Minimum Constant Acceleration for Aircraft Liftoff

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Homework Help Overview

The discussion revolves around calculating the minimum constant acceleration required for an aircraft to achieve liftoff speed after a specified distance. The aircraft's liftoff speed is given as 130 km/hr, and the distance for the run is 271 meters. The problem involves kinematics and unit conversion.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of kinematic equations to isolate acceleration. There are attempts to convert units from km/hr to m/s and to clarify the correct form of the equations used.

Discussion Status

Some participants have provided guidance on isolating the variable for acceleration and emphasized the importance of unit conversion. There is ongoing clarification regarding the correct application of the equations and the resulting units.

Contextual Notes

Participants are navigating through potential misunderstandings of the equations and the necessity of proper unit conversion, particularly from km/hr to m/s. There is also a focus on ensuring the correct mathematical representation of the equations involved.

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Homework Statement


aircraft has liftoff speed of 130km/hr. What minimum constant acceleration does it require if the aircraft is to be airborne after a run of 271m? answer is in m/s^2


Homework Equations


V^2 = Vi^2 + 2a(deltax)
a=(V^2)/Vi^2 + 2(deltax)

The Attempt at a Solution


(130^2)/O^2 + 2(271) = a
a = 28.79 km/hr = 230.24 m/s^2
 
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The 2nd equation under 2) is not correct. The first equation is ok, so you just need to be careful how you isolate a from it. You should also consider what unit of the values you will use before inserting them (hint: convert km/h to m/s before inserting the value).
 
is this right then

(V^2) - (Vi^2)/(deltax) = 2a
 
Close. You may mean the right thing, but as you have written it, it misses a set of parenthesis around the the two speeds. Alternatively you can insert vi=0 into the first equation before you start isolating for a.
 
so if Vi = 0 then the equation is (V^2)/(deltax) = 2a and 130km/hr = 36.11 m/s so
(36.11 m/s)^2/271m = 2a
4.812 = 2a
a = 2.41 m/s

I feel like i did it right but I need the final units in m/s^2 and isn't this in m^2/s^2
 
The square of velocity in unit m/s has unit m2/s2 which then is divided by distance in unit m, so the resulting unit is m/s2 as it should.
 

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