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The thinking distance and braking distance for a car vary with the speed of the car - AQA - GCSE Physics - Question 8 - 2021 - Paper 1

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The thinking distance and braking distance for a car vary with the speed of the car. 08.1 Explain the effect of two other factors on the braking distance of a car. ... show full transcript

Worked Solution & Example Answer:The thinking distance and braking distance for a car vary with the speed of the car - AQA - GCSE Physics - Question 8 - 2021 - Paper 1

Step 1

Explain the effect of two other factors on the braking distance of a car.

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Answer

  1. Tyre Condition: If the tyres are worn or in poor condition, the friction between the tyres and the road will decrease. This reduced friction compromises the car's ability to decelerate effectively, thereby increasing the braking distance. Poor grip means that even when the brakes are applied, the car may take longer to stop due to less force being applied to the ground, which affects the overall braking efficiency.

  2. Road Surface: A wet or icy road surface can drastically affect the car's braking distance. On slippery surfaces, the friction is significantly reduced, meaning that the brakes must work harder to bring the car to a stop. This results in a longer braking distance. Therefore, even with the same braking power, the car will not decelerate as quickly on an unfavorable surface.

Step 2

Which equation links acceleration (a), mass (m) and resultant force (F).

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Answer

The correct equation is:

resultant force = mass × acceleration.

Step 3

Calculate the deceleration of the car.

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Answer

Given

Mean braking force, F=7200NF = 7200 \, N
Mass, m=1600kgm = 1600 \, kg

Using the formula: F=mimesaF = m imes a
We can rearrange to find acceleration (deceleration in this case): a=Fm=72001600=4.5m/s2a = \frac{F}{m} = \frac{7200}{1600} = 4.5 \, m/s^2

Thus, the deceleration of the car is 4.5m/s24.5 \, m/s^2.

Step 4

Determine the stopping distance when the car is travelling at 80 km/h.

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Answer

From Figure 18, at a speed of 80 km/h, the stopping distance is indicated to be 53 meters.

Step 5

Which equation links area of a surface (A), the force normal to that surface (F) and pressure (p).

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Answer

The correct equation is:

p = \frac{F}{A}.

Step 6

Calculate the surface area of the piston.

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Answer

We know:
Pressure, p=120000Pap = 120000 \, Pa
Force, F=60NF = 60 \, N
Using the formula: p=FAp = \frac{F}{A}
Rearranging gives: A=Fp=60120000=0.0005m2A = \frac{F}{p} = \frac{60}{120000} = 0.0005 \, m^2

In standard form, this is: A=5.0×104m2A = 5.0 \times 10^{-4} \, m^2

The unit is m2m^2.

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