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

The total time can be calculated by determining the time spent in each phase.

1. First Phase: Acceleration to $V$: $t_1 = 20 ext{ seconds}$

2. Second Phase: Constant speed $V$ for 30 seconds: $t_2 = 30 ext{ seconds}$

3. Third Phase: Deceleration from $V$ to 8 m s$^{-1}$: Using $v = u + at$, setting $u = V$, $v = 8$, $a = -1.5$: $8 = 14 - 1.5t_3$ $t_3 = \frac{14 - 8}{1.5} = \frac{6}{1.5} = 4 ext{ seconds}$

4. Fourth Phase: Constant speed 8 m s$^{-1}$ for 15 seconds: $t_4 = 15 ext{ seconds}$

5. Fifth Phase: Deceleration from 8 m s$^{-1}$ to rest: Using $v = u + at$, setting $u = 8$, $v = 0$, $a = -1$: $0 = 8 - 1t_5$ $t_5 = 8 ext{ seconds}$

Now, summing all time intervals: $t_{ ext{total}} = t_1 + t_2 + t_3 + t_4 + t_5 = 20 + 30 + 4 + 15 + 8 = 77 ext{ seconds}$

The total distance can be calculated by summing the distances in each phase:

1. First Phase (acceleration): $d_1 = \frac{1}{2} a t^2 = \frac{1}{2} \cdot \frac{14}{20} \cdot 20^2 = 140 ext{ m}$

2. Second Phase (constant speed): $d_2 = V \cdot t_2 = 14 ext{ m s}^{-1} \cdot 30 ext{ s} = 420 ext{ m}$

3. Third Phase (deceleration): To find distance covered from $V$ to 8 m s$^{-1}$: Using the average speed: $d_3 = \frac{(V + 8)}{2} t_3 = \frac{(14 + 8)}{2} imes 4 = 44 ext{ m}$

4. Fourth Phase (constant speed): $d_4 = 8 ext{ m s}^{-1} \cdot 15 ext{ s} = 120 ext{ m}$

5. Fifth Phase (deceleration from 8 m s$^{-1}$ to rest): The average speed is: $d_5 = \frac{(8 + 0)}{2} t_5 = \frac{8}{2} \cdot 8 = 32 ext{ m}$

Now, summing all distance values: $d_{ ext{total}} = d_1 + d_2 + d_3 + d_4 + d_5 = 140 + 420 + 44 + 120 + 32 = 756 ext{ m}$