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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. b) The picture shows a railw... 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 it takes for the motorcyclist to travel 29 m.

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Answer

To find the time it takes, we can use the formula:

t=dvt = \frac{d}{v}

where:

  • d = distance = 29 m
  • v = speed = 13 m/s

Substituting the values, we get:

t=29132.23extsecondst = \frac{29}{13} \approx 2.23 ext{ seconds}

Step 2

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

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Answer

The design of the railway system is efficient because it utilizes gravitational potential energy. As one passenger cabin descends, it transfers its gravitational potential energy (GPE) to kinetic energy (KE) which helps to lift the other cabin. This means that the energy lost by one cabin is not wasted but rather transferred to the other cabin, reducing the overall energy required to operate the system.

Step 3

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

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Answer

The value of the kinetic energy of the car when 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

To find the velocity, we use the formula for kinetic energy:

KE=12mv2KE = \frac{1}{2} mv^2

Rearranging gives:

v=2KEm=25100001400Calculating,wefind:v = \sqrt{\frac{2 \cdot KE}{m}} = \sqrt{\frac{2 \cdot 510000}{1400}} \\ Calculating, we find:

v \approx 27 ext{ m/s}

Step 5

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

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Answer

Using the formula:

F=maF = ma

We can rearrange it to find acceleration:

a=Fm=15000140010.71extm/s2a = \frac{F}{m} = \frac{15000}{1400} \approx 10.71 ext{ m/s}^2

Now, using the equation of motion:

v2=u2+2asv^2 = u^2 + 2as

where:

  • u = initial velocity = 27 m/s
  • v = final velocity = 0 m/s
  • a = -10.71 m/s² (deceleration)

Rearranging gives:

s=v2u22a=0(27)22(10.71)Calculating,wefind:s = \frac{v^2 - u^2}{2a} = \frac{0 - (27)^2}{2(-10.71)} \\ Calculating, we find:

s \approx 34 ext{ m}

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