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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 Mechanics - 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 Mechanics - 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 analyze the velocity-time graph.

  1. Calculate the distance for each section of the graph:

    • From 0 to 6 seconds, the velocity is constant at 8 m/s:

      s1=vimest=8imes6=48 ms_1 = v imes t = 8 imes 6 = 48 \text{ m}

    • From 6 to 10 seconds, the velocity decreases to 0:

      s2=12×(0+8)×(10−6)=12×8×4=16 ms_2 = \frac{1}{2} \times (0 + 8) \times (10 - 6) = \frac{1}{2} \times 8 \times 4 = 16 \text{ m}

    • From 10 to 20 seconds, the train travels at a velocity of 0, so:

      s3=0 ms_3 = 0 \text{ m}

  2. Total distance travelled is the sum of these distances:

    st=s1+s2+s3=48+16+0=64 ms_t = s_1 + s_2 + s_3 = 48 + 16 + 0 = 64 \text{ 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

The train travels a total distance of 64 m in 20 seconds, starting from point A. Therefore, the distance of the front of the train from point A at the end of 20 seconds is:

Distance from point A = 64 m.

Step 3

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

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Answer

To calculate the maximum magnitude of the resultant force, we need to first find the maximum acceleration.

The maximum acceleration occurs from 6 seconds to 10 seconds:

  1. Calculate the change in velocity:

    Δv=vf−vi=0−8=−8 m/s\Delta v = v_f - v_i = 0 - 8 = -8 \text{ m/s}

  2. Calculate the time duration:

    Δt=10−6=4 s\Delta t = 10 - 6 = 4 \text{ s}

  3. Calculate the maximum acceleration using the formula:

    amax=ΔvΔt=−84=−2 m/s2a_{max} = \frac{\Delta v}{\Delta t} = \frac{-8}{4} = -2 \text{ m/s}^2

  4. Calculate the resultant force using Newton's second law:

    Fmax=m×amax=800×(−2)=−1600 NF_{max} = m \times a_{max} = 800 \times (-2) = -1600 \text{ N}

Thus, the maximum magnitude of the resultant force acting on the train is:

Resultant Force = 1600 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 may not be an accurate model of the motion of the train due to the abrupt changes and straight lines depicted in the graph. Changes in velocity in real-world scenarios are typically gradual rather than instantaneous. The graph's segments suggest that the train can immediately switch from one velocity to another, which is unlikely in practice. Additionally, operational factors such as friction, air resistance, and mechanical limitations will cause the train's acceleration and deceleration to be more gradual.

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