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.
