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.

Respuesta :

1) Acceleration: [tex]7.5 m/s^2[/tex]

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

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

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

[tex]a=\frac{v^2-u^2}{2s}=\frac{150^2-0}{2(1500)}=7.5 m/s^2[/tex]

2. Time: 20 s

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

[tex]v=u+at[/tex]

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)

[tex]a=7.5 m/s^2[/tex] (found in part 1)

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

[tex]t=\frac{v-u}{a}=\frac{150-0}{7.5}=20 s[/tex]

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