Physics questions~~
2012-09-28 8:01 am
圖片參考：http://imgcld.yimg.com/8/n/HA00098128/o/701209280000413873366360.jpg
2. The captain of a liner steaming due W at 16 knots sees that a launch appears to be approaching him from w N of W at 20 knots. What is the actual velocity of the launch?
5. A bomber flies due N at 1000 km/h and a fighter aircraft flies to intercept S60ﾟW at 1400 km/h. With what relative velocity is the fighter approaching the bomber?
8. A man who can row at 2.5 m/s in still water wishes to cross to the nearest point on the opposite bank of a river 400m wide. If the stream is running at 1.5 m/s, how many minutes does it take him to cross?
9. An observer in a car travelling at 50 km/h in a direction 30ﾟ E of N experiences a wind from the north. Relative to the earth, however, the wind is blowing from due W. Find
(i) the true speed of the wind,
(ii) its apparent velocity when the car is moving at the same speed in the opposite direction.
以圖為正確問題......
回答 (1)
2012-09-30 4:11 am
✔ 最佳答案
2. What is the value of w? It should be given.5. The velocities of the bomber, fighter and the relative velocity of the fighter to the bomber form a triagnle. The angle between the first two velocities is 120 degrees.
Hence, by consine rule,
Vr^2 = 1000^2 + 1400^2 - 2x1000x1400cos(120) (m/s)^2
where Vr is the velocity of the fighter relative to the bomber
Vr = 2088 m/s
Let a be the angle at which the relative velocity of the fighter makes with the South line, we have, using the sine rule,
sin(a)/1400 = sin(120)/2088
hence, a = 35.5 degrees west of south
6. Let a be the angle that makes with the bank at which the man rows. If the man wants to reach the point directly across the river.
Hence, 2.5cos(a) = 1.5
a = 53.13 degrees
Time neede to cross the river = 400/[2.5sin(53.13)] s = 200 s
9. (i) The velcoities of the car, the wind and the relative velocity of the winfd to the car form a right angle triangle. The hypothenus represents the velocity of the car.
Hence, true speed of the wind = 50sin(30) m/s = 25 km/h
(ii) The car is now moving at direction 30 degrees west of south.
The angle between the velocities of the car and wind is now 120 degrees
Using the cosine rule,
v^2 = 50^2 + 25^2 - 2x25x50.cos(120) (m/s)^2
where v is the apparent speed of the wind relative to the car.
v = 66.14 m/s
Let a be the angle between the apparent velocity of the wind and the velocity of the car. Using the sine rule,
sin(a)/25 = sin(120)/66.14
a = 19.11 degrees
Therefore, the apparent velocity of the wind = (19.11+30) degrees = 49.11 degrees west of south
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