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

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3 (a) (i) An aircraft starts from rest and accelerates along the runway for 36s to reach take-off velocity. Take-off velocity for this aircraft is 82 m/s. Show tha... show full transcript

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

Step 1

Show that the acceleration of the aircraft along the runway is about 2 m/s².

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Answer

To calculate the acceleration, we can use the formula for acceleration:
a=vuta = \frac{v - u}{t}
where:
v = final velocity (82 m/s),
u = initial velocity (0 m/s),
t = time (36 s).
Substituting the values:
a=82036a = \frac{82 - 0}{36}
a=8236a = \frac{82}{36}
a2.28 m/s²a \approx 2.28 \text{ m/s²}
This rounds to approximately 2 m/s², confirming the calculation.

Step 2

Calculate the distance the aircraft travels along the runway before take-off.

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Answer

We can apply the equation of motion:
v2=u2+2axv² = u² + 2ax
Rearranging the formula to find distance (x):
x=v2u22ax = \frac{v² - u²}{2a}
Substituting the known values:
x=822022×2.28x = \frac{82² - 0²}{2 \times 2.28}
Calculating gives:
x=67244.561470.4extmx = \frac{6724}{4.56} \approx 1470.4 ext{ m}
Thus, the aircraft travels approximately 1470.4 meters before take-off.

Step 3

Suggest one reason why the length of the runway used is always much longer than the calculated distance that the aircraft travels along the runway before take-off.

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Answer

One reason could be that real-life scenarios involve factors such as wind resistance and mechanical failures, which can affect the acceleration and distance requirements. Additionally, safety margins are included to ensure a safe take-off process.

Step 4

Calculate the kinetic energy of the aircraft as it lands.

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Answer

The formula for kinetic energy (KE) is:
KE=12mv2KE = \frac{1}{2} mv²
where:
m = mass of the aircraft (3.6 × 10² kg),
v = velocity (71 m/s).
Substituting the values into the equation:
KE=12×3.6×102×712KE = \frac{1}{2} \times 3.6 \times 10² \times 71²
Calculating gives:
KE9.1×105 JKE \approx 9.1 \times 10^5 \text{ J}
So, the kinetic energy of the aircraft as it lands is approximately 9.1 × 10⁵ joules.

Step 5

Give one way that the energy has been transferred to the surroundings.

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Answer

One way the energy has been transferred is through thermal dissipation due to air resistance and friction during landing.

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