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Figure 8 shows the distance-time graph for a car travelling at 15 m/s - AQA - GCSE Physics Combined Science - Question 6 - 2018 - Paper 2

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Figure 8 shows the distance-time graph for a car travelling at 15 m/s. When the driver is tired, his reaction time increases from 0.50 seconds to 0.82 seconds. Det... show full transcript

Worked Solution & Example Answer:Figure 8 shows the distance-time graph for a car travelling at 15 m/s - AQA - GCSE Physics Combined Science - Question 6 - 2018 - Paper 2

Step 1

Determine the distance travelled in the reaction time

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Answer

To find the extra distance the car travels while the driver is reacting, we first calculate the difference in reaction time:

0.82exts0.50exts=0.32exts0.82 ext{ s} - 0.50 ext{ s} = 0.32 ext{ s}

Next, we use the formula for distance:

extDistance=extSpeedimesextTime ext{Distance} = ext{Speed} imes ext{Time}

Substituting in the speed of the car (15 m/s) and the extra reaction time (0.32 s):

extDistance=15extm/simes0.32exts=4.8extm ext{Distance} = 15 ext{ m/s} imes 0.32 ext{ s} = 4.8 ext{ m}

Step 2

Explain why the temperature of the brakes increases

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Answer

When the brakes are used, they experience friction which generates heat. This friction is a result of the kinetic energy of the car being converted into thermal energy as it slows down. As the braking force is applied, the kinetic energy decreases, leading to an increase in thermal energy, thus raising the temperature of the brakes.

Step 3

Calculate the initial velocity of the lorry

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Answer

We use the equation of motion to relate the initial velocity, final velocity, acceleration, and distance:

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

Where:

  • Final velocity, v=19extm/sv = 19 ext{ m/s}
  • Acceleration, a=2.0extm/s2a = 2.0 ext{ m/s}^2
  • Distance, s=84extms = 84 ext{ m}

Rearranging for initial velocity (uu):

u2=v22asu^2 = v^2 - 2as

Substituting in the known values:

u2=1922imes2imes84u^2 = 19^2 - 2 imes 2 imes 84

Calculating:

u2=361336=25u^2 = 361 - 336 = 25

So, u=ext25=5extm/s.u = ext{√}25 = 5 ext{ m/s}.

Step 4

Describe the relationships shown in Figure 9

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Answer

The graph represents the relationship between thinking distance, braking distance, and stopping distance as the speed of the car increases.

  1. Thinking Distance: This distance increases linearly with the speed of the car, indicating that as speed increases, the distance the driver must cover to recognize a hazard also increases due to longer reaction times.

  2. Braking Distance: This distance increases at a greater rate than thinking distance; this indicates a quadratic relationship with the speed. As speed increases, the braking distance increases significantly, reflecting the greater stopping distance needed to bring the vehicle to a halt.

  3. Stopping Distance: Overall stopping distance is the sum of thinking distance and braking distance, thus also increasing as speed increases. Notably, the overall stopping distance grows at an increasing rate with speed, indicating that higher velocities lead to disproportionately greater stopping distances.

Factors Affecting the Gradients: Factors such as tire grip, road surface conditions, and vehicle mass could affect these relationships by altering the effective braking force and reaction times.

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