Question

✨✨HOLLA✨✨

❤Physics Genius Plzanswer it❤

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 3 m/s . The takeoff speed for this plane will be 65 m/s , Assuming this minimum acceleration , what is the minimum allowed length for the Runway?

⭐⭐I will mark u a Brainliest⭐⭐​

Answer 1

Answer:704m

Solution

U = 0

V = 65 m/s

a = 3 m/s²

To find out the distance, we will use the following equation:

S = Ut + ½ a t²

as U = 0, the equation becomes S = ½ a t² = ½ . 3. t² ……(1)

We will use the following eqn to find out the time required to attain this velocity.

V = u + at

As U = 0, the eqn becomes V = at

t = V/a = 65/3 sec……(2)

Putting the value of time t in equation 1 we get S = ½ . 3. (65/3)²

= 704 m

So the minimum allowed length of the airport is 704 m

Answer 2

Hey MATE!!!

We are provided with the information that :

initial velocity (u) = 0 m/s

final velocity (v) = 65 m/s

acceleration (a) = 3m/s^2

displacement (s) = ?

We are to find the minimum length of the runway, so we can apply 3rd equation of motion here :

=> v^2 = u^2 + 2as

(65)^2 = (0)^2 + 2(3)(s)

=> 4225 = 0 + 6s

s = 4225/6 => 704.1 ≈ 704m

$\huge\boxed{<strong><em>S </em></strong><strong><em>=</em></strong><strong><em> </em></strong><strong><em>704m </em></strong>\<strong><em>:</em></strong><strong><em>Answer</em></strong>}$

Hope it helps

HAKUNA MATATA :))