A car starts from rest and moves with constant acceleration along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 3 - 2014 - Paper 1

In this part, we must determine the time taken during each phase of the motion:

• During the first 20 seconds, the car accelerates to $V = 14 ext{ m s}^{-1}$.
• In the next 30 seconds, it continues at constant speed, so: total time = $20 + 30 + t_1 + t_2$, where $t_1$ is the time taken to decelerate to 8 m s$^{-1}$ and $t_2$ is the time for deceleration to rest. Using the formula for deceleration: $8 = V - \frac{1}{2}t_1$ Solving gives $t_1 = 12 ext{ seconds}$, and for the next part: $0 = 8 - \frac{1}{3}t_2$ Solving gives $t_2 = 24 ext{ seconds}$. Now, summing: $total time = 20 + 30 + 12 + 24 = 86 ext{ seconds}.$

To find the total distance, we consider each leg of the journey:

• Distance during acceleration to $V$: $140 ext{ m}$ (from part b).
• Distance at constant speed for 30 seconds: $D_1 = V imes 30 = 14 imes 30 = 420 ext{ m}.$
• Distance during deceleration to 8 m s$^{-1}$: $D_2 = \frac{1}{2}(V + 8)t_1 = \frac{1}{2}(14 + 8)(12) = 132 ext{ m}.$
• Distance at 8 m s$^{-1}$ for 15 seconds: $D_3 = 8 imes 15 = 120 ext{ m}.$
• Distance during the final deceleration: $D_4 = \frac{1}{2}(8 + 0)t_2 = \frac{1}{2}(8)(24) = 96 ext{ m}.$ Adding all these distances together gives: $Total ext{ distance} = 140 + 420 + 132 + 120 + 96 = 908 ext{ m}.$