Question

A Plane has a takeoff speed of 150 m/s and requires 1500m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed.

Answer 1

1) Acceleration:
`7.5 m/s^2`

The motion of the plane is a uniformly accelerated motion, so we can find its acceleration by using the suvat equation

`v^2-u^2=2as`

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Here we have

v = 150 m/s is the final velocity of the plane

u = 0 (it starts from rest)

a=?

s = 1500 m is the displacement

Solving for a, we find

`a=(v^2-u^2)/(2s)=(150^2-0)/(2(1500))=7.5 m/s^2`

2. Time: 20 s

For this part of the problem, we can use another suvat equation:

`v=u+at`

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

Here we already know:

v = 150 m/s is the final velocity of the plane

u = 0 (it starts from rest)

`a=7.5 m/s^2` (found in part 1)

Solving for t, we find the time taken for the plane to reach the final velocity of 150 m/s:

`t=(v-u)/(a)=(150-0)/(7.5)=20 s`