Question

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?

Answer 1

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

Step-by-step 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.

`a=(v-u)/(t)\\\\a=(75-0)/(15)\\\\a=5\ m/s^2`

Let the distance is d.

`v^2-u^2=2ad\\\\d=(v^2-u^2)/(2a)\\\\d=((75)^2-(0)^2)/(2* 5)\\\\d=562.5\ m`

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