A car moves along a straight horizontal road from a point A to a point B, where AB = 885 m - Edexcel - A-Level Maths Mechanics - Question 6 - 2012 - Paper 1

The speed-time graph will have three segments:

1. A straight line slope upward from (0, 0) to (( \frac{1}{3} T ), 15) representing acceleration.
2. A horizontal line from (( \frac{1}{3} T ), 15) to (( \frac{1}{3} T + T ), 15) where the car maintains a constant speed.
3. A downward slope from (( \frac{1}{3} T + T ), 15) to (T_total, 0) where the car decelerates to stop.

The total distance covered by the car is the sum of the distances during acceleration, constant speed, and deceleration. The distance is given by:

$\frac{1}{2} a (\frac{1}{3} T)^2 + 15T + \frac{1}{2} (2.5)(T_{deceleration})^2 = 885$

Let ( T_{deceleration} = 6 - T - \frac{1}{3} T ) and substitute known values to find T:

Using the previously established equation:

$a = \frac{15}{\frac{1}{3} T}$

From previous steps, we can calculate the value of a once T is known.

The acceleration-time graph will consist of three parts as well:

1. A horizontal line at a value of a for ( \frac{1}{3} T ).
2. A horizontal line at 0 for the time interval where the car maintains a speed of 15 m/s.
3. A horizontal line at -2.5 m/s² for the deceleration period.