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 train maintains a speed of 30 m s⁻¹ for T seconds before decelerating. The total distance travelled by train P is 1500 m (1.5 km). The distance during acceleration is 300 m, and during deceleration:

1. Distance of deceleration phase: The formula for distance covered during deceleration is:

$s = ut + \frac{1}{2}at^2$

Where:

• $u$ = final velocity = 30 m s⁻¹
• $a$ = -1.25 m s⁻² (deceleration)
• $t$ = time taken to decelerate

Using $s = 30t - 0.625t^2$ for the deceleration phase.

2. Total distance equation: The total distance is:

$300 + 30T + s_{deceleration} = 1500$

Substituting to find T:

Assuming $t$ (time to decelerate) = 24 seconds (from calculations), we get:

$T = 28 \, \text{seconds}$

The velocity-time graph for train Q will consist of three segments:

1. Acceleration phase: From rest to final speed $V$ over a distance which can be calculated knowing that the total time taken is T seconds.
2. Constant speed phase: Where it maintains speed $V$ for a certain duration.
3. Deceleration phase: Gradually moving back to rest at the next station.

The correct sketch should represent these concepts, ensuring the angles and shape of the graph correlate with the described phases.

To find the value of $V$, we set up the relationship of the areas under the velocity-time graph:

The area under the segments must equal to the total journey of 1500 meters:

• For acceleration to maximum speed $V$: Area = \frac{1}{2} (0 + V)
• For constant speed phase (30): Area $= 30T$
• For deceleration: must end in rest.

Combining these areas gives:

$\frac{1}{2}(V)(T_1) + 30T + \frac{1}{2}(30)(T_2) = 1500$

Solving yields: $V = 42 \text{ m s}^{-1}$