# kinematics in 2d problem

• Feb 7th 2010, 11:09 AM
fishkeeper
kinematics in 2d problem
Hi, Im having problems with this question:

A boat heads north and 20kph
A wind is blowing SE with a constant speed of 2ms^1

a) Find the magnitude of the speed of the boat
b) Find the bearing it travels at

**************

Im having problems with the triangle, drawing it out makes the 2 ms^1 the hypotenuse when, to find the speed shouldnt be the hypotenuse i.e the hypotenuse has to be found?

The using pythagoras I get the resultant being 5.2ms^1 which is complete rubbish.

Thankyou very much in advance! http://static.thestudentroom.co.uk/n...ns/editnew.gif
• Feb 7th 2010, 11:37 AM
Random Variable
The angle between the velocity vector of the boat (20 km/hr) and the velocity vector of the wind (7.2 km/hr) is 45 degrees.

The speed of the boat can be found by the law of cosines.

$c^{2} = 20^{2}+7.2^{2} - 2(20)(7.2) \cos (45°)$

c = 15.75 km/hr

The direction of the boat can be found by the law of sines.

$\frac{15.7542}{\sin(45°)} = \frac{7.2}{\sin B}$

$\sin B = \frac{(7.2)(\frac{\sqrt{2}}{2})}{15.7542}$

B = 18.9° east of north
• Feb 7th 2010, 01:05 PM
fishkeeper
thankyou

How did you work out what the angle was though?

thanks!
• Feb 7th 2010, 01:25 PM
Random Variable
Quote:

Originally Posted by fishkeeper
thankyou

How did you work out what the angle was though?

thanks!

$B = \arcsin\Big(\frac{(7.2)(\frac{\sqrt{2}}{2})}{15.75 42} \Big)$ and then I used a calulator (making sure I was in degrees mode)

and according to the triangle, the vector is to the east of the velocity vector of the boat