Question

What is the takeoff distance for an airplane on a runway at

300km/hr with acceleration of 1 m/s 2

Answer 1

x = 3472 [m]

Step-by-step explanation:

To solve this problem we must use the following expression of kinematics:

`v_(f)^(2)= v_(i)^(2)+(2*a*x)\\`

where:

Vi = initial velocity = 0

Vf = final velocity = 300 [km/h]

a = acceleration = 1 [m/s^2]

x = takeoff distance [m]

Note: The initial velocity is equal to zero as avion starts its movement from rest.

Now we have to convert units of kilometers per hour to meters per second, the final velocity.

`300 [(km)/(h)]*[(1h)/(3600s) ]*(1000m)/(1km) \\= 83.33 [(m)/(s) ]`

(83.33)^2 = 0 + (2*1*x)

2*x = 6943.88

x = 3472 [m]