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a) A motorcyclist is climbing a hill at a constant speed of 13 m/s - Edexcel - GCSE Physics - Question 3 - 2017 - Paper 1

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a) A motorcyclist is climbing a hill at a constant speed of 13 m/s. Calculate the time it takes for the motorcyclist to travel 29 m. (2) b) The picture shows a rai... show full transcript

Worked Solution & Example Answer:a) A motorcyclist is climbing a hill at a constant speed of 13 m/s - Edexcel - GCSE Physics - Question 3 - 2017 - Paper 1

Step 1

Calculate the time for the motorcyclist to travel 29 m

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Answer

Using the formula for time:

t=dv=29 m13 m/s=2.23extst = \frac{d}{v} = \frac{29 \text{ m}}{13 \text{ m/s}} \\ = 2.23 ext{ s}

The time taken is approximately 2.23 seconds.

Step 2

Explain how this design makes good use of energy transfers in the system

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Answer

The design employs gravitational potential energy (GPE) effectively. As the passenger cabin descends, it loses GPE, which is then transferred to kinetic energy (KE) of the other cabin ascending. This coupling allows for a more efficient energy transfer, as the energy lost by one cabin aids the ascent of another.

Step 3

State the value of the kinetic energy of the car when the brakes were first applied

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Answer

The kinetic energy of the car at the moment the brakes were first applied is 510 000 J.

Step 4

Calculate the velocity of the car when the brakes were first applied

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Answer

Using the kinetic energy formula:

KE=12mv2510000=12×1400×v2v2=2×5100001400v=10200001400=27extm/sKE = \frac{1}{2} mv^2 \\ 510000 = \frac{1}{2} \times 1400 \times v^2 \\ v^2 = \frac{2 \times 510000}{1400} \\ v = \sqrt{\frac{1020000}{1400}} = 27 ext{ m/s}

The velocity when the brakes were first applied is approximately 27 m/s.

Step 5

Calculate the distance it takes for the brakes to stop the car

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Answer

Using the work-energy principle:

W=F×d510000=15000×dd=51000015000=34extmW = F \times d \\ 510000 = 15000 \times d \\ d = \frac{510000}{15000} = 34 ext{ m}

The distance needed for the brakes to stop the car is 34 meters.

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