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

Now we need to find the total time for the journey.

1. Acceleration Phase (0 to 20 seconds):

   • Time = 20 seconds.

2. Constant Speed Phase (20 to 50 seconds):

   • Time = 30 seconds.

3. Deceleration Phase from V to 8 m s$^{-1}$: Using the formula $v = u + at$, where $v = 8$, $u = 14$, and $a = -\frac{1}{2}$: $8 = 14 - \frac{1}{2} t_1$ Solving for $t_1$ gives: $t_1 = 12 \, \text{seconds}$

4. Constant Speed Phase at 8 m s$^{-1}$ (15 seconds):

   • Time = 15 seconds.

5. Deceleration Phase to Rest: Using the same approach, where $u = 8$, $v = 0$, and $a = -\frac{1}{3}$: $0 = 8 - \frac{1}{3} t_2$ Solving for $t_2$ gives: $t_2 = 24 \, \text{seconds}$

Total Time: Combining all parts: $Total \, Time = 20 + 30 + 12 + 15 + 24 = 101 \, \text{seconds}$

To calculate the total distance, we will sum the distances from each segment of the journey:

1. Acceleration Phase:

   • Distance = 140 m (as given).

2. Constant Speed Phase:

   • Distance = speed × time = $14 imes 30 = 420$ m.

3. Deceleration Phase to 8 m s$^{-1}$: Using the equation: $s = ut + \frac{1}{2} at^2$ Here, we can replace $u$ with 14, $t$ with 12 seconds and $a$ with $-\frac{1}{2}$: Total distance in this phase: $s_1 = 14 \cdot 12 + \frac{1}{2}(-\frac{1}{2})(12)^2$ Which gives: $s_1 = 168 - 36 = 132 \, \text{meters}$

4. Constant Speed Phase at 8 m s$^{-1}$:

   • Distance = $8 imes 15 = 120$ m.

5. Deceleration Phase to Rest: Using: $s = ut + \frac{1}{2} at^2$ $u = 8$, $a = -\frac{1}{3}$, $t = 24$ seconds: $s_2 = 8 \cdot 24 + \frac{1}{2}(-\frac{1}{3})(24)^2$ This results in: $s_2 = 192 - 96 = 96 \, meters$

Total Distance: Combining all segments: $Total \, Distance = 140 + 420 + 132 + 120 + 96 = 908 \, ext{meters}$