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

The car is decelerating from 20 ms⁻¹ to 8 ms⁻¹. Using the formula for deceleration:

$v = u + at$

Where:

• Final speed, $v = 8$ m/s
• Initial speed, $u = 20$ m/s
• Acceleration, $a = -0.4$ m/s²

Rearranging:

$8 = 20 - 0.4t$

Solving for $t$ gives:

0.4t = 12 \ t = \frac{12}{0.4} = 30 ext{ seconds} $$

Given that the total distance from A to B is 1960 m, we can break this down into segments:

1. First Segment (0 to 25s):

   • Speed: $20$ ms⁻¹
   • Distance: $d_1 = 25 imes 20 = 500$ m.

2. Second Segment (25 to 30s):

   • Average Speed: rac{20 + 8}{2} = 14 ms⁻¹
   • Time: $5$ seconds; Distance: $d_2 = 14 imes 5 = 70$ m.

3. Third Segment (30 to 90s):

   • Speed: $8$ ms⁻¹; Time: $60$ seconds; Distance: $d_3 = 8 imes 60 = 480$ m.

4. Fourth Segment (90 to T seconds):

   • We need to find the unknown time $T$. Considering that the total distance must equal 1960 m:

   $1960 = d_1 + d_2 + d_3 + d_4$

   Thus,

   $d_4 = 1960 - (500 + 70 + 480) = 910$.

   Now for this segment, we have constant acceleration from 8 ms⁻¹ to 20 ms⁻¹:

   Using the formula for distance with uniform acceleration:

   $d = ut + \frac{1}{2}at^2$

   Here, the average speed is $14$ ms⁻¹, therefore:

   $910 = 14t$.

   Rearranging gives:

   $t = \frac{910}{14} = 65$.

   Therefore, total time $T = 25 + 5 + 60 + 65 = 155$ seconds.