At time $t$ hours after a high tide, the height, $h$ metres, of the tide and the velocity, $v$ knots, of the tidal flow can be modelled using the parametric equations $v = 4 - igg(rac{2t}{3} - 2igg)^{2}$ $h = 3 - 2igg(rac{t}{3}igg)^{rac{1}{2}}$ High tides and low tides occur alternately when the velocity of the tidal flow is zero - AQA - A-Level Maths Pure - Question 15 - 2019 - Paper 1

First, we find the time when the velocity $v = 0$. Using the velocity equation:

$0 = 4 - \left(\frac{2t}{3} - 2\right)^{2}$

This simplifies to:

$\left(\frac{2t}{3} - 2\right)^{2} = 4$

Taking the square root:

$\frac{2t}{3} - 2 = \pm 2$

From this, we get two cases:

1. $$$$

ightarrow \frac{2t}{3} = 4 ightarrow t = 6$$

2. $$$$

ightarrow \frac{2t}{3} = 0 ightarrow t = 0$$

Since $t=0$ corresponds to the high tide at 2 am, the next low tide occurs at 6 am.

To find the height at the first low tide, use the height equation with $t = 6$:

$h = 3 - 2\left(\frac{6}{3}\right)^{\frac{1}{2}}$

This simplifies to:

$h = 3 - 2(1) = 1$

Thus, the height of the first low tide is 1 metre.