The points P and Q lie on a straight level road - Leaving Cert Applied Maths - Question 1 - 2018

The total distance traveled, |PQ|, is the sum of three segments:

1. The distance during acceleration,

2. The distance at constant speed,

3. The distance during deceleration.

4. Distance during acceleration:

$d_1 = ut + \frac{1}{2} at^2 = 5 \times 7 + \frac{1}{2} \times 3 \times 7^2 = 35 + 73.5 = 108.5 ext{ m}.$

2. Distance at constant speed:

$d_2 = 234 ext{ m}.$

3. Distance during deceleration:

$d_3 = 52 ext{ m}.$

Therefore, the total distance is:

$|PQ| = d_1 + d_2 + d_3 = 108.5 + 234 + 52 = 394.5 ext{ m}.$

To find the average speed, we can use the formula:

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

The total distance is 394.5 m.

To find the total time, we calculate the time spent during each phase:

1. Time during acceleration: $t_a = 7$ s.
2. Time at constant speed:

$t_c = \frac{d_2}{v} = \frac{234}{26} = 9 \text{ s}.$ 3. Time during deceleration: $t_d = 4$ s (calculated earlier).

Thus, the total time:

$\text{Total Time} = 7 + 9 + 4 = 20 \text{ s}.$

Now substituting into the average speed formula gives:

$\text{Average Speed} = \frac{394.5}{20} = 19.725 \text{ m/s}.$

The speed-time graph for the motor-cyclist consists of:

1. An initial increase in speed from 0 to 30 m/s as she accelerates uniformly. (This will be a straight line segment from (0, 0) to (k, 30).)

2. A constant speed at 30 m/s until she begins to decelerate.

3. Then, a decrease from 30 m/s back to 0 m/s as she comes to a stop. The exact point where she starts to decelerate can be calculated by the distance covered.

You can plot the graph with the following points:

• Start at (0, 0)
• Reach (k, 30)
• Maintain (k + x, 30) where x is the time spent traveling at 30 m/s
• Finally, decrease back to (total_time, 0).

Using the equation of motion for constant acceleration:

$v = u + at,$ We have:

• Final velocity, $v = 30$ m/s,
• Initial velocity, $u = 0$ m/s,
• Acceleration, $a = \frac{30}{k}$ (uniform).

Then,

$30 = 0 + \frac{30}{k} \cdot k;$ This implies:

Using the formula for distance:

$s = \frac{1}{2} a t^2$ We set the total distance to 394.5 m. After calculating, we find that k = 26.3 seconds.