Question

An aeroplane accelerates down a runway at 3.20m/^{​​​​​​}s² for 32.8 seconds until lifts off from the ground. What is the distance travelled before takeoff?

Answer 1

Step 1: Given that:

The acceleration(a) in the aeroplane = $3.20m{s}^{-2}$

Time taken(t) until lift off = $32.8sec$

Initial velocity(u) of the aeroplane = 0

Step 2: Calculation of the distance travelled by aeroplane before takeoff:

According to the second equation of motion;

$s=ut+\frac{1}{2}a{t}^2$

Where s = distance travelled

u = Initial velocity

t = Time

a = acceleration

Putting the values, we get;

$s=0\times 32.8sec+\frac{1}{2}\times 3.2m{s}^{-2}\times {\left(32.8sec\right)}^2$

$s=1.6\times 32.8\times 32.8$

$s=1.6\times 1075.84$

$s=1721.344$

$s\approx 1721.3m$

Thus,

The distance covered by the aeroplane before takeoff is $1721.3m$ .