# Airplane on the runway

1. Sep 15, 2007

### cool_dude

1. The problem statement, all variables and given/known data
A plane accelerates from rest at a constant rate of 5.00 m/s^2 along a runway that is 1800 m long. Assume that the plane reaches the required takeoff velocity at the end of the runway. What is the time t_{TO} needed to take off?

2. Relevant equations
i think the formula i'm supposed to use is d = vt + 1/2at

3. The attempt at a solution
not exactly sure if this is the correct formula but thats what i thought i have to use. we know that d = 1800m and we know that initial velocity = 0 and a = 5 m/s^2 so then we isolate for t. thus we get t = 1800 / 2.5 = 720. however i think this might be wrong because i don't think its realistic for a plane to take 12 minutes on the run way to take off. any ideas?

Also part of the question was What is the distance d_last traveled by the plane in the last second before taking off?

I tried to use the same formula d = vt + 1/2at and since we figured out time from part a we sub (t - 1) into here and get distance traveled in the last second before taking off. not sure if any of this is right. can someone verify?

Thank you

2. Sep 15, 2007

### learningphysics

your equation is wrong. it should be:

d = v0*t + (1/2)at^2

you didn't have the t^2...

pluggin in t-1... gives the distance travelled in the first t-1 seconds... not the last 1 second. you're close...

3. Sep 15, 2007

### cool_dude

oops ya i forgot the square. so then i get 26.8 seconds. does that make sense for a plane to take amount of time for lift of? i think it makes sense to me. not sure. also does this mean that we don't care about the required velocity for takeoff?

hmmm the second part is tricky. i thoght that t - 1 would mean that lets say it took 10 seconds then t = 10 and the second before it is t - 1 = 9. i'm a bit confused on this part.

4. Sep 15, 2007