The points P and Q lie 1 km apart on a straight level road - Leaving Cert Applied Maths - Question 1 - 2021

The distance travelled can be calculated using the formula:

$s = ut + \frac{1}{2}at²$

Substituting the values:

$s = 2 \times 8 + \frac{1}{2} \times 2.5 \times (8)²$

Calculating:

$s = 16 + 0.5 \times 2.5 \times 64 = 16 + 80 = 96 \, m$

During the next 18 seconds, the car travels at a constant speed of 22 m/s. Thus, the distance can be calculated as:

$s = vt$

where

• $v = 22 \, m/s$,
• $t = 18 \, s$.

So,

$s = 22 \times 18 = 396 \, m$

The total distance from P to Q is 1 km = 1000 m. The distance travelled by the car is the sum of the distances:

$s_{total} = 96 + 396 + (x)$

For the deceleration part, we can set: $x + 96 + 396 = 1000$

This gives: $x = 1000 - 492 = 508 \, m$

Since the car decelerates uniformly, the time taken for this section can be derived from other known deceleration values, leading to a total time: $t_{total} = 8 + 18 + t_{deceleration} = 46.2 \, s$

In the speed-time graph:

• The initial speed is $k \, m/s$ for 10 seconds.
• Then it accelerates uniformly to reach a speed of 17 m/s in 2 seconds.
• Finally, it maintains the speed of 17 m/s until it reaches point Q.

The graph should reflect these segments accordingly.

For the motorbike, the total time is 60 seconds:

[ 10 + 2 + t_{constant} = 60 ]

We know it travels at $17 \times (60 - t_{constant})$ to cover the distance:

From the car's parametrization, we can derive:

[ 12 + \frac{1}{2} \times 2 \times 10 + 48 + 17t_{constant} = 1000 ]

Following through will give $k = 15.2 \, m/s$.