How does wind acceleration affect the speed and direction of a sailboat?

  • Thread starter Thread starter ere
  • Start date Start date
  • Tags Tags
    Wind
ere
Messages
9
Reaction score
0

Homework Statement


A sailboat is traveling east at 4.5ms-1. a suddent gust of wind gives the boat an acceleration a = 0.70 ms-2, 40 degress north to east. what are the boat's speed and direction 6s later when the gust subsides?

Homework Equations


V(@6s) = V(init.) + at

The Attempt at a Solution


i seriously have not a clue where to start. tried to look it up in books and online but i still don't know what to do with the acceleration.
so the a(x) = 0.70cos40 V(x) = 4.5
a(y) = 0.70sin40 V(y) = 0 am i right?​
thus, V(6s)(x) = 4.5 + .070cos40(6) = 7.7174
V(6s)(y) = 0 + 0.70sin40(6) = 3.8567​
therefore, direction n in degree:
tan(n) = 3.8567 / 7.7174, n = 26.55 degrees​
velocity = (7.7174^2 + 3.8567^2)^(1/2) = 8.6274​
and i am told that i am wrong. hope you guys can point out why. thanks!
 
Last edited:
on Phys.org
It does not appear to me that you have enough inforation to solve the problem. You need the time of the acceleration.

Unless it is a trick question and the answer is 4.5m/s, East...

Anyway, your analysis is certainly wrong as you are mixing velocity and acceleration.
 
oops sorry i left that out. just edited the thread.
 
Well, then the answer is that you mixed acceleration and speed. You need to use the acceleration and the time to calculate the magnitude of the second speed vector before adding the two vectors together.
 
ere said:
so the a(x) = 0.70cos40 V(x) = 4.5
a(y) = 0.70sin40 V(y) = 0 am i right?​
thus, V(6s)(x) = 4.5 + .070cos40(6) = 7.7174
V(6s)(y) = 0 + 0.70sin40(6) = 3.8567​
therefore, direction n in degree:
tan(n) = 3.8567 / 7.7174, n = 26.55 degrees​
velocity = (7.7174^2 + 3.8567^2)^(1/2) = 8.6274​
and i am told that i am wrong. hope you guys can point out why. thanks!

What is wrong? and Where? What do you mean by confusing velocity and acceleration? Find the effective acceleration for both Y and X axis, then use the
linear laws of motion to get the final velocity components? Why is this wrong?
 
ere said:
V(6s)(y) = 0 + 0.70sin40(6) = 3.8567[/indent]

This calculation seems to have error
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
3K
Replies
1
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
Replies
3
Views
12K
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K