Finding average acceleration with two sets of velocity vecotors

[mm13690]

Homework Statement

A jet plane is flying at a constant altitude. At time t1=0 it has components of velocity vx=95m/s, vy=115m/s. At time t2=33s the components are vx=172m/s, vy=35m/s.
Find average acceleration.

Homework Equations

avg acceleration=vfinal-vinitial/change in time

The Attempt at a Solution

I have no idea if this attempt at the solution makes sense but I took the first set of velocity components and by using the pythagrium theorum found the velocity at t=0. I then did the same thing for t=33. I took the initial velocity number I got and subtracted it from the final velocity number I got and dividied that number by 33. The answer I got was incorrect. I have no idea where to go from here.

 

[frogjg2003]

Acceleration is also a vector quantity. What are the components of the acceleration?

 

[cbasst]

Since velocity is a vector, you can't necessarily just take the difference between the initial and final vectors directly and come up with the correct answer. With vectors, you can only add them together if they point in the same direction. So in this question, it would be wise to keep the vector components separate while adding/subtracting them. Once you have found the changes in the x and y velocity vectors, you can add them together with the Pythagorean theorem to get the net change in velocity. Make sense? Then you can divide that answer by the change in time.

 

[mm13690]

Oh now I understand it. Thank you that really helped!

 

[Nickie]

Your approach to finding the average acceleration is on the right track, but there are a few key steps missing. First, you need to find the magnitude of the velocity vector at both t=0 and t=33s using the pythagorean theorem, as you mentioned. Then, you need to subtract the initial velocity vector from the final velocity vector to get the change in velocity. Finally, you need to divide this change in velocity by the change in time (33s) to get the average acceleration. So your final equation should look like this:

avg acceleration = (vfinal-vinitial)/change in time

= (172m/s, 35m/s) - (95m/s, 115m/s) / (33s)

= (77m/s, -80m/s) / (33s)

= (2.33m/s^2, -2.42m/s^2)

So the average acceleration of the jet plane is 2.33m/s^2 in the x-direction and -2.42m/s^2 in the y-direction. This means that the plane is accelerating at a rate of 2.33m/s^2 in the direction of its initial velocity (t=0) and decelerating at a rate of 2.42m/s^2 in the direction perpendicular to its initial velocity. This could be due to factors such as air resistance or wind.