How Much Runway Does a Jet Need for Takeoff?

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The discussion revolves around calculating the runway distance needed for a jet to reach its minimum takeoff speed of 285 km/h, given its thrust and weight. Participants clarify that the weight of the jet is 2.75*10^6 N, which can be converted to mass by dividing by the acceleration due to gravity. The thrust of 6.35*10^6 N is also discussed, leading to confusion about whether to add forces or use them separately. The key equation for acceleration, F=ma, is highlighted as essential for solving the problem, along with the kinematic equation for distance. Ultimately, the participant successfully figures out the necessary calculations to determine the runway distance.
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Homework Statement


Jet A 2.75*10^6 N catapult jet plane is ready for takeoff. If the jets engines supply a constant thrust of 6.35*10^6 N, how much runway will it need to reach its minimum takeoff speed of 285 km/h.

velocity initial = 0 km/hr
velocity final = 285 km/hr
F = 2.75*10^6 N
F= 6.35*10^6 N
Distance of runway = ?

Homework Equations


? See below

The Attempt at a Solution


Ok so here's my problem. I can't figure out what equation(s) to use. I'm not even sure if we've been taught it yet (I'm doing the homework a few days in advance cause I'm doing really bad so far in the class) but I can't find the equation in the book. So here's what I've done so far:
Vf = 285 km/hr * 1000m/km = 285000 m/hr * 1 hr/3600 s = 79.2 m/s
Velocity final = 79.2 m/s
Velocity initial = 0 m/s

So I need help with what equation to use and I have one other question. Why are there 2 forces? Are you supposed to add them and get the net force or am I way off base? I'm so confused.
 
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What is the acceleration?
 
It doesn't give me one. :( What I typed out is the problems exact wording:

A 2.75*10^6 N catapult jet plane is ready for takeoff. If the jets engines supply a constant thrust of 6.35*10^6 N, how much runway will it need to reach its minimum takeoff speed of 285 km/h.

I honestly cannot figure it out. I still want to know why they gave me 2 forces.
 
No, they didn't give you an acceleration, that was my hint to you :)
On the one hand, the acceleration is related to the change in velocity (which is an important factor in the problem) and on the other hand acceleration has to do with forces (which are given), so it is the missing link in finding the answer.
Do you know of any relationships or formulas involving acceleration? Can you quote one or more of Newtons laws?
 
Sorry about not picking up on the hint. I thought you wanted more information and I was sad cause I didn't have any more.

The only one of Newtons laws we've covered in detail is force = mass times acceleration (f=ma). So you could do:
F=m*(v/t)
But you don't have the mass or the time. I'm so confused :( Can I have another hint?

And why do I have 2 forces? I understand they're different pieces of information but do I add them for net force or just use one or... I don't know. (btw, I'm not trying to get the answer from you guys, I just want to know how to solve it so I can understand it better in class. Please don't tell me the answer.)
 
Just a thought. Is the 2.75*10^6 N a force that propels the jet, or is it the weight? Because if that number is the weight of the jet, you could figure out the mass.
 
roz77 is spot on.
A 2.75*10^6 N catapult jet plane is ready for takeoff.
In this sentence, the weight of the plane is given in terms of Newtons. Remember that weight is a consequence of gravity, because the Earth pulls on a mass of 1 kg with a force of g \approx 10 Newtons. So instead of saying; the mass is 1000 kg, you can also say: the weight is 10 000 N.
 
Ok sweet I finally figured it out. You would divide the 2.75*10^6 by -9.8 to get the weight in kg. You would then use that to figure out the acceleration with F=ma. And then use an equation (forgot which one but I managed to do the problem) to solve for distance. Thanks for your help everyone!
 
OxxCloudxxO said:
Ok sweet I finally figured it out. You would divide the 2.75*10^6 by -9.8 to get the weight in kg. You would then use that to figure out the acceleration with F=ma. And then use an equation (forgot which one but I managed to do the problem) to solve for distance.
You forgot s = \frac12 a t^2 - and yes, that was the solution to the problem.

OxxCloudxxO said:
Thanks for your help everyone!
You are welcome.
 
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