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

Since the front of the train travels 64 m in the first 20 seconds:

Distance from point A = 64 m.

Using Newton's second law, we calculate the maximum acceleration (deceleration) of the train between 6 to 10 seconds:

1. Acceleration (a) = Change in Velocity / Time = (0 - 8) m/s / 4 s = -2 m/s² (deceleration)

2. Now, using F = ma, where m is the mass of the train = 800 kg:

   • Maximum Force (F) = 800 kg × 2 m/s² = 1600 N.

In reality, the graph may not accurately represent the motion of the train because:

• Abrupt changes in velocity, as shown in the graph, are unlikely in real-world scenarios. Trains typically accelerate and decelerate gradually.
• The straight lines indicate constant velocity or instantaneous changes which do not account for the natural dynamics of train motion.