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The graph below models the velocity of a small train as it moves on a straight track for 20 seconds - AQA - A-Level Maths Pure - Question 14 - 2017 - Paper 2

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The graph below models the velocity of a small train as it moves on a straight track for 20 seconds. The front of the train is at the point A when t = 0. The mass ... show full transcript

Worked Solution & Example Answer:The graph below models the velocity of a small train as it moves on a straight track for 20 seconds - AQA - A-Level Maths Pure - Question 14 - 2017 - Paper 2

Step 1

Find the total distance travelled in the 20 seconds.

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Answer

To calculate the total distance travelled, we need to find the area under the velocity-time graph. The graph consists of three distinct segments:

  1. From 0 to 6 seconds (constant velocity of 8 m/s):

    Area = Base × Height = 6 imes 8 = 48 ext{ m}

  2. From 6 to 10 seconds (deceleration to 0 m/s):

    This segment is a triangle with:

    • Base = 10 - 6 = 4 s
    • Height = 8 m/s (decreasing to 0)

    Area = ( \frac{1}{2} \times 4 \times 8 = 16 \text{ m})

  3. From 10 to 20 seconds (constant velocity of 0 m/s):

    Area = 0 (no distance travelled)

Adding these areas together gives:

Total Distance = 48 + 16 + 0 = 64 ext{ m}

Step 2

Find the distance of the front of the train from the point A at the end of the 20 seconds.

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Answer

At the end of the 20 seconds, the front of the train has travelled a total distance of 64 m from point A.

Step 3

Find the maximum magnitude of the resultant force acting on the train.

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Answer

First, we need to calculate the maximum acceleration, which occurred during the deceleration phase (from 8 m/s to 0 m/s) between 6 and 10 seconds:

Acceleration = ( \frac{\Delta v}{\Delta t} = \frac{0 - 8}{10 - 6} = -2 \text{ m/s}^2 ).

Using Newton's second law, the resultant force is given by:

( F = m \cdot a = 800 \text{ kg} \times | -2 | = 1600 \text{ N} ).

Step 4

Explain why, in reality, the graph may not be an accurate model of the motion of the train.

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Answer

In reality, the graph's abrupt changes and straight lines indicate instant changes in velocity, which are unlikely to occur. Real trains experience gradual acceleration and deceleration due to inertia, friction, and mechanical limits, and would typically not have the sharp transitions depicted in the graph.

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