A car starts from rest and moves with constant acceleration along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 3 - 2014 - Paper 2

In this journey, there are three phases:

1. Acceleration Phase: In the first 20 seconds, the car accelerates to $V$.
2. Constant Velocity Phase: The car moves at $V$ for 30 seconds.
3. Deceleration Phases:
   • From $V$ (14 m/s) to 8 m/s with deceleration of $1/2$ m/s²:
     Using $v = u + at$, 8 = 14 - rac{1}{2}t_1 $t_1 = 12 ext{ seconds}$
   • At 8 m/s for 15 seconds.
   • Decelerating from 8 m/s to rest at $1/3$ m/s²: 0 = 8 - rac{1}{3}t_2 $t_2 = 24 ext{ seconds}$

Thus, total time: $ext{Total Time} = 20 + 30 + 12 + 15 + 24 = 101 ext{ seconds}$

To calculate the total distance travelled, we sum up the distances for all phases:

1. Acceleration Phase: From rest to $V$ in 20 seconds:

   • $140$ m (as given)

2. Constant Velocity Phase: Distance for 30 seconds at $V$:

   • Distance = $V imes t = 14 ext{ m/s} imes 30 ext{ s} = 420 ext{ m}$

3. Deceleration from $V$ to 8 m/s:

   • Distance = rac{(14 + 8)}{2} imes t_1 = rac{(14 + 8)}{2} imes 12 = 132 ext{ m}

4. Distance at 8 m/s for 15 seconds:

   • Distance = $8 imes 15 = 120 ext{ m}$

5. Deceleration from 8 m/s to rest:

   • Distance = rac{(8 + 0)}{2} imes t_2 = rac{(8 + 0)}{2} imes 24 = 96 ext{ m}

Thus, total distance: $ext{Total Distance} = 140 + 420 + 132 + 120 + 96 = 908 ext{ m}$