Constant Acceleration Plane Problem

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SUMMARY

The constant acceleration required for a jetliner to take off from a 2100 m runway at a speed of 360 km/h (100 m/s) is calculated to be 2.38 m/s². This conclusion is derived using the kinematic equation v² = initial V² + 2a(x - initial x), where the final velocity (v) is 100 m/s, the initial velocity is 0, and the distance (x - initial x) is 2100 m. The calculations confirm that the acceleration is indeed 2.38 m/s², which is appropriate for the given parameters.

PREREQUISITES
  • Understanding of kinematic equations
  • Basic knowledge of unit conversions (e.g., km/h to m/s)
  • Familiarity with acceleration concepts
  • Ability to perform algebraic manipulations
NEXT STEPS
  • Study kinematic equations in detail
  • Learn about unit conversion techniques for speed
  • Explore real-world applications of constant acceleration in aviation
  • Investigate factors affecting takeoff distances for different aircraft
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Aerospace engineers, physics students, and anyone interested in the mechanics of jetliner takeoff performance will benefit from this discussion.

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The takeoff speed of a jetliner is 360 km/h. If the jetliner is to take off from a 2100 m long runway, what must be the (constant) acceleration along the runway?

Velocity of jetliner=(360 km/h)(1000 m/ 1 km)(1 h/3600 s)=100 m/s

v^2=initial V^2+2a(x-initial x)
(v^2-(initial V^2))/(x-intial x)=2a
((100 m/s)^2-0)/(2100-0)=2a
4.76 m/s^2=2a
a=2.38 m/s^2

Look right?
Anybody?
 
Last edited:
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yup looks fine to me.
 

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