Question

Calculate the time when an aircraft arrives in an airport and brakes from 160knots to 20knots on a distance of 1500meters. Please help.. :(

Answer 1

Try using this kinematics formula:
\[v_f ^2=v_i ^2 + 2ad\] You have the distance and velocities, so solve for the acceleration. From there, you can solve for the time, since you know:
\[a=\frac{ v_f - v_i }{ t }\]

Answer 2

I can't seem to derive this formula. Can i use this formula to get the time, can you help out please?.

s=a/2 *t^2+v0*t+s0
t=s/((v2+v1)/2)

Answer 3

\[1 \text{ knot} = 1 \text{ (naut. mile / hr)} / 1 \text{ knot} * 1852 \text{ m} / 1 \text{ naut. mile} * 1 \text{ hr}/ 3600 \text{ s}\]\[=\frac{463}{900} \text{ m/s}\]
\[v_f = 160 \text{ knots} * (\frac{463}{900}\text{ m/s})/1 \text{ knot} = 82.3 \text{ m/s}\]\[v_i = 20 \text{ knots} * (\frac{463}{900}\text{ m/s})/1 \text{ knot} = 10.3 \text{ m/s}\]
\[d = 1500m\]
\[v_f^2=v_i^2+2ad\]Solve for \(a\):
\[v_f^2-v_i^2 = 2ad\]\[\frac{v_f^2-v_i^2}{2d} = a\]
\[\frac{v_f-v_i}{t} = a\]Substitute in prior equation:
\[\frac{v_f-v_i}{t} = \frac{v_f^2-v_i^2}{2d} \]Solve for \(t\)
\[t = \frac{2d(v_f-v_i)}{v_f^2-v_i^2}\]Plug in the numbers...

Answer 4

thank u :)

Answer 5

by the way, the acceleration is constant..

Answer 6

Yes, that formula assumes constant acceleration.

Answer 7

***I can't seem to derive this formula. Can i use this formula to get the time, can you help out please?.

s=a/2 *t^2+v0*t+s0
t=s/((v2+v1)/2)
****

yes you can use the 2nd formula, though you must do as whpalmer and change to consistent units. Working in meters/second you have (moving the 2 up top):
\[ t= \frac{2\cdot 1500 \text { m }}{ 82.3 \text{ m/s} + 10.3 \text{ m/s}} \\t= \frac{3000 \text { m }}{92.6 \text{ m/s} }= 32.4 \text{ s}\]

Answer 8

what does v0 and s0 mean? how will I determine if it's v0 or s0?

Answer 9

\(v_0\) is the initial velocity, \(s_0\) is the initial position. You can tell which one it is by examination of the units.

Answer 10

Thank you guys so much :)