The points P and Q lie on a straight level road - Leaving Cert Applied Maths - Question 1 - 2017

To find the deceleration, we use: v^2 = u^2 + 2as where: v = final velocity = 0, u = initial velocity = 24 m s⁻¹, a = deceleration, s = distance during deceleration = 72 m.

Setting up the equation:

$$0 = (24)^2 + 2a(72)$$

Solving for a:

a = -rac{(24)^2}{2(72)} = -4 ext{ m s}^{-2}

To find the total distance |PQ|, we need to consider all parts of the journey.

1. Distance during acceleration: (s_1 = ut + \frac{1}{2}at^2) (s_1 = (12)(4) + \frac{1}{2}(3)(4^2) = 48 + 24 = 72 \text{ m})

2. Distance at constant speed: (s_2 = v \cdot t = 24 \cdot 14 = 336 \text{ m})

Total distance:

$$|PQ| = s_1 + s_2 + s_3$$

where s3 (distance during deceleration) is 72 m:

$$|PQ| = 72 + 336 + 72 = 480 \text{ m}$$

The total time taken is given for the journey from P to Q.

Average Speed = Total Distance / Total Time

Total distance = 480 m Total time = 24 s (4s of acceleration + 14s of constant speed + 6s of deceleration)

Average Speed:

$$\text{Average Speed} = \frac{480}{24} = 20 \text{ m s}^{-1}$$

The speed-time graph for the van is as follows:

• The van starts from rest, accelerates uniformly to a maximum speed of k m s⁻¹.
• It takes the same time as the car (24 seconds), with equal acceleration and deceleration phases.
• The area under the graph will equal the total distance of 480 m.

Since the distance equals the area:

$$\frac{1}{2}(k)(24) = 480 \ ightarrow k = 40 ext{ m s}^{-1}$$