You swim at 3 km/hr with your body perpendicular to a stream. Because of the stream's flow, your actual velocity vector makes an angle of 25 degrees with the direction you are heading.
a. How fast is the current of the stream?
b. What is the magnitude of your actual velocity?
2007-05-27 09:03:16 · 3 answers · asked by Dr. Richard Anderson, M.D., M.S. 3 in Science & Mathematics ➔ Mathematics
If you swim at 3km/hr in the perpendicular direction
Consider the problem in distances swimed in 1 hour
In 1 hour you will swim 3 km deep
Considering a right triangle with a 25° angle and 3 km length
................
......../.|..
..b../...|..
..../.....|.3 ....... <===========
../.......|.. .......... Stream Current
/____|..
.....a......
tan 25 = a/3
=> a = 1.40
cos 25 = 3/b
=> b = 3.31
so the stream is 1.40 km/hr fast
your velocity is 3.31 km/hr
.
2007-05-27 09:18:19 · answer #1 · answered by Anonymous · 1⤊ 0⤋
a) well I could take the length of the opposite side of the right triangle described by this 25 degree angle and the staight line you are swimming to be the magnitude of the stream's velocity, using basic trig
b) since magnitude and direction of the stream's velocity is know, and yours is known, add them.
2007-05-27 09:12:56 · answer #2 · answered by Anonymous · 0⤊ 2⤋
x = current speed
x = 3 tan25 = 1.4 km/h
actual velocity = 3/cos 25 = 3.3 km/h
2007-05-27 09:16:56 · answer #3 · answered by ironduke8159 7 · 0⤊ 1⤋