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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 A cyclist travels down a slope - Edexcel - GCSE Physics - Question 6 - 2019 - Paper 1

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Question 6

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--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 A cyclist travels dow... 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 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

To determine a typical speed for a racing cyclist on a steep straight 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 relatively slow)
  • C: 20 m/s (this speed is reasonable for a racing cyclist on a slope)
  • D: 200 m/s (this speed is implausible and extremely fast)

Conclusion: Option C (20 m/s) is typical for a racing cyclist.

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=75kgm = 75 \, \text{kg} (mass of the cyclist)
  • g=10N/kgg = 10 \, \text{N/kg} (gravitational field strength)
  • h=20mh = 20 \, \text{m} (height of the slope)

Substituting the values, we get:

ΔGPE=75kg×10N/kg×20m\Delta GPE = 75 \, \text{kg} \times 10 \, \text{N/kg} \times 20 \, \text{m}

Evaluating this gives:

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

Step 3

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

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Answer

To find the distance travelled by the aircraft during acceleration, we use the formula:

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

Where:

  • v=80m/sv = 80 \, \text{m/s} (final speed)
  • u=0m/su = 0 \, \text{m/s} (initial speed)
  • a=4m/s2a = 4 \, \text{m/s}^2 (acceleration)

Substituting the values, we have:

x=(80)2(0)22×4x = \frac{(80)^2 - (0)^2}{2 \times 4}

Calculating gives:

x=64008=800mx = \frac{6400}{8} = 800 \, \text{m}

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