1. (a) A car passes four collinear markers A, B, C, and D while moving in a straight line with uniform acceleration - Leaving Cert Applied Maths - Question 1 - 2017

To find the value of k, we start by calculating the distances between the markers:

• The distance |AB| is given by the formula:

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

• The distance |AC| can be expressed as:

  $|AC| = u(2t) + \frac{1}{2}a(2t)^2 = 2ut + 2at^2$

• Similarly, the distance |AD| is:

  $|AD| = u(3t) + \frac{1}{2}a(3t)^2 = 3ut + \frac{9}{2}at^2$

• The distance |BC| can be found as:

  $|BC| = u(t) + \frac{1}{2}a(t)^2 = ut + \frac{1}{2}at^2$

• The distance |CD| is:

  $|CD| = u(t) + \frac{1}{2}a(t)^2 = ut + \frac{1}{2}at^2$

Now, substituting into the equation |AB| + |CD| = k|BC|:

$|AB| + |CD| = (2ut + 3at^2) = k(2ut + 2at^2)$

From here, we simplify to find that k = 2.