# Math Help - Vectors

1. ## Vectors

1. Find the magnitude and direction of the resultant velocity vector for the following perpendicular velocities.

a. a fish swimming at 3.0 m/s relative to the water across a river that moves at 5.0 m/s

b. a surfer traveling at 1.0 m/s relative to the water across a wave that is traveling at 6.0 m/s

2. Find the resultant displacement of a fox that heads 55 degrees north of west for 10.0 meters, then turns and heads west for 5.0 meters.

2. Originally Posted by rod789
1. Find the magnitude and direction of the resultant velocity vector for the following perpendicular velocities.

a. a fish swimming at 3.0 m/s relative to the water across a river that moves at 5.0 m/s

b. a surfer traveling at 1.0 m/s relative to the water across a wave that is traveling at 6.0 m/s

2. Find the resultant displacement of a fox that heads 55 degrees north of west for 10.0 meters, then turns and heads west for 5.0 meters.
to #a):

Use a coordinate system: Let the direction of the current be the positive x-axis, then the direction in which the fish is swimming forms the positive y-axis:

$\vec c = (5,0)$ .... and .... $\vec f = (0,3)$

Then the resulting velocity is

$\vec r = \vec c + \vec f$....I've got $\vec r = (5,3)$

Calculate the length of the vector. If $\vec v = (a,b)$ then $|\vec v| = \sqrt{a^2+b^2}$....I've got $|\vec r| = \sqrt{34}$

to #b): This question has to be done in exactly the same way.

to #2:

Do you mean W55°N or do you mean N55°W? I'll take the first version. If this is wrong then you know at least how to do the problem.

Use a coordinate system according the geographical directions. The positive x-axis is pointing East, the positive y-axis is pointing North.

Then the first part of the way of the fox is:

$\overrightarrow{w_1} = 10(-\cos(55^\circ)\ ,\ \sin(55^\circ))$ ........... and the second part is:

$\overrightarrow{w_2} = 5(-1, 0)$

$\vec r = \overrightarrow{w_1} + \overrightarrow{w_2} = \left( -10\cos(55^\circ)-5\ ,\ 10\sin(55^\circ) \right)$

The displacement is $d = |\vec r| = 5\cdot \sqrt{(-10\cos(55^\circ)-5)^2+(10\sin(55^\circ))^2}$ ..........I've got $|\vec r| \approx 13.5$