powell s speed is just a little less than that o a worldclass sprinter. One part o being a long-jumper, in act, is excellent speed (a) The ball was caught at a higher level than where it was launched. You can tell this rom the asmmetr in the graph o the -component o the velocit. I the ball came down to the same level rom which it was launched, the inal speed would have matched the initial speed. Because the inal speed is lower than the initial speed, the ball must be higher than the launch point. (b) 30 m (c) At t =.0 s. The ball s -component o velocit is zero at the highest point, and t =.0 s is when the -velocit passes through zero in this case. Essential Phsics, Answers to selected Chapter 4 Problems Page 5

6 With these numbers, the ball is.8 m below the launch point. 55. (a) The shortest time is zero. The longest time is 4.0 seconds. (b) 0 m (c) 40 m 57. (a) 13.9 m/s (b) 16.0 m/s (c) You can t do it! A launch angle o 45 gets the closest to our riend, but the range in that case is 19.6 m, a little short o reaching our riend. You can t reach our riend at an angle unless ou increase the speed o the ball rom what it is in (a) and (a) 1.4 m (b) 1.35 m (c) 1. m Essential Phsics, Answers to selected Chapter 4 Problems Page 6