Question

An aircraft has a landing velocity of 180 km/h . What length of runway is needed given that the deceleration of aircraft along the runway is 2.5 m/s 2?

Answer 1

Given:

An aircraft has a landing velocity of 180 km/h

The deceleration of aircraft along the runway is 2.5 m/s²

To find:

The length of the runway needed

Solution:

We will first covert the speed in km/hr to speed in m/s.

∴ Landing velocity, u = 180 km/hr = $180 \times \frac{5}{18} = 10 \times 5 = 50 \:m/s$

Since the aircraft is decelerating so the value of "a" will be negative

i.e., a = - 2.5 m/s²

As the plane will come to rest after landing so, the final velocity of the plane, v = 0 m/s

Let "s" meter represent the required length of the runway.

Now, to find the required length of the runway, we will use the following 3rd equation of motion:

$\boxed{\bold{v^2 - u^2 = 2as}}$

By substituting the value of v, u and a in the above formula, we get

$(0)^2 - (50)^2 = 2\times (- 2.5) \times s$

$\implies - 2500 = -5 \times s$

$\implies 2500 = 5 \times s$

$\implies s = \frac{2500}{5}$

$\implies \bold{s = 500\:m}$

Thus, the required length of the runway is 500 m.

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