Question

A boat can travel 2.20 m/s in still water.
(a) If the boat points its prow directly across a stream whose current is 1.50 m/s, what is the velocity (magnitude and direction) of the boat relative to the shore?
(b) What will be the position of the boat, relative to its point of origin, after 3.00 s? (See Figure 3-30.)
(Distance downstream and distance across the river.)

Answer 1

Figure 3-30
http://www.webassign.net/giancoli/3-30fig.gif

Answer 2

@heylainey what's prow?

Answer 3

I have no idea, honestly. That's just what the problem says. :/

Answer 4

Is it a word or is it a letter with subscript?

Answer 5

I think it's an actual word, like a part of the boat.

Answer 6

It's the boat's noze, i think.

Answer 7

The first one you use a right triangle vector.
use Pythagorean theorem formula
C^2 = a^2 + b^2
C = the resultant vector which is the hypotenuse of the triangle = BS
a = adjacent side which is the direction of bat crossing at 2.2 m/s = BW
b = river direction perpendicular = WS
C^2 = (2.2)^2 + (1.5)^2
C = sq rt [ (2.2)^2 + (1.5)^2 ]
C = 2.66 m/s magnitude

direction of vector C = theta angle = 1/tan * ( 2.2/ 1.5)
theta angle = 55.7 degrees toward the NW

So resultant is boat goes 2.66 m/s in NW direction 55 deg,( left of 90 degree)
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(b) What will be the position of the boat, relative to its point of origin, after 3.00 s?

1.5 m/s * 3s = 4.5 m
2.2 m/s * 3s = 6.6 m

C^2 = a^2 + b^2
C^2 = (6.6)^2 + (4.5)^2
C = 8.0 meters from origin heading 55.7 degree NW