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

AI Thread Summary
To calculate a plane's average acceleration from velocity components, the average acceleration is defined as the change in velocity over time. For the x component, the formula is applied using the initial and final velocities along with the time interval, resulting in a value that needs unit conversion from km/h to m/s. The y component is similarly calculated, and both components are combined using the Pythagorean theorem to find the magnitude of average acceleration. Direction is determined using trigonometric functions based on the calculated components. Proper unit conversion is crucial for accurate results, as errors in units can lead to incorrect answers.
Abarak
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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
 
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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...
 
yes:biggrin:
 
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
 
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
 
Argh, I caught that on a few other problems. Dang it...

Thanks for the help, it's really appreciated.

Abarak
 
can u now do the other two parts? C and D?
 
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
 
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  • #10
after u calculate both of the accelerations (ax and ay) use pythagorean theorem to find the resultant acceleration and its direction
 
  • #11
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?
 
  • #12
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
 
  • #13
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!
 
  • #14
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
 
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