Question

A jet must reach a velocity of 75 m/s for takeoff. If the runway is

2100 meters long, what must the constant acceleration be?

Answer 1

`a=1.33\ m/s^2`

Step-by-step explanation:

Given that,

Initialy, the jet is at rest, u = 0

Final velocity of the jet, v = 75 m/s

Distance, d = 2100 m

We need to find the acceleration of the jet. It is based on the concept of equation of kinematics. Using third equation of motion, w get :

`v^2-u^2=2ad`

a is acceleration

`75^2=2a* 2100\\\\a=(75^2)/(2* 2100)\\\\a=1.33\ m/s^2`

So, the acceleration of the jet is
`1.33\ m/s^2`.