The points P and Q lie on a straight level road - Leaving Cert Applied Maths - Question 1 - 2013

To find the acceleration during the last phase (from 16 ms⁻¹ to 24 ms⁻¹), we again use:

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

Where:

• Final velocity, $v = 24 \, \text{ms}^{-1}$
• Initial velocity, $u = 16 \, \text{ms}^{-1}$
• Distance travelled while accelerating, $s = 40 \, \text{m}$

Plugging in:

$(24)^2 = (16)^2 + 2a(40)$

This leads us to:

$a = \frac{(24)^2 - (16)^2}{2(40)} = 4 \, \text{ms}^{-2}$

Thus, the acceleration is $4 \, \text{ms}^{-2}$.

To calculate |PQ|, we must sum the distances:

1. Distance during deceleration: $s_1 = ut + \frac{1}{2}at^2$ $s_1 = 28(6) + \frac{1}{2}(-2)(6^2) = 168 - 36 = 132 \, \text{m}$

2. Distance at constant speed: $s_2 = v imes t = 16(8) = 128 \, \text{m}$

3. Distance during acceleration: $s_3 = 40 \, \text{m}$

Thus, the total distance is:

$|PQ| = s_1 + s_2 + s_3 = 132 + 128 + 40 = 300 \, \text{m}$

12 seconds before passing Q includes the last 4 seconds of acceleration (6 seconds of deceleration + 8 seconds at constant speed gives 14 seconds total).

So, we use the speed after 2 seconds of acceleration:

$v = u + at = 16 + 4(2) = 24 \, \text{ms}^{-1}$

Thus, the speed of the car is $20 \, \text{ms}^{-1}$.

The average speed is calculated as total distance divided by total time:

Total time = 6s (deceleration) + 8s (constant speed) + (time during acceleration)

Using the formula for acceleration to find the time:

$t = \frac{v - u}{a} = \frac{24 - 16}{4} = 2 \, \text{s}$

Total time = 6 + 8 + 2 = 16 seconds.

Average speed = $\frac{300 \, \text{m}}{16 \, \text{s}} = 18.75 \, \text{ms}^{-1}$.