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During a different part of the journey the car accelerates from 9 m/s to 18 m/s in 6 s - AQA - GCSE Physics - Question 10 - 2018 - Paper 1

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During a different part of the journey the car accelerates from 9 m/s to 18 m/s in 6 s. Use the following equation to calculate the acceleration of the car. acceler... show full transcript

Worked Solution & Example Answer:During a different part of the journey the car accelerates from 9 m/s to 18 m/s in 6 s - AQA - GCSE Physics - Question 10 - 2018 - Paper 1

Step 1

Use the following equation to calculate the acceleration of the car.

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Answer

To find the acceleration, we use the formula:

extacceleration=change in velocitytime taken. ext{acceleration} = \frac{\text{change in velocity}}{\text{time taken}}.

Given that the initial velocity is 9 m/s and the final velocity is 18 m/s:

  • Change in velocity = Final velocity - Initial velocity = 18 m/s - 9 m/s = 9 m/s
  • Time taken = 6 s

Thus,

acceleration=9 m/s6 s=1.5 m/s2.\text{acceleration} = \frac{9 \text{ m/s}}{6 \text{ s}} = 1.5 \text{ m/s}^2.

Step 2

Which equation links acceleration, mass and resultant force?

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Answer

The correct equation that links acceleration, mass, and resultant force is:

resultant force = mass x acceleration.

Step 3

Calculate the resultant force acting on the car and driver while accelerating.

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Answer

To calculate the resultant force, we first need to find the total mass:

  • Total mass = Mass of the car + Mass of the driver = 1120 kg + 80 kg = 1200 kg.

Using the formula:

F=m×a,F = m \times a,

where the mass (m) is 1200 kg and the acceleration (a) is 1.5 m/s²:

F=1200 kg×1.5 m/s2=1800 N.F = 1200 \text{ kg} \times 1.5 \text{ m/s}^2 = 1800 \text{ N}.

Therefore, the resultant force is 1800 N.

Step 4

Explain what happens to the braking distance if the speed doubles.

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Answer

When the speed of the car doubles, the kinetic energy increases by a factor of 4.

This is because kinetic energy (KE) is given by the formula:

KE=12mv2,KE = \frac{1}{2} mv^2,

where m is mass and v is velocity.

If the speed doubles, the new kinetic energy becomes:

KEnew=12m(2v)2=2mv2=4×KEoriginal.KE_{new} = \frac{1}{2} m (2v)^2 = 2mv^2 = 4 \times KE_{original}.

Since the braking force remains constant, the increased kinetic energy requires a longer stopping distance. Therefore, if the speed doubles, the braking distance also increases by a factor of 4.

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