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The graph represents the motion of a cyclist at the start of an Olympic race - Edexcel - GCSE Physics - Question 6 - 2016 - Paper 1

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The graph represents the motion of a cyclist at the start of an Olympic race. (i) Calculate the initial acceleration. (ii) Another cyclist has a smaller initial ac... show full transcript

Worked Solution & Example Answer:The graph represents the motion of a cyclist at the start of an Olympic race - Edexcel - GCSE Physics - Question 6 - 2016 - Paper 1

Step 1

Calculate the initial acceleration.

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Answer

To calculate the initial acceleration, we need to determine the change in velocity and the time taken for that change.

From the graph, at time 0 seconds, the velocity is 0 m/s, and at time 2 seconds, the velocity is 16 m/s.

The change in velocity is:

Δv=vfvi=16m/s0m/s=16m/s\Delta v = v_f - v_i = 16 \, \text{m/s} - 0 \, \text{m/s} = 16 \, \text{m/s}

The time taken for this change is:

Δt=2s0s=2s\Delta t = 2 \, \text{s} - 0 \, \text{s} = 2 \, \text{s}

Now, we can calculate the acceleration using the formula:

a=ΔvΔt=16m/s2s=8m/s2a = \frac{\Delta v}{\Delta t} = \frac{16 \, \text{m/s}}{2 \, \text{s}} = 8 \, \text{m/s}^2

Step 2

Draw her motion on the graph above.

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Answer

To represent the second cyclist's motion, we can draw her line starting similarly to the first cyclist but with a smaller initial acceleration. Let's say her initial acceleration is approximately 4 m/s².

  1. From time 0 to 2 seconds: The velocity will increase from 0 m/s to: v=u+at=0+4×2=8m/sv = u + at = 0 + 4 \times 2 = 8 \, \text{m/s}

  2. From 2 to 4 seconds: She reaches a constant velocity of 17 m/s. Therefore, the graph will show a straight horizontal line at 17 m/s after the initial acceleration phase, starting after 2 seconds.

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