The velocity-time graph in Figure 4 represents the journey of a train P travelling along a straight horizontal track between two stations which are 1.5 km apart - Edexcel - A-Level Maths Mechanics - Question 5 - 2013 - Paper 1

The total distance covered by train P is 1500 m (1.5 km). We can break this down into segments:

1. First 300 m (acceleration phase)
2. T seconds at constant speed (30 m/s)
3. Deceleration phase until coming to rest.

The time taken to cover the first 300 m when accelerating is:

• Time (t₁) = \frac{300}{u + v} \cdot 2 = \frac{300}{0 + 30} \cdot 2 = 20 , \text{seconds}

For the deceleration phase, let the time taken be t₂:

• Using the formula, we have: $0 = 30 - 1.25t_2$
• Which gives us: $t_2 = \frac{30}{1.25} = 24 \, \text{seconds}$

Now, we can find T as: $T = 1500 - (20 + 24) = 28 \, \text{seconds}$

The velocity-time graph for train Q should be a triangle. It starts from the origin (0,0), rises to the maximum speed V, and then drops back to rest. The area under the graph must equal the total distance of 1.5 km. Ensure that the maximum speed reaches line y = 30 and meets the horizontal line at some point, showing the triangular shape.

To calculate V, consider the total time taken for train Q. The total distance covered is also 1500 m. The time for the travel parts for Q:

• First part with acceleration to V: let time = t_1
• Second part with uniform speed = t_2 (from speed V to rest)

We can use the relation of distance: $\frac{1}{2}a t^2 + Vt_2 = 1500$ Substituting appropriate values to solve for V while taking into account that the distance covered by each segment must sum to the total distance of 1500 m. After setting up the equations, re-arranging gives:

$V = \frac{750}{(20 + 28 + 24)}$ yielding V = 41.67 , \text{m/s or better.}$$