Question

An aeroplane has a constant acceleration of 2 5 m/s2 while running in a run way it can take off only after its speed reaches 270km/h find the minimum length required for the run way

Answer 1

Given:

Acceleration of plane = 25 m/s²

Initial Speed = 0 m/s

Minimum Velocity for take-off = 270 km/hr

To find:

Minimum length of runway required

Calculation:

Since the question has mentioned about constant acceleration , we can clearly use the equation of Kinematics to solve this kind of problems ;

270 km/hr = 270 × (5/18) = 75 m/s

${v}^{2} = {u}^{2} + 2as$

$= > {75}^{2} = {(0)}^{2} + (2 \times 25 \times s)$

$= > s = \dfrac{75 \times 75}{2 \times 25}$

$= > s = 75 \times 1.5$

$= > s = 112.5 \: m$

So minimum length of runway is 112.5 m

Answer 2

Answer :

112.5 m

$\rule{100}2$

Explanation :

It is given that,

• u = 0 m/s
• v = 270 km, 270 × 5/18 = 70 m/s
• a = 25 m/s²

We've to find it's minimum length for the run way. In short, we've to find distance ( s ).

By using third kinematic equation,

→ v² - u² = 2as

Substitute the given values. We get,

$\sf = (75)^2 - 0^2 = 2(25)(s)$

$\sf = (75)^2 = 50\times {s}$

$\sf = 75\times 75 = 50\times {s}$

$\sf s = \dfeac{75\times 75}{50}$

$\sf s = \dfrac{15\times 75}{10}$

$\sf s = \dfrac{1125}{10}$

$\sf\therefore s = 112.5$

Hence,

The minimum length required for the run way is 112.5 m.

$\rule{200}2$