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

Using the formula for uniform acceleration, the velocity at the end of deceleration can be expressed as:

$v = u + at$

Where:

• $v = 8$ ms⁻¹ (final velocity)
• $u = 20$ ms⁻¹ (initial velocity)
• $a = -0.4$ ms⁻² (deceleration rate)

Plugging in the values gives us:

$8 = 20 - 0.4t$

Solving for $t$:

t = \frac{20 - 8}{0.4} = 30 \text{ s}.

The total distance from A to B is 1960 m. We can break this distance into sections:

1. From A to the end of the 25 s at 20 ms⁻¹:

   $d_1 = vt = 20 \times 25 = 500 \text{ m}$

2. From 25 s to the end of deceleration:

   The average speed during deceleration can be calculated as: $\text{Average speed} = \frac{u + v}{2} = \frac{20 + 8}{2} = 14 \text{ ms}^{-1}$

   Thus, the distance covered is:

   $d_2 = 14 \times 30 = 420 \text{ m}$

3. From the end of deceleration (t = 60 s) the car moves at 8 ms⁻¹:

   $d_3 = 8 \times 60 = 480 \text{ m}$

4. Let $T$ be the time taken to accelerate back to 20 ms⁻¹. The distance for this segment can be expressed as:

   $d_4 = \frac{1}{2} (8 + 20) T = 14T$

The total distance is:

$1960 = d_1 + d_2 + d_3 + d_4$

Substituting the known distances gives:

$1960 = 500 + 420 + 480 + 14T$

Simplifying:

$1960 = 1400 + 14T$

Thus, we have:

$T = \frac{1960 - 1400}{14} = 40 \text{ s}$

Finally:

$\text{Total time} = 25 + 30 + 60 + 40 = 155 \text{ s}.$