Runway Length for Takeoff: Solving for x

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Homework Help Overview

The problem involves calculating the required runway length for a 727 jet to take off, given its acceleration to a speed of 200 mph over a period of 30 seconds. The context is centered around kinematics and the application of equations of motion under constant acceleration.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the integration of velocity to find distance and question the correctness of their unit conversions. There is an exploration of the relationship between speed, time, and distance, with some participants expressing confusion over the setup of the problem and the use of units.

Discussion Status

The discussion is ongoing, with participants providing feedback on unit conversions and the mathematical setup. Some guidance has been offered regarding the importance of consistent units, but there is no explicit consensus on the correct approach yet.

Contextual Notes

Participants are grappling with mixed time units (hours, minutes, seconds) and the implications of these on their calculations. There is a recognition that the estimates provided for runway length are significantly off, prompting further examination of the problem setup.

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Homework Statement



A 727 jet needs to attain a speed of 200 mph to take off. If it can accelerate from 0 to 200 mph in 30 seconds, how long must the runway be? (Assume constant acceleration). Show all your work and include units in your answer.

Homework Equations



The Attempt at a Solution



Since v = dx/dt = 200t

I took the integral of dx = integral of 200t dt

and I got x = 100t^2 + C

So I plugged in 0.5 as t and got x = 25.

So x = 25 miles? But.. 25 miles sounds wrong for an airplane runway.. where did I go wrong?
 
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You have time units of both hours, minutes and seconds in your problem. I would suggest before mixing them (and not indicating units) that you convert your numbers to a common time unit.
 
Last edited:
Oh, so instead of 0.5, it should've been

( (30sec) / (3600sec/hr) ) = t

so it comes out to 0.0069 miles?

Thanks again
 
0.0069 miles=about 35 feet. About as bad an estimate as 25 miles. Just convert everything to seconds. You are doing everything correctly mathematically - except the grasp of units is horrible.
 
Last edited:
If you've ever taken off in a plane you know the answer is roughly on the order of a mile. Don't rest until you get something in that ballpark.
 
Actually, I don't even think I set it up right after I look at it again, so I'm stuck now.

v is not 200t (is it?). v is just 200. so dx = 200 dt,
so, x = 200t

so (30/3600) hour = t

plugged it in, got x = 1.6666666666667

so about 2 miles, that seems right. So did I actually set the problem up wrong?
 
Still wrong. Please start using units, ok? v=at. d=(1/2)*a*t^2. You had it set up just fine. The units are killing you. What it 200 mph in miles/sec?
 

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