# Finding length of airport runway

1. Sep 21, 2009

### NDbogan

the landing speed of a jumbo jet is 300Kn/hr. After landing, the jet comes to the safe taxiing speed of 30Km/hr in 12 seconds. The airport authorities would like to keep a saftey margin of 250 metres of runway. How long should the runway be for the safe landing of jumbo jets?

Now, I remember doing this in class. I just can't find my notes.

I was thinking that it might have something to do with:

Final velocity(v)= initial velocity(u)+acceleration(a) x time(t)

maybe re-arranging it then being able to find the distance travelled.
so:
a=(u-v)/t

all help appreciated

2. Sep 21, 2009

### rl.bhat

Your formula for a is correct. Convert km/h to m/s and find the acceleration.
There is a kinematic equation which contains vi,a,t and x. Using that equation find x.To it add safety margin to find the length of the run way.

3. Sep 21, 2009

### NDbogan

ok. from what you've said this is what i've tried
u=300km/hr=1080m/s
v=30km/hr=108m/s
t=12s
safety margin=250m

to get a
a=(u-v)/t
=(1080-108)/12
=81m/s

so: distance covered=u.t + 1/2.a.t^2
=(1080 x 12) + (1/2 x 81 x 144)
=12960 + 5832
=18792m
=18792 + 250
= 19042m
= 19.042km

am I any closer to the answer do you think?

4. Sep 21, 2009

### rl.bhat

Your conversion of km/hr to m/s is wrong.
The plane is slowing down. So the acceleration must be negative.

5. Sep 21, 2009

### NDbogan

ok so it would be
a=(v-u)/t
=(108-1080)/12
=-81m/s

so: distance covered=u.t + 1/2.a.t^2
=(1080 x 12) + (1/2 x -81 x 144)
=12960 + -360
=12600m
=12600 + 250
= 12850m
= 12.85km
maybe?

6. Sep 21, 2009

### rl.bhat

300km/hr = 300*1000/3600 = ? m/s

7. Sep 21, 2009

### NDbogan

had that the first time and thought i was wrong.
that being the case
so: distance covered=u.t + 1/2.a.t^2
=(83.33 x 12) + (1/2 x -6.25 x 144)
=999 + -450
=549m
=549 + 250
= 799m
= 0.799km
you must hate me by now =]

8. Sep 21, 2009