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6 (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 Combined Science - Question 6 - 2020 - Paper 1

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6-(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 Combined Science-Question 6-2020-Paper 1.png

6 (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. (ii) A... show full transcript

Worked Solution & Example Answer:6 (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 Combined Science - Question 6 - 2020 - Paper 1

Step 1

Which of these is a vector?

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Answer

The correct answer is B. Force is a vector quantity, as it has 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 that relates acceleration (a) to change in velocity (Δv) and time (t) is given by:

a=Δvta = \frac{\Delta v}{t}

where Δv is the change in velocity, calculated as final velocity minus initial velocity.

Step 3

Calculate the acceleration of the van.

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Answer

To calculate the acceleration of the van, we can use the formula:

a=Δvta = \frac{\Delta v}{t}

where:

  • Initial velocity (u) = 2 m/s
  • Final velocity (v) = 20 m/s
  • Time (t) = 12 s

First, we calculate the change in velocity:

Δv=vu=20m/s2m/s=18m/s\Delta v = v - u = 20 \, \text{m/s} - 2 \, \text{m/s} = 18 \, \text{m/s}

Now, substituting the values into the acceleration formula:

a=18m/s12s=1.5m/s2a = \frac{18 \, \text{m/s}}{12 \, \text{s}} = 1.5 \, \text{m/s}^2

Step 4

Calculate the distance the cyclist travels in the 15 s.

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Answer

To find the distance traveled by the cyclist using the velocity-time graph:

The area under the graph represents the distance.

The graph appears to be a triangle with:

  • Base (time) = 15 s
  • Height (velocity) = 7 m/s

The area of a triangle is given by:

Area=12×Base×Height\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}

Substituting the values:

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

Step 5

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

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Answer

The stopping distance of a car can be significantly affected by various driver-related factors:

  1. Tiredness: A tired driver may have delayed reaction times leading to increased stopping distances.

  2. Distractions: If a driver is distracted (e.g., using a mobile phone), this may result in slower response times.

  3. Experience: An experienced driver may react more swiftly than a novice, thereby reducing stopping distance.

  4. Type of footwear: Wearing inappropriate shoes can hinder the driver's ability to apply the brakes effectively.

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 factors related to the car or the road that affect stopping distance include:

  1. Mass/Weight of Car: Heavier cars will require a longer distance to stop compared to lighter ones due to inertia.

  2. Condition of Tyres: Worn-out or under-inflated tyres decrease traction, leading to longer stopping distances.

  3. Road Surface: Wet or icy roads can significantly increase the distance needed to stop due to decreased friction.

  4. State of Brakes: Well-maintained brakes will respond better, reducing stopping distance while faulty brakes may lead to much longer distances.

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