The points P and Q lie on a straight level road - Leaving Cert Applied Maths - Question 1 - 2014

Using the formula for acceleration:

$a = \frac{v - u}{t}$

Where:

• v = final velocity = 28 m s⁻¹
• u = initial velocity = 13 m s⁻¹
• t = time = 5 s

The calculation is:

$a = \frac{28 - 13}{5} = 3 \, \text{m s}^{-2}$

To find the deceleration when the car comes to rest from 28 m s⁻¹ over a distance of 98 m, we can use the formula:

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

Where:

• v = final velocity = 0 m s⁻¹ (at rest)
• u = initial velocity = 28 m s⁻¹
• a = acceleration (we're solving for deceleration, so it will be negative)
• s = distance = 98 m

Plugging in values, we arrange the equation:

$0 = (28)^2 + 2a(98)$

This simplifies to:

$0 = 784 + 196a$

Solving for a gives:

$a = -\frac{784}{196} = -4 \, \text{m s}^{-2}$

To find the total distance from P to Q, we calculate:

1. Distance during constant speed (from P to the end of the first segment): s_1 = ut = 13 imes 9 = 117 \, \text{m} \

2. Distance during acceleration (from 13 m s⁻¹ to 28 m s⁻¹):
   Using the formula :
   $s_2 = ut + \frac{1}{2}at^2$
   Here, u = 13 m s⁻¹, t = 5 s and a = 3 m s²:
   $s_2 = 13(5) + \frac{1}{2}(3)(5^2) = 65 + 37.5 = 102.5 \, \text{m}$

3. Distance during deceleration:
   Total distance = $|PQ| = s_1 + s_2 + 98$ m = $117 + 102.5 + 98 = 317.5 \, \text{m}$.

The average speed can be calculated by dividing total distance by total time:

1. Total distance calculated previously: 317.5 m
2. Total time:
   • Time during constant speed: 9 s
   • Time during acceleration: 5 s
   • Time during deceleration can be calculated using: $v = u + at$, with $v = 0, u = 28, a = -4$: $0 = 28 - 4t \Rightarrow t = 7 \text{ s}$
   • Therefore, total time = 9 + 5 + 7 = 21 seconds.

Thus, the average speed:

$\text{Average speed} = \frac{317.5 \text{ m}}{21 \text{ s}} = 15.12 \text{ m s}^{-1}$