# Homework Help: A trig problem? Need help with bearings and shotty wording.

1. Oct 10, 2009

### roger911

1. The problem statement, all variables and given/known data

A bird take 8.5 seconds to travel from point A to point B. At point A, the velocity of the bird is 4.4m/s [31 degrees S of E], and at point B, the bird's velocity is 7.8m/s [25 degrees N of E]. What is the average rate of acceleration?

2. Relevant equations

A=V/T, Ax-Vx/T, Ay=Vy/t, Tan(angle)=Ay/Ax, V=D/t, Viy=Vi sin(angle i), Vix=Vi Cos(angle i), repeat vi(x), vi(y) for vf variables.

3. The attempt at a solution

A=0.4 m/s^

The part i don't understand is the bearing. What i DO understand is that to get the final bearing you use the Tan(angle)=Ay/Ax equation. I can get both initial velocity bearings for x and y, but I am having problems figuring out how to get the final velocity's x and y. I know that #1, the difference in degree's between point A and point B is 56 degrees. I also know that when making a right angle triangle with the points, the angle of point b is 59 degrees. My problem is finding the bearing of acceleration. I've been stumped on this one for a day or 2 now, and my math skills definitely need some practice.

2. Oct 10, 2009

### tiny-tim

Welcome to PF!

Hi roger911! Welcome to PF!

(try using the X2 tag just above the Reply box )

I'm not sure what you're doing, so let's start again …

velocities and accelerations are vectors, so they add (or subtract) like vectors, and average acceleration = (difference in velocity)/time.

ie a (the average acceleration) = (vf - vi)/t.

What did you get for vi and vf ?

3. Oct 10, 2009

### rl.bhat

Hi roger911, welcome to PF.Draw a co-ordinate axis at O. Draw the position vector OA and OB. The change velocity is equal to (OB - OA). By taking the components find the change in velocity and hence the acceleration.

4. Oct 10, 2009

### roger911

Re: Welcome to PF!

The part of the course I'm working on is "Motion in two dimensions". I obtained A = 0.4m/s^2 (where ^ is an exponential value and 2 gives the variable hence squared), by using Vi = 4.4m/s, and Vf= 7.8 m/s. Now 7.8 - 4.4 = 3.4, and the time is 8.5 seconds, so divide 3.4 by 8.5 and you get 0.4 m/s^2.

The problem I'm having is with the direction. In order to get the direction, you must break down acceleration and/or velocity into 2 directions, using X and Y.

Looking above : Vi = 4.4m/s [31 degrees S of E], Vf = 7.8 m/s [25 degrees N of E].

For instance, Vi(y)=Vi Sin(angle). Therefor Vi(y) = 4.4m/s Sin(31 degrees). Which equates to 2.26 or 2.3m/s. We can apply the same equation, only using Cos to obtain the X value which is 3.77 or 3.8m/s.

The issue I am having is figuring out the values for Vf(y) and Vf(x), because in relation to Trig, I don't know what angle to input when doing the calculations. This is because of the way that they measure the angles.

I know that if you were to draw a triangle, since all sides = 180 degrees, you subtract 90 (since it would be a right angle triangle) and 31, leaving you with 59 degrees.

Would i input the 59 degrees into the equation, or would I input the 25 degrees given to me? Or would I notice that 25 degrees N of E is not the same as 25 degrees S of E, then realizing the difference between the 2 angles is 56 degrees. At this point I get confused.

5. Oct 10, 2009

### tiny-tim

Sorry, roger911, that's completely wrong.

Vectors don't add (or subtract) like that.

4.4 and 7.8 are the lengths of the vectors from the origin to the two velocities.

The difference in velocity is the third side of the triangle, the vector joining the two velocities.

That third side is not 3.4 long.