Physics 1 HW #8: Chapter 3

Transcription

1 Phsics 1 HW #8: Chapter 3 Problems 6-9, 31, 3, 49, 54. For EVERY one of the problems, draw ector diagrams, labeling the ectors and write out the ector equation! Hae fun! 3.6 We use the following notation: and elocit of boat relatie to the shore elocit of boat relatie to the water, elocit of water relatie to the shore. If we take downstream as the positie direction, then 1.5 m s for both parts of the trip. Also, 10 m s while going downstream and 10 m s for the upstream part of the trip. The elocit of the boat relatie to the shore is gien b While going downstream, 10 m s 1.5 m s and the time to go 300 m downstream is 300 m tdown 6 s m s When going upstream, 10 m s 1.5 m s 8.5 m s and the time required to moe 300 m upstream is t up 300 m 35 s 8.5 m s The time for the round trip is t t t 6 35 s 61 s down up

2 3.7 Prior to the leap, the salmon swims upstream through water flowing at speed 1.50 m s relatie to Earth. The fish swims at 6.6 m s relatie to the water in such a direction to make its elocit relatie to Earth,, ertical. Since FE as shown in the diagram at the right, we find that FE FW FW m s cos cos m s FW and the ertical elocit of the fish as it leaes the water is 0 FE FW sin 6.6 m s sin m s The height of the salmon aboe the water at the top of its leap (that is, when 0 ) is gien b m s a 9.80 m s 1.88 m m s, directed northward, is the elocit of the boat relatie to the water. 1.5 m s, directed eastward, is the elocit of the water relatie to shore. is the elocit of the boat relatie to shore, and directed at an angle of, relatie to the northward direction as shown. The northward component of is cos 10 m s (1) The eastward component is sin 1.5 m s () (a) Diiding equation () b equation (1) gies tan tan 8.53 From equation (1), 10 m s 10.1 m s cos8.53 Therefore, 10.1 m s at 8.53 E of N

3 300 m 300 m (b) The time to cross the rier is t 30.0 s cos 10.0 m s and the downstream drift of the boat during this crossing is drift sin t 1.50 m s 30.0 s 45.0 m 3.9 elocit of boat relatie to the water, elocit of water relatie to the shore and elocit of boat relatie to the shore. as shown in the diagram. The northward (that is, cross-stream) component of is sin mi h sin mi h north The time required to cross the stream is then mi t h.93 mi h The eastward (that is, downstream) component of east cos6.5 is 3.30 mi h cos mi h 0.74 mi h Since the last result is negatie, it is seen that the boat moes upstream as it crosses the rier. The distance it moes upstream is 580 ft d t 0.74 mi h0.173 h mi 49 ft east 1 mi

4 3.31 Choose the positie direction to be the direction of each car s motion relatie to Earth. The elocit of the faster car relatie to the slower car is gien b FS, where 60.0 km h is the FE ES FE elocit of the faster car relatie to Earth and ES SE 40.0 km h is the elocit of Earth relatie to the slower car. Thus, 60.0 km h 40.0 km h 0.0 km h and the time required for the faster car to moe 100 FS m (0.100 km) closer to the slower car is d km 3600 s t 0.0 km h 1 h FS h 18.0 s 3.3 BC the elocit of the ball relatie to the car elocit of the car relatie to Earth 10 m s CE the elocit of the ball relatie to Earth These elocities are related b the equation as illustrated in the diagram. Considering the horizontal components, we see that BC CE CE 10.0 m s cos60.0 CE or BC 0.0 m s cos60.0 cos60.0 From the ertical components, the initial elocit of the ball relatie to Earth is sin m s BC, with 0 when the ball is at maximum height, we find Using a m s max a g 9.80 m s 15.3 m as the maximum height the ball rises.

5 3.49 The elocit of the boat relatie to the shore is, where is the elocit of the boat relatie to the water and is the elocit of the water relatie to shore. In order to cross the rier (flowing parallel to the banks) in minimum time, the elocit of the boat relatie to the water must be perpendicular to the banks. That is, Hence, the elocit of the boat relatie to the shore must be 1 km h 5.0 km h 13 km h must be perpendicular to km h at tan tan km h same as the line of the rierbank). to the direction of the current in the rier (which is the The minimum time to cross the rier is width of rier 1.5 km 60 min t 1 km h 1 h 7.5 min During this time, the boat drifts downstream a distance of 3 1 h 10 m 5.0 km h 7.5 min m d t 60 min 1 km 3.54 We are gien that: 0 knots due north (elocit of boat relatie to Earth) and 17 knots due east (elocit of wind relatie to Earth)

6 The elocit of the wind relatie to the boat is WB EB where 0 knots south is the elocit of Earth relatie to the boat. The ector diagram EB aboe shows this ector addition. Since the ector triangle is a 90 triangle, we find the magnitude of to be WB WB EB 17 knots 0 knots 6 knots and the direction is gien b 0 knots 17 knots 1 EB 1 tan tan 50 Thus, WB 6 knots at 50 south of east From the ector diagram aboe, the component of this elocit parallel to the motion of the boat (that is, parallel to a north-south line) is seen to be 0 knots south EB