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9 (a) Which of these is a vector? A energy B force C mass D work (b) (i) State the equation that relates acceleration to change in velocity and time taken - Edexcel - GCSE Physics - Question 9 - 2020 - Paper 1

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9-(a)-Which-of-these-is-a-vector?---A-energy---B-force---C-mass---D-work----(b)-(i)-State-the-equation-that-relates-acceleration-to-change-in-velocity-and-time-taken-Edexcel-GCSE Physics-Question 9-2020-Paper 1.png

9 (a) Which of these is a vector? A energy B force C mass D work (b) (i) State the equation that relates acceleration to change in velocity and time taken... show full transcript

Worked Solution & Example Answer:9 (a) Which of these is a vector? A energy B force C mass D work (b) (i) State the equation that relates acceleration to change in velocity and time taken - Edexcel - GCSE Physics - Question 9 - 2020 - Paper 1

Step 1

Which of these is a vector?

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Answer

The correct answer is B: force, as it possesses both magnitude and direction.

Step 2

State the equation that relates acceleration to change in velocity and time taken.

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Answer

The equation is given by: a=Δvta = \frac{\Delta v}{t}
where aa is acceleration, Δv\Delta v is the change in velocity, and tt is the time taken.

Step 3

Calculate the acceleration of the van.

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Answer

To calculate the acceleration, we first determine the change in velocity:
Δv=vfvi=20m/s2m/s=18m/s\Delta v = v_f - v_i = 20 m/s - 2 m/s = 18 m/s
Now, we use the acceleration formula:
a=Δvt=18m/s12s=1.5m/s2a = \frac{\Delta v}{t} = \frac{18 m/s}{12 s} = 1.5 m/s^2
Thus, the acceleration of the van is 1.5m/s21.5 m/s^2.

Step 4

Calculate the distance the cyclist travels in the 15 s.

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Answer

The distance traveled can be found using the area under the velocity-time graph. Since the graph is a triangle, the area can be calculated as follows:

The formula for the area of a triangle is:
Area=12×base×height\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
Here, the base (time) is 15 s and the height (velocity) is 7 m/s (the maximum velocity at 15 s).
Thus,
Distance=12×15s×7m/s=52.5m\text{Distance} = \frac{1}{2} \times 15 s \times 7 m/s = 52.5 m
Therefore, the cyclist travels a distance of 52.5m52.5 m.

Step 5

Explain how the stopping distance of a car is affected by factors involving the driver.

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Answer

The stopping distance of a car can be increased by several driver-related factors:

  • Change in reaction time: A tired or distracted driver may react more slowly, increasing stopping distance.
  • Tiredness: A fatigued driver may take longer to respond to hazards.
  • Effect of drugs: Impairment from substances can severely reduce driver reaction time.
  • Type of footwear: Wearing inappropriate footwear (like flip-flops) can hinder proper braking.
  • Pedal pressure: How hard the driver presses the brake pedal can also influence stopping ability.

Step 6

Explain how the stopping distance of a car is affected by factors involving the car or the road.

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Answer

Several car and road factors that impact stopping distance include:

  • Weight of the car: Heavier vehicles require more distance to stop due to increased momentum.
  • Speed of the car: Higher speeds result in longer stopping distances due to the greater kinetic energy that must be dissipated.
  • State of brakes: Worn brakes can lead to longer distances as they may not effectively slow down the vehicle.
  • State of tyres: Poorly maintained tyres can reduce grip and increase stopping distances.
  • State of road: Slippery or uneven road surfaces can significantly lengthen the distance a car needs to stop.

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