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Boat crossing river question?
A boat can go at 10 m/s.
a) in what direction (angle) must the motorboat be pointed to directly cross the river flowing 3.0m/s east.
b) what is the velocity of the boat directly across?
a) I used pythagorean theorem: Tan(thetha) = 3/10 -> (Theta) = tan^-1(0.3) = 16.7 degrees
but the answer says the angle is 17.5 and I only get that if used sin instead of tan.
b) since there's a right angle I used the pythagorean theorem to get 10.4m but that's about all I got.
3 Answers
- Old Science GuyLv 71 decade agoFavorite Answer
start with a diagram
it will only be an approximation until you do some calculations
but it will tell you about some basic relationships
draw a line straight across (resultant)
draw a line angled a bit upstream, maybe 20 deg or so (the boat)
draw a line perpendicular to the resultant so that it crosses the boat line (stream flow)
now you have a rt. triangle with boat as hypotenuse and boat and stream as sides.
the upstream angle is indeed found by
sin(angle) = 3/10
angle = 17.46 deg
and
resultant^2 + stream^2 = boat^2
X^2 + 9 = 100
X = 9.54 m/s
so
the boat must sail 10 m/s at 17.45 deg upstream
to realize 9.54 m/s straight across
- KnowallLv 51 decade ago
As 10m/s is the max speed, hence 10 is the Hypotenuse. So Sin Inv (Theta) = 17,5 deg. is correct.
To understand this vector,
It has to be pointed due West of North so that its projection in West direction would cancel velocity of the flow 3m/sec.
The resultant velocity will be 10 Cos (17.5) = 9.5 m/s due North
