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
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.
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 then add the safety margin =18792 + 250 = 19042m = 19.042km am I any closer to the answer do you think?
Your conversion of km/hr to m/s is wrong. The plane is slowing down. So the acceleration must be negative.
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 then add the safety margin =12600 + 250 = 12850m = 12.85km maybe?
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 then add the safety margin =549 + 250 = 799m = 0.799km you must hate me by now =]