The points P and Q lie on a straight level road - Leaving Cert Applied Maths - Question 1 - 2015

To find the acceleration from 8 m s⁻¹ to 26 m s⁻¹, we again use the formula:

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

Where:

• Final speed, $v = 26$ m s⁻¹
• Initial speed, $u = 8$ m s⁻¹
• Distance, $s = 102$ m

Rearranging gives:

$26^2 = 8^2 + 2a(102)$

Solving for $a$:

$676 - 64 = 204a$

$a = \frac{612}{204} = 3 \, \text{m s}^{-2}$

Thus, the acceleration is $3 \, \text{m s}^{-2}$.

To find the total distance |PQ|, we sum the distance traveled during deceleration, acceleration, and the constant speed segment:

1. Distance during deceleration (from P to the end of deceleration):

   • Using the formula:

   $s = ut + \frac{1}{2}at^2$

   $s = 24(4) + \frac{1}{2}(-4)(4^2) = 96 - 32 = 64 \, \text{m}$

2. Distance during acceleration:

   • Already given as 102 m.

3. Distance during constant speed:

   • Speed = 26 m s⁻¹ for 10 seconds, hence:

   $s = 26(10) = 260 \, \text{m}$

Therefore, the total distance is:

$|PQ| = 64 + 102 + 260 = 426 \, ext{m}$

The average speed is calculated by dividing the total distance by the total time taken:

• Total distance = 426 m (from part (iii))
• Total time = 4 s (deceleration) + 6 s (acceleration) + 10 s (constant speed) = 20 s

Therefore, average speed:

$v_{avg} = \frac{426}{20} = 21.3 \, \text{m s}^{-1}$

The legal speed limit is 100 km h⁻¹. To convert this to m s⁻¹:

$100 \, \text{km h}^{-1} = \frac{100 \times 1000}{3600} = 27.7 \, \text{m s}^{-1}$

The maximum speed the car reaches is 26 m s⁻¹ during its motion from P to Q, which does not exceed the speed limit of 27.7 m s⁻¹. Therefore, the car does not exceed the speed limit.