# Homework Help: Average acceleration problem

1. Dec 1, 2007

### Imperil

Hi! I am new here and can't for the life of me figure out what I am doing wrong solving this problem, if someone could help that would be great!

QUESTION:
A bird takes 8.5 s to fly from position A (va= 4.4 m/s (31 ° S of E) to position B(vb= 7.8 m/s ( 25° N of E) along the path. Find the bird's average acceleration.

vax/va = cos31
vax = 3.8 m/s

vay/va = sin31
vay = -2.3 m/s

vbx/vb = cos25
vbx = 7.1 m/s

vby/vb = sin32
vby = 3.3 m/s

ax = vbx - vax / t
ax = 7.1 m/s - 3.8 m/s / 8.5s
ax = 0.4 m/s(2)

ay = vby - vbx / t
ay = 3.3 m/s + 2.3 m/s / 8.5s
ay = 0.7 m/s(2)

a(2) = ax(2) + ay(2)
a(2) = 0.16 + 0.49
a = 0.8 m/s(2)

tan(theta) = ay / ax
tan(theta) = 0.7 m/s(2) / 0.4 m/s(2)
tan(theta) = 1.75
theta = 60 degrees

Therefore the bird's acerage acceleration is 0.8 m/s 60 degrees N of E.

2. Dec 2, 2007

3. Dec 2, 2007

### Imperil

The correct answer isn't actually listed so I'm not sure what it is, my answer just does not seem right to me as the angle seems wrong.

4. Dec 2, 2007

### Joshrk22

Hmm, I came up with .8 m/s/s also, and I did that even before I looked at your answer, so I don't know.

5. Dec 2, 2007

### Imperil

Well I thought I was correct with the .8 m/s/s but what I really wasn't sure about was the 60 degrees N of E. Thanks for trying it out btw!

6. Dec 2, 2007

### Joshrk22

I also used v=at and x=.5at^2. I now came up with 1.5 m/s/s at 60 degrees. So now I'm stuck between which answer it is.. But I do believe it's 60 degrees.

7. Dec 2, 2007

### Imperil

Hmm how did you come up with 1.5 m/s/s if I may ask?

8. Dec 2, 2007

### Joshrk22

I took the x and y components of each vector and used x=vt. I then found the x and y components of the distance. I found the change in distances of each vector to get the resultant vector (distance). I then used x = .5at^2 and used the time above to get 1.5 m/s/s at 60 degrees. But I think I did something wrong with the math because you'd think it'd come out to be .8 m/s/s again.

9. Dec 2, 2007

### Imperil

Great thanks so much for the help, I'll do it over again just to see if I get 0.8 or 1.5, but I'm glad we come out with the same angle.