# Homework Help: Powerboat Velocity Question

1. Nov 9, 2007

### helpneeded01

This question is confusing, can someone please help me on this question.
A powerboat heads due northwest at 10 m/s relative to the water across a river that flows due north at 4.2 m/s. What is the velocity (both magnitude and direction) of the motorboat relative to the shore?
How would I find the velocity and the degrees west of north?

2. Nov 9, 2007

### helpneeded01

This question is confusing, can someone please help me on this question.
A powerboat heads due northwest at 10 m/s relative to the water across a river that flows due north at 4.2 m/s. What is the velocity (both magnitude and direction) of the motorboat relative to the shore?
How would I find the velocity and the degrees west of north?

3. Nov 9, 2007

### Astronuc

Staff Emeritus
This is a vector sum problem.

One adds the two vectors velocity of the boat 10 m/s at 45° from north with respect to water, and 4.2 m/s (velocity with respect to shoreline) due north (river flows north). Find the resultant vector and the angle with respect to north, then component of velocity due north. Think scalar product.

4. Nov 9, 2007

### helpneeded01

im still not getting this...

5. Nov 9, 2007

### Astronuc

Staff Emeritus
Please refer to these examples of relative motion and please read about vector addition if that concept is not clear.

http://hyperphysics.phy-astr.gsu.edu/hbase/relmot.html#c2

Then please write the velocity vector for the boat with respect to water, i.e. 10 m/s due NW. Then write the velocity vector for the river flow, 4.2 m/s due N.

Then find the sum (resultant vector). Let unit vector i point due E, so -i points due west, and take unit vector j as pointing due north.

Then knowing the resultant vector, find its magnitude and the angle with respect to north.

6. Nov 9, 2007

### helpneeded01

thank you very much :)

7. Nov 9, 2007

### jdogg0075

Assuming when you say "due northwest" meaning 45 degrees west of north. Using this then we can set up vectors to solve this. One of your vectors is 10 m/s at 45 degrees north of west and another is at 4.2 m/s north or 0 degrees. Using law of cosines you can use this information to solve for your resulting vector. This was my equation for law of cosines
C=$$\sqrt{10^{2}+4.2^{2}-2*10*4.2*cos(135)}$$

8. Nov 13, 2007

### helpneeded01

how would you find the angle west of north?