# QUESTION:What is the bird's average acceleration along its flight path?

• Imperil
In summary, the question asked for the average acceleration of a bird flying from position A to position B along a given path. By calculating the x and y components of the bird's velocity at each position and using the formulas v=at and x=.5at^2, two possible answers were obtained: 0.8 m/s(2) at 60 degrees N of E and 1.5 m/s(2) at 60 degrees. Further calculations may be needed to determine the correct answer.
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.

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.

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.

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!

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.

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

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.

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.

## What is average acceleration?

Average acceleration is a measure of how quickly the velocity of an object changes over a specific period of time. It is calculated by dividing the change in velocity by the change in time.

## How is average acceleration different from instantaneous acceleration?

Average acceleration is calculated over a specific period of time, while instantaneous acceleration is the acceleration at a specific moment in time. Average acceleration gives an overall idea of the change in velocity over a period of time, while instantaneous acceleration gives the exact acceleration at a particular moment.

## What are some real-life examples of average acceleration?

Some examples of average acceleration in everyday life include a car accelerating from 0 to 60 mph in a certain amount of time, a rollercoaster accelerating down a hill, or a person running and gradually increasing their speed.

## How is average acceleration represented graphically?

Average acceleration is represented graphically by a line on a velocity vs. time graph. The slope of this line represents the average acceleration, with a steeper slope indicating a higher acceleration and a flatter slope indicating a lower acceleration.

## What are the units of average acceleration?

The units of average acceleration are distance over time squared, such as meters per second squared (m/s^2) or kilometers per hour squared (km/h^2). This is because acceleration is a measure of how much velocity changes over time.

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