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

The distance of the front of the train from point A at the end of the 20 seconds can be calculated as follows:
Using the total distance calculated earlier, the distance from point A is:

$\text{Distance from A} = 74 \text{ m}$

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

• The maximum acceleration occurs during the interval from 6 seconds to 10 seconds when the train slows down from 8 m/s to 0 m/s.
• The change in velocity is:
  $\Delta v = 0 - 8 = -8 \text{ m/s}$
• The time over which this change occurs is 4 seconds. Therefore, the maximum acceleration is:
  $a_{max} = \frac{\Delta v}{\Delta t} = \frac{-8}{4} = -2 \text{ m/s}^2$
• Using Newton's second law, the resultant force can be calculated as:
  $F_{max} = m \cdot a_{max} = 800 \cdot -2 = -1600 \text{ N}$
  Thus, the maximum magnitude of the resultant force is 1600 N.

In reality, the graph does not accurately model the motion of the train for several reasons:

1. Abrupt Changes in Velocity: The graph shows abrupt changes in velocity, with straight lines representing instant acceleration and deceleration. In real scenarios, these changes are smoother and involve various factors such as friction and inertia.
2. Constant Velocity: The graph depicts a constant velocity for a prolonged duration. However, due to factors like resistance and track conditions, this is often unrealistic.
3. Curvature in Motion: Instead of straight lines, real motion is likely to produce curves, particularly during acceleration and deceleration phases. Thus, we would expect to see smoother transitions rather than sharp changes.