Answered

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A plane starting from rest accelerates
to takeoff velocity of 75 m/s in 15
seconds. What was the plane’s
acceleration and how far did it travel
before takeoff?

Sagot :

Answer:

The acceleration is 5 m/s² and the distance is 562.5  m.

Explanation:

Given that,

Initial velocity of the plane, u = 0 (at rest)

Final speed, v = 75 m/s

Time, t = 15 s

We need to find the acceleration of the plane and distance it travel before takeoff.

[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{75-0}{15}\\\\a=5\ m/s^2[/tex]

Let the distance is d.

[tex]v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{(75)^2-(0)^2}{2\times 5}\\\\d=562.5\ m[/tex]

So, the acceleration is 5 m/s² and the distance is 562.5  m.

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