# Calculate the components of the vector

1. Dec 1, 2007

1. The problem statement, all variables and given/known data
Calculate the components of the vector representing a velocity of 40m/s and an 55 degree angle.

2. Relevant equations
none

3. The attempt at a solution
180-55=125
40(sin125)=33
40(cos125)=-23
(-23,33)m/sec

correct?

2. Dec 1, 2007

### rock.freak667

Why did you take 180-55?

you could have just used the 55 degree angle...but it is still correct...

3. Dec 1, 2007

not really..

40cos125=-23
40cos55=23

4. Dec 1, 2007

### rock.freak667

...you would get the same resultant....but the vectors are the same, just in opposite directions

5. Dec 1, 2007

### amolv06

I suppose it depends on what orientation you use, but generally I would have thought that cos(55) would be what was being asked for.

6. Dec 1, 2007

### FedEx

The answers would be definitely coming different. As you are using two different vectors. Why are you doing 180 - 55.Its just 55. And doing so you will get its x and y components.
When you take 180 - 55 then you are taking a vector which is in the second quadrant which is different from the vector in the first quadrant.And we want to find the component of the vector which is in the first quadrant.

7. Dec 1, 2007

so I should use 55? In my textbook it showed 180-55, thats why I used 125. But that example involved a car route or something.

8. Dec 2, 2007

### cristo

Staff Emeritus
Yes, you should use 55.

This is a prime example of focusing too much on equations and trying to follow previous methods without actually thinking physically about what you are trying to do.

Draw a right angled triangle, with the bottom angle of 55 degrees, and the length of the hypotenuse as 40. Now, the x and y components are the horizontal and vertical sides of the triangle, respectively. You should be able to use trigonometry on this triangle in order to find the components.

9. Dec 2, 2007