# Motion in Two Dimensions

1. May 8, 2015

### Mary1910

Question:
In an anniversary celebration of Marilyn Bell's 1954 feat- she was the first person to swim across Lake Ontario- a swimmer set out from the shores of New York and maintained a velocity of 4m/s (N). As the swimmer approached the Ontario shore, she encountered a cross-current of 2 m/s (E 25 degrees S) - 25 degrees South of East. Calculate her velocity with respect to the crowd observing from the beach (with respect to the ground).

My Attempt at a Solution:

First I drew a triangle. 4m/s (N) is the left side of the triangle. 2m/s (E 25 degrees S) is the top of the triangle. And the right side is the side that I think that I am solving for.

Next I divided the triangle into two right angled triangles.

Solving for the top triangle:

sin=Opposite/Hypotenuse
=2sin(65)
=1.81 m/s

=2cos(65)
=0.845 m/s
=0.85m/s

To solve for the bottom triangle:

4m/s - 0.85m/s = 3.15m/s

Use Pythagorean theorem to solve for the unknown side:

3.15^2 + 1.81^2
=3.63m/s

Calculate the direction of the velocity:

tan=ay/ax
tan=1.81/3.15
=29.88
=30 degrees

Therefore her velocity, with respect to the crowd observing from the beach is 3.63m/s (N 30 degrees E)

Could someone please let me know if my calculations are correct? I would appreciate the help. :)

2. May 9, 2015

### paisiello2

Looks right to me.