An aircraft starts from rest and accelerates along the runway for 36s to reach take-off velocity - Edexcel - GCSE Physics - Question 5 - 2022 - Paper 1

We can use the equation:
v² - u² = 2ax to find the distance (x).
Substituting the known values:
$82^2 - 0^2 = 2 \times 2.28 \times x$
This simplifies to:
$6724 = 4.56x$
Solving for x:
$x = \frac{6724}{4.56} \approx 1470.39 \, m$
Thus, the distance the aircraft travels along the runway is approximately 1470.39 m.

One reason is that the runway length must accommodate additional safety margins that account for variables such as wind conditions, pilot reaction time, and the need for take-off clearance.

The kinetic energy (KE) can be calculated using the formula:
$KE = \frac{1}{2} mv²$
where:
m = mass of the aircraft = 3.6 \times 10^4 kg,
v = velocity = 71 m/s.
Substituting these values into the formula, we have:
$KE = \frac{1}{2} \times 3.6 \times 10^4 \times 71^2$
Calculating this gives:
$KE = \frac{1}{2} \times 3.6 \times 10^4 \times 5041 \approx 9.1 \times 10^8 \, J$
Thus, the kinetic energy of the aircraft as it lands is approximately 9.1 × 10^8 J.

One way that the kinetic energy has been transferred to the surroundings is through friction between the aircraft's brakes and the runway, which converts kinetic energy into thermal energy.