What Is the Minimum Acceleration Needed for a Jumbo Jet to Take Off?

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SUMMARY

The minimum acceleration required for a jumbo jet to take off, reaching a speed of 300 km/hr on a 2 km runway, is calculated using the equation V^2 = Vo^2 + 2a(X-Xo). The correct acceleration needed is 2250 km/hr², although an arithmetic error was noted in the initial calculations. The variables used include initial velocity (Vo), final velocity (V), and distance (X-Xo). This analysis confirms the necessity of precise arithmetic in physics calculations.

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cheerspens
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1. Homework Statement
A jumbo jet needs to reach a speed of 300 km/hr on a runway for take off. Assuming a constant acceleration and a runway 2 km long, what minimum acceleration from rest is required to get the jet in the air?


2. Homework Equations
I tried using the equation:
2a(X-Xo)=V^2+Vo^2

But there is also:
X=Xo+volt+(1/2)at^2
V=Vo+at

3. The Attempt at a Solution
I set up the following variable list:
Xo=1.5 Vo=?
X=21.5 V=0
t=? a=-9.8

I plugged these values into the equations mentioned above to get a final answer of 2250km/hr^2 for the acceleration needed. Would this be correct?
 
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cheerspens said:
I tried using the equation:
2a(X-Xo)=V^2+Vo^2
This should be:
V^2 = Vo^2 + 2a(X-Xo)

And it's the only one you need.I plugged these values into the equations mentioned above to get a final answer of 2250km/hr^2 for the acceleration needed. Would this be correct?
Almost; I think you made an arithmetic error with the decimal point.

(I merged this thread by mistake--then unmerged it. So I might have changed its title. My apologies.)
 

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