A car travels along a straight level road - Leaving Cert Applied Maths - Question 1 - 2010

To find the deceleration, we can also apply the relevant kinematic equation:

$v^2 = u^2 + 2as$ where:

• final velocity $v = 0$ (at rest)
• initial velocity $u = 30 \text{ ms}^{-1}$
• distance $s = 45 \text{ m}$

Plugging in the values: $0 = (30)^2 + 2a(45)$ This simplifies to:

To compute the distance |PQ|, we sum the distances traveled in each segment:

1. Distance while accelerating: $s_1 = ut + \frac{1}{2}at^2$ Here: $u = 12 \text{ ms}^{-1}$, $t = 6 \text{ s}$, and $a = 3 \text{ ms}^{-2}$:

$$s_1 = 12(6) + \frac{1}{2}(3)(6^2) = 72 + 54 = 126 \text{ m}.$

2. Constant speed distance: $s_2 = vt = 30 \text{ ms}^{-1} \times 15 \text{ s} = 450 \text{ m}.$

3. Distance while decelerating: $s_3 = 45 \text{ m}.$

Thus, the total distance is: $$|PQ| = s_1 + s_2 + s_3 = 126 + 450 + 45 = 621 \text{ m}.$

The average speed is calculated by dividing the total distance by the total time taken:

1. Total distance: $|PQ| = 621 \text{ m}$.

2. Total time taken from P to Q:

• Time during acceleration: $t_1 = 6 \text{ s}$.
• Time at constant speed: $t_2 = 15 \text{ s}$.
• Time during deceleration: $t_3 = \frac{v - u}{a} = \frac{0 - 30}{-10} = 3 \text{ s}$.

Thus, total time = $t_1 + t_2 + t_3 = 6 + 15 + 3 = 24 \text{ s}.$

Average speed: $\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{621}{24} \approx 25.875 \text{ ms}^{-1}.$