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

To find the total time for the journey, we consider each phase of the motion:

1. Acceleration phase until reaching speed $V$: 20 seconds

   • The car takes 20 seconds to reach the speed $V = 14 ext{ m/s}$.

2. Constant speed phase: 30 seconds

   • The car travels at 14 m/s for 30 seconds.

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

   • We can calculate the time taken to decelerate from 14 m/s to 8 m/s using: $v = u + at$
     where:
   • $u = 14 ext{ m/s}$,
   • $v = 8 ext{ m/s}$,
   • a = -rac{1}{2} ext{ m/s}^2.
   • Rearranging gives: 8 = 14 - rac{1}{2} t
     $t = 12 ext{ seconds}$

4. Constant speed phase at 8 m/s: 15 seconds

   • The car travels at 8 m/s for 15 seconds.

5. Final deceleration phase from 8 m/s to rest:

   • Using the formula again, for deceleration: 0 = 8 - rac{1}{3} t Solving gives: $t = 24 ext{ seconds}$

So the total time is: $20 + 30 + 12 + 15 + 24 = 101 ext{ seconds}$

To find the total distance:

1. Distance during acceleration (0 to $V$): d = rac{1}{2} V t = rac{1}{2} imes 14 imes 20 = 140 ext{ m}

2. Distance during constant speed for 30 seconds: $d = V imes t = 14 imes 30 = 420 ext{ m}$

3. Distance during deceleration from $V$ to 8 m/s:

   • We can calculate the distance: d = (u + v) rac{t}{2} = (14 + 8) rac{12}{2} = 132 ext{ m}

4. Distance at constant speed of 8 m/s for 15 seconds: $d = 8 imes 15 = 120 ext{ m}$

5. Distance during deceleration from 8 m/s to rest: d = (u + v) rac{t}{2} = (8 + 0) rac{24}{2} = 96 ext{ m}

Now, adding all these distances gives: $D = 140 + 420 + 132 + 120 + 96 = 908 ext{ m}$