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 1,500 m. We can break this down into segments:

1. Distance during acceleration (300 m)
2. Distance at constant speed (30 m/s for T seconds) = $30T$
3. Distance during deceleration (using the formula: $s = ut + \frac{1}{2}at^2$): The train comes to rest, having a final speed of 0. So, using an average speed of 15 m/s during the deceleration phase (since it decelerates uniformly), let $t_2$ be the time taken to decelerate:
   • The deceleration is 1.25 m/s². Thus, we can find the time taken to decelerate:
   • Using $v = u + at$: $0 = 30 - 1.25t_2$ gives $t_2 = 24$ seconds.
   • The distance covered during deceleration: $s = \frac{1}{2} (30 + 0)t_2 = \frac{1}{2}(30)(24) = 360 ext{ m}$

Combining these:

$300 + 30T + 360 = 1500$

Solving for T:

$30T = 1500 - 660$ $30T = 840$ $T = 28 ext{ seconds}$

The velocity-time graph for train Q will also have three segments:

1. An initial segment representing uniform acceleration from rest until it reaches speed V m/s.
2. A constant velocity segment of V m/s until it begins to decelerate.
3. A final segment where the graph slopes down to rest. Ensure that the area under the graph equals the total distance of 1.5 km.

• The total time taken for the journey should match that of train P, which is $28 + t_2$. Adjust the deceleration phase to accurately reflect the total distance.

Draw the triangle that reflects this relationship and ensure the top vertex shows the maximum velocity at V.

To find V, we need to use the same principles of motion: The total distance for train Q is 1.5 km (or 1500 m).

The distances are:

1. From rest to V m/s (acceleration for time $t_1$).
2. From V m/s to rest, which we can analyze using the same area concept:

• The time for deceleration is given by the average velocity. The equation representing this area in terms of distance can be laid out as:

$\frac{1}{2} * V * t_2 = D_{deceleration}$ The total distance also calculates to:

$D_{acceleration} + D_{constant} + D_{deceleration} = 1500$

Given the timing constraints, we solve for V, ensuring all proportions of time are considered:

• After calculations following the conditions: V can be found to equal approximately 42 m/s or 41.67 m/s.