Actually a resultant vector can't be obtained from two different objects!
The answer is zero or none of the above!
@markwbird you get half credit for the right answer "none of the choices is correct" but wrong presumption.
Jim...
Calculus professor who has clearly never been sailing
- Thread starter Phil Herring
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The answer is zero or none of the above!
@markwbird you get half credit for the right answer "none of the choices is correct" but wrong presumption.
Jim...
I'm still wondering if the "1 km = 1000 mm is a joke. And if I recall my Physics correctly the perpendicular vectors would not have any effect on each other only if there are no sails up. But at a velocity of 200 km/hr there are likely no sails left! Given the westward velocity of 200 km/hr one must also assume the sailboat is surfing a tsunami so why worry about a 50 km/hr wind heading north?
And I might add, there are 10 kinds of people who understand binary: Those who do and those who don't. But that is a whole 'nother dimension in math.
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Nerds of a feather... I was thinking none are correct either but along different lines of reason.... the keel produces a lift and drag vector that allows us to point.... likewise the the sail has lift and drag vectors as well... and the rudder has lift and drag vectors....so a simple two vector analysis is not going to cut it... (ha ha "cut it") Never a good sign when you laugh at your own puns.
I just watched that again over the Holidays. Awesome. My kids totally did not get the hare krishna jokes. But I remember when I was a kid having to scream at one to leave me alone....My kids grew up post 9-11 and to them airports are sterile places.
What two different objects? There only one object. You are reading too much into a poorly worded questions. He just wants you to add the two vectors to get the resultant vector. 200 km/h West plus 50 km/h North. Don't over complicate it.
Maybe not as crazy as it sounds, with Sailrocket having achieved 120km/h and 2.4 times the wind speed...
And Mark, you are right, velocity is a vector, but if you are to be pedantic then the resulting direction is not the same thing as the heading..the heading is due west..
And Mark, you are right, velocity is a vector, but if you are to be pedantic then the resulting direction is not the same thing as the heading..the heading is due west..
One of my favorite film snippets. We actually used this as the basis of a promotion a couple years ago. It bombed, as I recall, but we had fun anyway.
If he is heading west and is hit by a wind from the south at 50 km/h, he would be on a beam reach. What was his point of sail and the wind speed before the wind shifted from the south? Is the south wind faster or slower than what was blowing when he was heading west at 200 km/h? I am not sure you can figure an answer.
You are missing the point entirely.... of course that is what he wanted but that still does not make it a correct answer
...if you think I understand anything about calc other than "a sailboat wouldn't be moving that fast," you're reading too much into my title.
I couldn't begin to figure that out even if all the necessary data was there, and I had no idea it might not be.Now, that said, I just posted this for some winter fun, I didn't think we'd end up with an argument among the engineers!