Question

An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3m/s2. The takeoff speed for this plane will be 65m/s. assuming this minimum acceleration, what is the minimum allowed length for the runway

Answer 1

vi = 0 m/s

vf = 65 m/s

a = 3 m/s2

d = ??
vf2 = vi2 + 2*a*d
(65 m/s)2 = (0 m/s)2 + 2*(3 m/s2)*d

4225 m2/s2 = (0 m/s)2 + (6 m/s2)*d

(4225 m2/s2)/(6 m/s2) = d

d = 704 m

Answer 2

Answer:

The minimum allowed length for the runway is 704.1 m

Explanation:

Acceleration of the plane, $a=3\ m/s^2$  

The takeoff speed for this plane will, v = 65 m/s

Initially the plane is at rest, u = 0

We have to find the minimum allowed length for the runway. It can be calculated using third equation of motion as :

$v^2-u^2=2as$

s is the length of the runway.

$s=\dfrac{v^2}{2a}$

$s=\dfrac{65^2\ m/s}{2\times 3\ m/s^2}$

s = 704.1 m

Hence, the minimum allowed length for the runway is 704.1 m