Velocity component questions

1. The problem statement, all variables and given/known data
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.

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...
 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?

Velocity component questions

yes
 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/air...nslations.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
 after u calculate both of the accelerations (ax and ay) use pythagorean theorem to find the resultant acceleration and its direction
 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?
 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
 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!
 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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