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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 ti... 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 answer is B: force. A vector quantity has both magnitude and direction, while energy, mass, and work are scalar quantities.

Step 2

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

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Answer

The equation that relates acceleration to change in velocity and time taken 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 of the van, we first determine the change in velocity:

Δv=vfvi=20 m/s2 m/s=18 m/s\Delta v = v_f - v_i = 20 \text{ m/s} - 2 \text{ m/s} = 18 \text{ m/s}

Now we use the equation for acceleration:

a=Δvt=18 m/s12 s=1.5 m/s2a = \frac{\Delta v}{t} = \frac{18 \text{ m/s}}{12 \text{ s}} = 1.5 \text{ m/s}²

Thus, the acceleration of the van is 1.5m/s21.5 \, \text{m/s}².

Step 4

Calculate the distance the cyclist travels in the 15 s.

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Answer

To calculate the distance traveled by the cyclist, we can find the area under the velocity-time graph:

  1. The area of the triangle can be calculated as:

    Area=12×base×height\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}

  2. Here, the base is 15 s and the height is 7 m/s (the final velocity):

    Area=12×15 s×7 m/s=52.5 m\text{Area} = \frac{1}{2} \times 15 \text{ s} \times 7 \text{ m/s} = 52.5 \text{ m}

Thus, the distance the cyclist travels in the 15 s is 52.5m52.5 \, \text{m}.

Step 5

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

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Answer

Factors involving the driver that can affect the stopping distance include:

  • Change in Reaction Time: A tired driver or one who is not paying attention may take longer to react to a situation, increasing the stopping distance.
  • Effect of Drugs: Drivers under the influence of drugs or alcohol may experience impaired judgment and slower reactions.
  • Type of Footwear: Inappropriate footwear (e.g., high heels) may reduce the driver's ability to apply the brakes effectively.
  • Driver Behavior: How firmly the driver presses the brake pedal can dramatically affect stopping distance; a stronger application results in shorter distances.

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

Factors concerning the car or the road affecting stopping distance include:

  • Mass/Weight of Car: Heavier cars require a longer stopping distance than lighter ones due to greater inertia.
  • Speed of Car: Higher speeds increase the stopping distance as kinetic energy is proportional to the square of speed.
  • State of Brakes: Worn brakes may not provide the required stopping force, leading to increased distances.
  • State of Tires: Bald or poorly inflated tires reduce traction, increasing stopping distances on wet or slick surfaces.
  • State of Road: Conditions such as ice, rain, or gravel can drastically increase stopping distances, requiring careful consideration by the driver.

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