How Do You Calculate a Plane's Average Acceleration from Velocity Components?

[Abarak]

Homework Statement

At an air show, a jet plane has velocity components v_x = 625 km/h and v_y = 415 km/h at time 3.85 s and v_x2 = 838 km/h and v_y2 = 365 km/h at time 6.52 }.

2. Questions
Question A.
For this time interval, find the x component of the plane's average acceleration.
{I have already used all of my tries on this one and have moved onto question B}

Question B.
For this time interval, find the y component of the plane's average acceleration.

I tried 81.9 but it says it's wrong. ((415/3.85)+(362/6.52))/2 = 81.9??

Question C.
For this time interval, find the magnitude of its average acceleration.

Question D.
For this time interval, find the direction of its average acceleration.

Please help, I have no idea how to go about solving the above questions. I'm not looking for the answer, just a little guidance.

Thanks,

Abarak

 

[esalihm]

first define average acceleration.
then, state the formula for average acceleration explicitly, then plug in the values

can't understand what u are trying here ((415/3.85)+(362/6.52))/2 = 81.9 ?

why divide the speed by time? acceleration is something different.

actually first define acceleration here...

 

[Abarak]

Ave. Acceleration = (Vf - Vi) / t (http://www.physicsclassroom.com/Class/1DKin/U1L1e.html)

would this do the trick?

Aave = (365-415)/(6.52-3.85) = -18.73?

 

[esalihm]

yes

 

[Abarak]

When I enter "-18.73" it tells me it's the wrong answer. That was the last attempt at the question so I need to move on to part C.

In part C I have no idea where to begin as the answer to part B was wrong...

I'm lost...

Abarak

 

[esalihm]

but u forgot the units, ur equation is correct but the value at the end would not be 18,73 because the units do not correspond. one is km/h and the other is seconds

 

[Abarak]

Argh, I caught that on a few other problems. Dang it...

Thanks for the help, it's really appreciated.

Abarak

 

[esalihm]

can u now do the other two parts? C and D?

 

[Abarak]

Well, I am doing some research on them right now, trying to figure out how I can find the magnitude and direction from the two time's. I did find this list (http://www.grc.nasa.gov/WWW/K-12/airplane/translations.html) but it seems to just tell how to calculate average acceleration.

I know the velocity is a vector, indicating it has both a magnitude and a direction. Part C is asking to find the magnitude in km/h indicating a straight velocity. I think I know everything to finish part C and D but something is not clicking.

I'm a little lost. Could you possibly point me in the right direction?

After converting my units (_s>_hr) I came up with the following answers:
Part A: 287191.01 km/hr^2
part B: -67415.73 km/hr^2

 

[esalihm]

after u calculate both of the accelerations (ax and ay) use pythagorean theorem to find the resultant acceleration and its direction

 

[Abarak]

Hmmm... let me see if I got this right.

To find the magnitude:
x^2 = (287191.01 _km/_hr^2)^2 + (-67415.73 _km/_hr^2)^2
x = .036172991 _m/_s^2 or -.036172991 _m/_s^2
x = 486.802 _km/_hr^2 or -486.802 _km/_hr^2
Which one should I use? I'm thinking 486.802 _km/_hr^2 because it's (+).

Also, to find the direction I used:
sin(AavgX/AavgY) = sin(287191.01/-67415.73) = -.07 deg.

Are the above correct or am I way off?

 

[esalihm]

x comp of average acc:
((838-625)/3,6)/(6,52-3,85)=22,16 m/s^2 in (+x)

y comp. of average acc.:
((365-415)/3,6)/(6,52-3,85)=5,202 m/s^2 in (-y)

magnitude of average acc.:
sqrt[(22,16^2)+(5,202^2)]=22,76 m/s^2

direction of average acc.:
arccos (22,16/22,76)=13,2 degrees
13,2 degrees S of E

 

[Abarak]

Wow, I do suck at this stuff...

Thanks for all the help and time. Just one little question, when calculating the x and y average acceleration, you used 3.6. Where did this number come from?

Again, thank you for all of your help!

 

[esalihm]

when converting km/h to m/s u divide by 3,6
1 km =1000 m
and 1 h= 3600 sec.
1000/3600=1/3,6