winterbone143 winterbone143
  • 14-10-2020
  • Physics
contestada


A jet must reach a velocity of 75 m/s for takeoff. If the runway is
2100 meters long, what must the constant acceleration be?

Respuesta :

Muscardinus Muscardinus
  • 21-10-2020

Answer:

[tex]a=1.33\ m/s^2[/tex]

Explanation:

Given that,

Initialy, the jet is at rest, u = 0

Final velocity of the jet, v = 75 m/s

Distance, d = 2100 m

We need to find the acceleration of the jet. It is based on the concept of equation of kinematics. Using third equation of motion, w get :

[tex]v^2-u^2=2ad[/tex]

a is acceleration

[tex]75^2=2a\times 2100\\\\a=\dfrac{75^2}{2\times 2100}\\\\a=1.33\ m/s^2[/tex]

So, the acceleration of the jet is [tex]1.33\ m/s^2[/tex].

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