3 (a) (i) An aircraft starts from rest and accelerates along the runway for 36 s to reach take-off velocity - Edexcel - GCSE Physics Combined Science - Question 3 - 2022 - Paper 1

We can use the equation for motion:

$v^2 = u^2 + 2ax$

Rearranging gives:

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

Substituting the known values:

• $v = 82 \, \text{m/s}$
• $u = 0 \, \text{m/s}$
• $a \approx 2.3 \, \text{m/s}²$

Calculating:

$x = \frac{82^2 - 0^2}{2 \times 2.3} = \frac{6724}{4.6} \approx 1460.87 \, \text{m}$

Therefore, the distance traveled by the aircraft before take-off is approximately $1461 \, \text{m}$.

One reason could be that factors such as wind resistance, take-off safety protocols, and varying weights of the aircraft (due to passengers, cargo, etc.) contribute to the need for a longer runway.

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

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

Where:

• $m = 3.6 \times 10^5 \, \text{kg}$ (mass of the aircraft)
• $v = 71 \, \text{m/s}$ (velocity)

Substituting the values:

$KE = \frac{1}{2} \times 3.6 \times 10^5 \times 71^2$

Calculating gives:

$KE \approx 9.1 \times 10^8 \, \text{Joules}$

The energy can be transferred to the surroundings mechanically (e.g., through sound energy during landing) or due to heating (e.g., friction between the aircraft and the runway).