Question about average acceleration

In summary, a bird takes 8.5 seconds to fly from position A to position B with velocity A being 4.4 m/s [31 Degrees S of E] and velocity B being 7.8 m/s [25 Degrees N of E]. The bird's average acceleration is 0.41 m/s^2 [17 Degrees N of E]. This is calculated by finding the x and y components of both velocities and subtracting them separately, then finding the tangent of the resulting angle.
  • #1
highc
18
0
A bird takes 8.5s to fly from position A to Position B. Determine the birds average acceleration.

Velocity A = 4.4 m/s [31 Degrees S of E]
Velocity B = 7.8 m/s [25 Degrees N of E]
 
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  • #2
highc said:
A bird takes 8.5s to fly from position A to Position B. Determine the birds average acceleration.

Velocity A = 4.4 m/s [31 Degrees S of E]
Velocity B = 7.8 m/s [25 Degrees N of E]

?? Do you know the definition of "average acceleration"??
 
  • #3
Change in velocity/change in time = 0.4 m/s^2. I'm having a hard time believing that this is all that is expected as an answer, seems too simple. (This if for a grade 12 physics course). The question is accompanied by a diagram showing the birds curved flight path.
 
  • #4
highc said:
Change in velocity/change in time = 0.4 m/s^2. I'm having a hard time believing that this is all that is expected as an answer, seems too simple. (This if for a grade 12 physics course). The question is accompanied by a diagram showing the birds curved flight path.
That's the definition except that you are forgetting that velocity is a vector .So the average acceleration is given by
[tex] { {\vec v_f} - {\vec v_i} \over \Delta t} [/tex].
You need to do a vector subtraction and divide by delta t. Your answer will be a vector! (the average acceleration is a vector, as is the instantaneous acceleration)

Patrick
 
  • #5
I am totally lost on this one. I don't know how to even start to do a vector subtraction with this one. I think my confusion stems from the fact that the bird is flying slightly upward on a curved path. The other examples in the book are nothing like this one. I just started this course, getting started is always the hardest part. Hoping someone can help.
 
  • #6
Your first error was thinking of the velocities as "positions".

a vector has 3 components for example: (East, North, Up).
Each component is added or subtracted *separately* from the others
using regular algebra (with signs ... South is NEGATIVE North!).
Just keep each resulting component separate, in a list.

Do you know enough "triangle trig" to figure out how much of A is East?

Each component is divided by a scalar (like time) SEPARATELY.
Just keep the resulting components separate, in a list.
 
  • #7
highc said:
I am totally lost on this one. I don't know how to even start to do a vector subtraction with this one. I think my confusion stems from the fact that the bird is flying slightly upward on a curved path. The other examples in the book are nothing like this one. I just started this course, getting started is always the hardest part. Hoping someone can help.
lightgrav already helped out a lot..

Just to add my two cents: you do NOT need to worry about the path of the bird. You just need to know the initial velocity and the final velocity and these you gave in your first post. Just find the x and y components of both velocities and subtract them separately.

To be more precise
[tex] A_{average,x} = { {v_{f,x} - v_{i,x} \over \Delta t} [/tex]
and the same for the y component.
 
  • #8
Thank for the help. If anybody is ambitious enough to work this one out, can you confirm if the correct answer is:

0.41 m/s^2 [17 Degrees N of E]

(I arrived at this using for B, v(X) = 7.8cos25, v(y) = 7.8sin 25. For A, v(x) = 4.4cos31, v(y) = 4.4sin31. Then, a(x) = (7.1-3.8)/8.5 = 0.39, and a(y) = (3.3-2.3)/8.5 = 0.12. Then I found the tangent, to arrive at 17 Degrees)
 
  • #9
Anybody?
 
  • #10
highc said:
Thank for the help. If anybody is ambitious enough to work this one out, can you confirm if the correct answer is:

0.41 m/s^2 [17 Degrees N of E]

(I arrived at this using for B, v(X) = 7.8cos25, v(y) = 7.8sin 25. For A, v(x) = 4.4cos31, v(y) = 4.4sin31. Then, a(x) = (7.1-3.8)/8.5 = 0.39, and a(y) = (3.3-2.3)/8.5 = 0.12. Then I found the tangent, to arrive at 17 Degrees)
Watch out. For A the angle is given as being *south of east*. That affects how you calculate the x and y components. Do you see how?
 

Related to Question about average acceleration

What is average acceleration?

Average acceleration is the change in velocity divided by the change in time. It measures the rate at which an object's velocity is changing over a certain period of time.

How do you calculate average acceleration?

Average acceleration can be calculated by taking the final velocity of an object and subtracting it from the initial velocity, then dividing that value by the change in time. The formula for average acceleration is: a = (vf - vi) / t

What is the difference between average acceleration and instantaneous acceleration?

Average acceleration is calculated over a period of time, while instantaneous acceleration is the acceleration at a specific moment in time. Average acceleration takes into account any changes in velocity over a given time frame, while instantaneous acceleration only considers the acceleration at one point in time.

What are some examples of average acceleration in everyday life?

One example of average acceleration is when a car speeds up from a stoplight. Another example is when a ball is thrown into the air and then falls back down. In both cases, the velocity of the object is changing over time, leading to an average acceleration.

How does average acceleration relate to Newton's Second Law of Motion?

Newton's Second Law of Motion states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. Average acceleration is a measure of an object's change in velocity, which is affected by the net force acting on the object. Therefore, average acceleration is directly related to Newton's Second Law of Motion.

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