Minimum Constant Acceleration for Aircraft Liftoff

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The discussion centers on calculating the minimum constant acceleration required for an aircraft to achieve liftoff speed of 130 km/h over a distance of 271 meters. The correct approach involves using the equation V^2 = Vi^2 + 2a(deltax) and converting the speed from km/h to m/s. After applying the formula, it is determined that the acceleration should be 2.41 m/s². Participants clarify the importance of unit conversion and proper equation manipulation to ensure accurate results. The final units confirm that the calculations yield the expected acceleration in m/s².
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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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