The points P and Q lie on a straight level road - Leaving Cert Applied Maths - Question 1 - 2016

To calculate deceleration, we use a similar formula:

$v = u + at$

Where:

• Initial velocity, $u = 30 \, \text{m/s}$
• Final velocity, $v = 22 \, \text{m/s}$
• Distance, $s = 52 \, \text{m}$

From the equation of motion:

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

Substituting the values gives:

$22^2 = 30^2 + 2a(52)$

Solving for $a$ results in:

$a = -4 \, \text{m/s}^2\text{ (deceleration)}$

The total distance from P to Q can be calculated in parts:

1. Distance during acceleration: $s_1 = ut + \frac{1}{2}at^2 = (14)(8) + \frac{1}{2}(2)(8^2) = 112 + 64 = 176 \, \text{m}$
2. Distance during deceleration is already given as 52 m.
3. Distance during constant speed: $s_3 = vt = (22)(10) = 220 \, \text{m}$

Thus, the total distance is:

$|PQ| = s_1 + s_2 + s_3 = 176 + 52 + 220 = 448 \, \text{m}$

The average speed can be computed using the formula:

$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$

Total distance = 448 m.

Total time = 20 s (8 s acceleration + 10 s constant speed + 2 s deceleration time, which we can find using: $t = \frac{v - u}{a} = \frac{22 - 30}{-4} = 2.0 \, \text{s}$)

Thus, average speed becomes:

$\text{Average Speed} = \frac{448}{20} = 22.4 \, \text{m/s}$

The average speed is 22.4 m/s. The car moves at its average speed for:

1. Constant speed phase: 10 seconds at 22 m/s (not at average speed), contributes 0 seconds.
2. Constant speed of 22 m/s: During this time, the car is below average speed.
3. Acceleration phase: The car reaches a maximum speed of 30 m/s which is above average speed for 8 seconds.

Since the car does not exceed average speed during deceleration or the later constant speed phase, the total time moving at or above average speed is 8 s.