Question

An aircraft is travelling along a runway at velocity of 25ms-1. It accelerates at a rate of 4ms-2 for a distance of 750m before taking off. Calculate its take-off speed

Answer 1

Answer:

The take off speed is 81.39 m/s.

GIVEN :

Initial velocity, u = 25 m/s

Acceleration, a = $4 m/s^{2}$

Distance, d = 750 m

TO FIND :

The take off speed.

FORMULA :

$2as \: = \: v^{2} \: - \: u^{2}$

SOLUTION :

• $2as \: = \: v^{2} \: - \: u^{2}$

Put the value in above formula,

→ $2 \times 4 \times 750 \: = \: v^{2} \: - \: 25^{2}$

→ $8 \times 750 \: = \: v^{2} \: - \: 625$

→ $6000 \: = \: v^{2} \: - \: 625$

→ $6000 \: + \: 625 \: = \: v^{2}$

→ $6625 \: = \: v^{2}$

→ $v \: = \: \sqrt {6625}$

$\sf {\boxed {v \: = \: 81.39 \: m/s}}$