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Which of these would be a typical speed for a racing cyclist travelling down a steep straight slope? A 0.2 m/s B 2 m/s C 20 m/s D 200 m/s (iii) A cyclist travels down a slope - Edexcel - GCSE Physics - Question 6 - 2019 - Paper 1

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Which-of-these-would-be-a-typical-speed-for-a-racing-cyclist-travelling-down-a-steep-straight-slope?--A---0.2-m/s-B---2-m/s-C---20-m/s-D---200-m/s--(iii)-A-cyclist-travels-down-a-slope-Edexcel-GCSE Physics-Question 6-2019-Paper 1.png

Which of these would be a typical speed for a racing cyclist travelling down a steep straight slope? A 0.2 m/s B 2 m/s C 20 m/s D 200 m/s (iii) A cyclist t... show full transcript

Worked Solution & Example Answer:Which of these would be a typical speed for a racing cyclist travelling down a steep straight slope? A 0.2 m/s B 2 m/s C 20 m/s D 200 m/s (iii) A cyclist travels down a slope - Edexcel - GCSE Physics - Question 6 - 2019 - Paper 1

Step 1

Which of these would be a typical speed for a racing cyclist travelling down a steep straight slope?

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Answer

When selecting a typical speed for a racing cyclist on a steep slope, we analyze the options:

  • A: 0.2 m/s - This speed is too slow for a racing cyclist.
  • B: 2 m/s - This speed is also considered slow for racing purposes.
  • C: 20 m/s - This speed is realistic for a racing cyclist on a steep slope.
  • D: 200 m/s - This speed is unrealistic.

Therefore, the most appropriate answer is C: 20 m/s.

Step 2

Calculate the change in gravitational potential energy of the cyclist between the top and the bottom of the slope.

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Answer

To calculate the change in gravitational potential energy (GPE), we use the formula:

ΔGPE=m×g×h\Delta GPE = m \times g \times h

Where:

  • m = mass of the cyclist = 75 kg
  • g = gravitational field strength = 10 N/kg
  • h = height difference = 20 m

Substituting the values:

ΔGPE=75×10×20\Delta GPE = 75 \times 10 \times 20

Calculating this yields:

ΔGPE=15000 J\Delta GPE = 15000 \text{ J}

Step 3

Calculate the distance, x, travelled by the aircraft while it is accelerating.

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Answer

We will utilize the equation to find distance:

x=v2u22ax = \frac{v^2 - u^2}{2a}

Given:

  • Final speed, v = 80 m/s
  • Initial speed, u = 0 m/s
  • Acceleration, a = 4 m/s²

Now substituting these values:

x=802022×4=64008=800 m x = \frac{80^2 - 0^2}{2 \times 4} = \frac{6400}{8} = 800 \text{ m}

Thus, the distance travelled by the aircraft while accelerating is 800 m.

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