# Thread: kinematics in 2d problem

1. ## 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.

2. 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.

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

c = 15.75 km/hr

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

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

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

B = 18.9° east of north

3. thankyou

How did you work out what the angle was though?

thanks!

4. Originally Posted by fishkeeper
thankyou

How did you work out what the angle was though?

thanks!
$\displaystyle B = \arcsin\Big(\frac{(7.2)(\frac{\sqrt{2}}{2})}{15.75 42} \Big)$ 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