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 for the motion of the car will consist of three segments:

1. Acceleration Phase: From 0 to ( \frac{1}{3} T ), the speed increases linearly from 0 to 15 m/s.
2. Constant Speed Phase: From ( \frac{1}{3} T ) to ( \frac{1}{3} T + T ), the speed remains constant at 15 m/s.
3. Deceleration Phase: From ( \frac{1}{3} T + T ) to the final time, the speed decreases linearly from 15 m/s to 0 m/s over a specific duration.

The exact shape will depend on the actual values of T and the deceleration phase.

To find the value of T, consider the total distance covered:

Total distance = distance during acceleration + distance at constant speed + distance during deceleration = 885 m.

1. Distance during acceleration:

   • Using ( s = ut + \frac{1}{2} at^2 ):
   • For the acceleration phase, ( s_1 = 0 + \frac{1}{2} a \left( \frac{15}{a} \right)^2 = \frac{15^2}{2a} = \frac{225}{2a} ).

2. Distance during acceleration:

   • For the constant speed phase, ( s_2 = 15T. )

3. Distance during deceleration:

   • If deceleration lasts for ( d ) seconds, the formula provides:
   • ( s_3 = 15d - \frac{1}{2} (2.5)d^2 = 15d - 1.25d^2. )

Setting up the equation:

$\frac{225}{2a} + 15T + (15d - 1.25d^2) = 885$

From this, if we can solve for T as given.

Using the earlier derived relation between ( T ) and ( a ):

From the acceleration phase:

$a = \frac{15}{\frac{1}{3}T} \implies a = \frac{45}{T}$

Once we have T from the previous step, substitute the value into this formula to find the value of acceleration a.

The acceleration-time graph will consist of three segments as follows:

1. Acceleration Phase: The graph starts at ( a ) (the initial acceleration) and remains at this constant value until ( \frac{1}{3} T ).
2. Constant Speed Phase: From ( \frac{1}{3} T ) to ( \frac{1}{3} T + T ), the acceleration is 0.
3. Deceleration Phase: The graph then drops to ( -2.5 , \text{m/s}^2 ) during the deceleration until it returns to 0 at the final point.

Clearly mark the regions with horizontal lines to represent each phase.