Question

A motorboat heads due west at 13 m/s relative to a river that flows due north at 5.0 m/s. What is the velocity (both magnitude and direction) of the motorboat relative to the shore?

I got 16.9 m/s at 57 degrees. I'm not sure if I'm right. Any help would be greatly appreciated :).

Answer 1

given:
\(v_{mc}=13\ m/s\\v_{cg}=5.0\ m/s\\v_{mg}=?\)
\(v_{mg}=v_{mc}+v_{cg}\)
\(|v_{mg}|=\sqrt{(v_{mc})^2+(v_{cg})^2}\\|v_{mg}|=\sqrt{(13)^2+(5)^2}\\|v_{mg}|=13.9\ m/s\)

Answer 2

Really? That's interesting. So far, everyone has gotten the same velocity as me. I used the Pythagorean Theorem, @Data_LG2.

Answer 3

\(\Large tan\ \theta =(\frac{v_{cg}}{v_{mc}})\\\theta =tan^{-1}(\frac{v_{cg}}{v_{mc}})\\\theta =tan^{-1}(\frac{5}{13})\\\theta =21°\)

Answer 4

I also used the Pythagorean theorem

Answer 5

Huh, I used different numbers. I squared 13 and got 169. Then divided it to get two components. And got 84.5. Then I took the square root and got 9.19. Then, for one of the components I added 5.0 to 9.19 = 14.19. I then used 9.19 and 14.19 as my A^2 and B^2.

Answer 6

it says on the problem that the motorboat is going WEST, so it means no need to get the components because it's already on x-component and the river/current flows NORTH, which means that there's also no need to get the components since it also lies on the y-component.. so just apply the pythagorean theorem for this problem.


for your solution, where did you divide 169 to get two components??

Answer 7

Ohhhh. Wow. That makes so much more sense. I just, wow. You are completely right! I didn't take into account the direction, wow, @Data_LG2.

Answer 8

I'm glad that you understand it ^.^

Answer 9

Thank you so much, @Data_LG2! :)

Answer 10

np :)