What Is the Average Acceleration of a Bird Flying from Point A to B?

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In summary, the bird takes 8.5 seconds to fly from position A to position B along the given path. Using the components of the bird's initial and final velocities, it can be determined that the bird's average acceleration was 0.8 m/s(2) at an angle of 60 degrees north of east.
  • #1
lanvin
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A bird takes 8.5s to fly from position A to position B along the path in the figure shown. Determine the bird's average acceleration.

http://i299.photobucket.com/albums/mm286/lanvin12/physics.jpg

Vax = (4.4m/s)sin31° = 2.26616m/s
Vbx = Vbsin25° = (7.8m/s)sin25° = 3.2964m/s
Vx = Vbx - Vax = 3.2964m/s - 2.26616m/s = 1.030262m/s

Vay = Vacos31° = (4.4m/s)cos31° = 3.771536123m/s
Vby = Vbcos25° = (7.8m/s)cos25° = 7.069200739m/s
Vy = Vby + Vay = 3.771m/s + 7.0692m/s = 10.840736m/s

V² = Vx² + Vy² = (10.840m/s)² + (1.030m/s)² = 118.530155m/s
V = √118.530155m/s = 10.8895m/s

a = v/t = 10.8895m/s) / (8.5s) = 1.2811m/s²

tan theta = 10.84 / 1.030 = 10.522
theta = tan^(-1) 10.522 = 84.57 degrees

therefore...
the bird's acceleration was 1.3m/s² [85 degrees N of E]
 
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  • #2
lanvin said:
A bird takes 8.5s to fly from position A to position B along the path in the figure shown. Determine the bird's average acceleration.

http://i299.photobucket.com/albums/mm286/lanvin12/physics.jpg

Vax = (4.4m/s)sin31° = 2.26616m/s
Vbx = Vbsin25° = (7.8m/s)sin25° = 3.2964m/s
Vx = Vbx - Vax = 3.2964m/s - 2.26616m/s = 1.030262m/s

Vay = Vacos31° = 4.4m/s)cos31° = 3.771536123m/s
Vby = Vbcos25° = 7.8m/s)cos25° = 7.069200739m/s
Vy = Vby + Vba = 3.771m/s + 7.0692m/s = 10.840736m/s

V² = Vx² + Vy² = (10.840m/s)² + (1.030m/s)² = 118.530155m/s
V = √118.530155m/s = 10.8895m/s

a = v/t = 10.8895m/s) / (8.5s) = 1.2811m/s²

tan theta = 10.84 / 1.030 = 10.522
theta = tan^(-1) 10.522 = 84.57 degrees

therefore...
the bird's acceleration was 1.3m/s² [85 degrees N of E]

There is no problem with the physics. However, I believe you made a mistake with angle of the components of the final velocities of the bird. It should be 35 deg insted of 25 deg.
 
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  • #3
In addition to the error of using 25 degrees instead of 35 degrees that horatio89 pointed out, I think you are using the wrong trig functions. Several places it seems clear that you are using the x direction as the east and y direction as the north (which makes sense).

The problem is, for example, in your first line, 4.4 sin(31 degrees) does not give the component in the east direction.
 
  • #4
horatio89 said:
There is no problem with the physics. However, I believe you made a mistake with angle of the components of the final velocities of the bird. It should be 35 deg insted of 25 deg.

oops! That was my mistake... the angle in the diagram should have read 25°.
Is the answer fine otherwise?
 
  • #5
see alphysicist's post. It affects the calculation of the angle of the acceleration.
 
  • #6
alphysicist said:
In addition to the error of using 25 degrees instead of 35 degrees that horatio89 pointed out, I think you are using the wrong trig functions. Several places it seems clear that you are using the x direction as the east and y direction as the north (which makes sense).

The problem is, for example, in your first line, 4.4 sin(31 degrees) does not give the component in the east direction.

I don't think I understand. What do you mean? I wrote x where I was supposed to write y, and vice versa...??
 
  • #7
lanvin said:
I don't think I understand. What do you mean? I wrote x where I was supposed to write y, and vice versa...??

I was saying that you used the sine function to find the eastward components (and the cosine function to find the north-south components) which is not correct.
 
  • #8
is this correct?

Vx = 4.4sin31 + 7.8cos25 = 9.335
Vy = -4.4cos31 + 7.8cos25 = -0.4745
V^2 = 9.3^2 + (-0.47)^2
V = 9.3m/s
 
  • #9
lanvin said:
is this correct?

Vx = 4.4sin31 + 7.8cos25 = 9.335
Vy = -4.4cos31 + 7.8cos25 = -0.4745
V^2 = 9.3^2 + (-0.47)^2
V = 9.3m/s

No; when the angles say 25 degrees N of E or 31 degrees S of E, that means both angles are measured away from the eastward direction. So the east component is the adjacent side to both angles (so you have to use cosine), and the north-south component is the opposite side from both angles (so you have to use sine).
 
  • #10
wow I'm good (sarcasm)
maybe i'll get this right before 2009
next attempt:

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.
 
Last edited:
  • #11
lanvin said:
wow I'm good (sarcasm)
maybe i'll get this right before 2009
next attempt:

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.

That looks right to me. (If the percent accuracy is important, you might want to keep more digits in your calculations, but the procedure is correct.)
 
  • #12
cool, thanks for the help
 

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