Calculating Takeoff Time and Distance for a Plane on a Runway

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A plane accelerates from rest at 5.00 m/s² on a 1800 m runway, requiring the correct formula d = v₀t + (1/2)at² to calculate takeoff time. The initial calculation of time was incorrect, leading to a conclusion of approximately 26.8 seconds for takeoff, which is deemed realistic. For the distance traveled in the last second before takeoff, the correct approach involves calculating the difference in distances covered in the last two seconds. The discussion clarified the importance of accurately applying the equations of motion. Ultimately, the participants confirmed the calculations and expressed gratitude for the assistance received.
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Homework Statement


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?


Homework Equations


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


The Attempt at a Solution


not exactly sure if this is the correct formula but that's 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
 
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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 traveled in the first t-1 seconds... not the last 1 second. you're close...
 
oops you 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 let's 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.
 
cool_dude said:
oops you 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 let's 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.

Your time looks correct.

Suppose you know the distance traveled in the first 10 seconds... and the distance traveled in the first 9 seconds... you can get the distance in the last second by subtracting the two...
 
Wow thanks for the fast reply. And thank you very much for all your help. i got both answers correct thanks to you. :)
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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