5 (a) (i) 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:

$v^2 - u^2 = 2ax$

where:

• $v = 82 \text{ m/s}$
• $u = 0 \text{ m/s}$
• $a = 2.28 \text{ m/s}²$
• $x$ is the distance we need to calculate.

Rearranging the equation gives:

$x = \frac{v^2 - u^2}{2a}$

Substituting in the values:

$x = \frac{82^2 - 0^2}{2 \times 2.28} = \frac{6724}{4.56} \approx 1470.4 \text{ m}$

The distance traveled along the runway is approximately 1460 m to 1482 m.

One reason is that pilots require additional runway length for safety margins and to account for variables such as weather conditions, unexpected mechanical issues, and the need for longer distances during take-off to ensure a smooth ascent.

The kinetic energy (KE) can be calculated using the formula:

$KE = \frac{1}{2}mv^2$

where:

• $m = 3.6 \times 10^{2} \text{ kg}$
• $v = 71 \text{ m/s}$

Substituting in the values:

$KE = \frac{1}{2} \times 3.6 \times 10^{2} \times 71^2 = \frac{1}{2} \times 3.6 \times 10^{2} \times 5041 \approx 90761.5 \text{ J}$

Thus, the kinetic energy of the aircraft as it lands is approximately 90,761.5 J.

One way the energy has been transferred to the surroundings is through friction between the aircraft's wheels and the runway, which converts kinetic energy into thermal energy, causing heat to be dissipated.