A car is moving on a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 4 - 2012 - Paper 1

The initial speed $u = 20$ ms$^{-1}$ and the final speed $v = 8$ ms$^{-1}$. The deceleration $a = -0.4$ ms$^{-2}$. We use the formula:

$v = u + at$

Substituting the known values:

$8 = 20 + (-0.4)t$

Rearranging gives us:

thus, \\ t = \frac{12}{0.4} = 30 \text{ s}$$ Thus, the time for which the car is decelerating is 5 s.

To find the total time, we need to sum the distances covered during each phase of motion.

1. Distance during constant speed (0 to 25s):

   $d_1 = v imes t = 20 imes 25 = 500 ext{ m}$

2. Distance during deceleration (25 to 30s):

   The average speed during deceleration:

   $\text{average speed} = \frac{(20 + 8)}{2} = 14 ext{ ms}^{-1}$

   Therefore, the distance:

   $d_2 = 14 imes 5 = 70 ext{ m}$

3. Distance during constant speed (30 to 90s):

   $d_3 = 8 imes 60 = 480 ext{ m}$

Now we calculate total distance from A to the point just before accelerating:

d_total = d_1 + d_2 + d_3 = 500 + 70 + 480 = 1050 ext{ m}

4. Remaining distance to B:

   Distance from A to B = 1960 m, so:

   $d_4 = 1960 - 1050 = 910 ext{ m}$

5. For the acceleration phase (90 to t):

   Assign uniform acceleration $a$ to find time, knowing that final speed $v=20$ ms$^{-1}$ and starting from $v=8$ ms$^{-1}$ with:

   (20 = 8 + a(T - 90))$$ Compute the time spent accelerating: $$910 = \frac{(20 + 8)}{2} \cdot (T - 90) \\ 910 = 14(T - 90) \\ T - 90 = \frac{910}{14} = 65 \\ T = 155\text{ s}$$

Overall, the total time taken for the car to move from A to B is 155 s.