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

From the previous calculation, the total distance travelled is 64 m. Therefore, the distance of the front of the train from point A at the end of 20 seconds is:

Distance = 64 m.

To calculate the maximum acceleration, we find the change in velocity over the time period over which it occurs:

• The maximum acceleration occurs from 8 m/s to 0 m/s between 6 to 10 seconds:

  $a_{max} = \frac{\Delta v}{\Delta t} = \frac{0 - 8}{10 - 6} = \frac{-8}{4} = -2 \, m/s^2$

Now, using Newton's second law of motion, we find the force:

$F_{max} = m \cdot a = 800 \, kg \cdot (-2 \, m/s^2) = -1600 \, N$

The magnitude of the resultant force is thus 1600 N.

In reality, the graph may not be an accurate model of the motion of the train because abrupt changes in velocity are unlikely to result in straight lines. Actual motion would exhibit smoother transitions due to factors like inertia, friction, and the gradual application of force.