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Feb 21st, 2003, 08:19 PM
#1
Thread Starter
New Member
A Jet Airplane Problem...
Here's another one:
N, S, E, W, stands for North, South, East, and West.
A Boeing 777 is flying at 720 km/h [N 25 degrees E] and then executes a turn that brings it to a new velocity of 770 km/h [S 50 degrees E]. The turn took 10.0s. Determine the acceleration. Answer in m/s^2.
Anyone? Thanks...
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Feb 23rd, 2003, 01:44 PM
#2
New Member
speaking of components... can't this problem be done that way???
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Feb 23rd, 2003, 02:24 PM
#3
Fanatic Member
Linear Acceleration:
a = vv/vt;
a = 1.4 m/s^2

prog_tom
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Feb 23rd, 2003, 04:01 PM
#4
Fanatic Member
Rotational Acceleration,
If not mistaken, it's
-4.3m/s^2
Last edited by prog_tom; Mar 1st, 2003 at 01:03 AM.

prog_tom
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Feb 23rd, 2003, 05:00 PM
#5
New Member
prog_tom i dont know what u did....maybe cause im tryig to use high school physics.
Anyway I am nt guru at this and I may have this completey wrong, this is what i did:
Can anyone confirm this????
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Feb 24th, 2003, 12:18 AM
#6
New Member
please ??
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Feb 24th, 2003, 12:56 AM
#7
Registered User
I don't think so.
You have: c = 1182.5 kmh-1 then you convert it to ms-1.
You say that it is 328.47 m/s
But to convert to m/s wouldn't it be: 1182.5 / 1000 (-> meters / hour) then * 3600 (-> meters / second), giving 4257 ms-1?
I'm not sure, but I think the answer is 288 ms-2.
Acceleration = delta v (displaced distance) / delta t (displaced time)
So, 2592 cos 65 + 2772 sin 40 = 1095 + 1782 (approx.) = 2877
2877 / 10 = 288 ms-2
Of course you could do it with greater accuracy and I am not 100% sure - we just started doing vectors today.
Let me know how it goes.
s.
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Feb 24th, 2003, 03:18 AM
#8
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Feb 24th, 2003, 06:53 AM
#9
Registered User
Hey, you talking to me? 
It's asking for the acceleration throughout the turn.
We did a similar problem for school but I've handed it in...
s.
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Feb 25th, 2003, 03:53 AM
#10
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Feb 25th, 2003, 08:15 PM
#11
Registered User
122.65 ms^-2
i think
i'll post tonight
s.
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Feb 26th, 2003, 01:21 AM
#12
Registered User
Where's this question from?
Anyway, using vectors, components etc:
xfinal = 770 sin 50
yfinal = -770 cos 50
xinitial = 720 sin 25
yinitial = 770 cos 25
/\x = (770 sin 50 - 720 sin 25)
/\y = (770 cos 25 + 770 cos 50)
Äv = root(/\x2 + /\y2)
a = /\v / /\t
/\t = 10s
/\v = 1226.511ms-1
.'. a = 122.65ms-2
I think that's right.
Let me know how it goes.
s.
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Mar 1st, 2003, 12:17 AM
#13
Fanatic Member
Originally posted by silva
Where's this question from?
Anyway, using vectors, components etc:
xfinal = 770 sin 50
yfinal = -770 cos 50
xinitial = 720 sin 25
yinitial = 770 cos 25
/\x = (770 sin 50 - 720 sin 25)
/\y = (770 cos 25 + 770 cos 50)
Äv = root(/\x2 + /\y2)
a = /\v / /\t
/\t = 10s
/\v = 1226.511ms-1
.'. a = 122.65ms-2
I think that's right.
Let me know how it goes.
s.
so, according to your answer, after 10 seconds, the velocity would be 1227 m/s + 720km/h, or 1227 m/s + 200m/s, which would be 1427m/s????

prog_tom
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Mar 1st, 2003, 12:21 AM
#14
Fanatic Member
Re: Nah.
Originally posted by sql_lall
Nah silva. Soz, i was just looking at the picture by mykg4orce1, how he had v2-v1. I was just intrigued.
Oh, and what is the difference between turning acceleration and normal acceleration? Cos we haven't done that yet in fizix, only normal acceleration in 1 direction
(*so* boring so far, it's like '"and everyone, this is a vector...")
So yeah, is it just something like the triangles mykg4orce1 set up?
Rotational Acceleration is a little different from Linear Acceleration:
Rotational accel = change in rotational velocity / time
rotational velocity = 2pi * frequency = angle travelled /time

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Mar 1st, 2003, 12:58 AM
#15
Fanatic Member
Originally posted by silva
Acceleration = delta v (displaced distance) / delta t (displaced time)
Isn't it suppose to be change in VeLOCITY? Can you do Algebra 1?

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