- #1
Mary1910
- 31
- 1
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
cos=Adjacent/Hypotenuse
=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. :)
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
cos=Adjacent/Hypotenuse
=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. :)
