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A free-fall parachutist is falling straight down - Edexcel - GCSE Physics - Question 5 - 2018 - Paper 1

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A free-fall parachutist is falling straight down. This is a velocity/time graph of the first part of the jump, before the parachute is opened. (a) Which of these po... show full transcript

Worked Solution & Example Answer:A free-fall parachutist is falling straight down - Edexcel - GCSE Physics - Question 5 - 2018 - Paper 1

Step 1

Which of these points on the graph shows the largest acceleration?

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Answer

To determine which point shows the largest acceleration, we need to examine the slope of the velocity-time graph. The acceleration is represented by the gradient (slope) of the graph. In this case:

  • Point A (0 to 5s): The slope is increasing but is mild.
  • Point B (5 to 10s): The slope is getting steeper, indicating increasing acceleration.
  • Point C (10 to 15s): The graph begins to plateau, indicating lesser acceleration.
  • Point D (15 to 30s): The slope is almost horizontal, indicating very low or no acceleration.

Thus, point B (Q) shows the largest acceleration, and the answer is B.

Step 2

Estimate the distance that the parachutist falls in the first 2.5 s.

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Answer

To estimate the distance fallen, we refer to the velocity-time graph. In the first 2.5 seconds (between points A and up to point B), we can calculate the distance as the area under the curve. The velocity at 2.5 seconds is approximately 12.5 m/s. Assuming constant acceleration:

The formula for distance is: d=vtd = vt Where:

  • dd = distance,
  • vv = average velocity,
  • tt = time.

The average velocity for the first 2.5 seconds is approximately: vavg=0+12.52=6.25extm/sv_{avg} = \frac{0 + 12.5}{2} = 6.25 ext{ m/s} Thus, the estimated distance is: d=6.25extm/s×2.5exts=15.625extmd = 6.25 ext{ m/s} \times 2.5 ext{ s} = 15.625 ext{ m} Therefore, the parachutist falls approximately 16 meters in the first 2.5 seconds.

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