The points P and Q lie on a straight level road - Leaving Cert Applied Maths - Question 1 - 2019

To calculate the distance from P to Q, we consider each segment of the car's journey:

1. Acceleration phase:

   • Distance covered during acceleration can be calculated using: $s_1 = ut + \frac{1}{2}at^2$
   • Here, $u = 7$, $t = 3.5$, and $a = 4$: $s_1 = 7(3.5) + \frac{1}{2}(4)(3.5^2)$
   • This results in: $s_1 = 24.5 + 49 = 73.5 \, \text{m}$

2. Constant speed phase:

   • Distance for the constant speed of 21 m s⁻¹ for 9.5 seconds: $s_2 = 21 imes 9.5 = 199.5 \, \text{m}$

3. Deceleration phase:

   • The car decelerates for a distance of 98 m. Therefore, the total distance |PQ| is: $|PQ| = s_1 + s_2 + 98 \ = 73.5 + 199.5 + 98 = 371 \, \text{m}$

The average speed can be calculated using the formula:

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

Total Distance from P to Q:

• Total distance traveled is 346.5 m (as calculated from parts (i) and (ii)).

Total Time taken:

• Total time $= 3.5 + 9.5 + 2 = 15$ seconds.
• Thus:

$\text{Average Speed} = \frac{346.5}{15} = 17.325 \, \text{m s}^{-1}$

For the van:

1. Initial speed = 7 m s⁻¹, then it accelerates to maximum speed $k$.
2. It takes the same time to travel from P to Q, which is derived from the car’s journey:
   • Given that the time taken is 20 seconds total.
3. Using the equation for uniform acceleration: $s = ut + \frac{1}{2}at^2$
4. Solving for distance:
   • Distance for the van gives: $20k - 7(20) = 346.5$ $20k = 346.5 + 140 \rightarrow k = 27.65 \, \text{m s}^{-1}$