This preview shows page 1. Sign up to view the full content.
depends on
t
and on the boat’s speed (relative to the water), and we set it equal to the
Pythagorean “sum” of the triangle’s sides:
4 0
200
82
11
2
2
..
b
g
b
g
tt
=+
+
which leads to a quadratic equation for
t
46724
180 4
14 8
0
2
+−=
...
We solve this and find a positive value:
t
= 62.6 s.
The angle between the northward (200 m) leg of the triangle and the hypotenuse (which
is measured “west of north”) is then given by
θ
=
+
F
H
G
I
K
J
=
F
H
G
I
K
J
=°
−−
tan
.
tan
.
11
82
11
200
151
200
37
t
82. We construct a right triangle starting from the clearing on the
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

Click to edit the document details