A swimmer is on a starting block at the beginning of a race - Leaving Cert Mathematics - Question 8 - 2020

To show that $h(12) = 0$, we substitute $x = 12$ into the function:

$$\begin{align*} h(12) & = \frac{1}{60}(12)^3 - \frac{1}{4}(12)^2 + \frac{3}{5} \\ & = \frac{1}{60}(1728) - \frac{1}{4}(144) + \frac{3}{5} \\ & = 28.8 - 36 + 0.6 \\ & = 0\end{align*}$$

Thus, the swimmer is on the surface of the water when she is 12 metres from the starting block.

To find the horizontal distance where the swimmer enters the water, we solve:

$h(x) = 0$

Substituting into the equation:

$\frac{1}{60}x^3 - \frac{1}{4}x^2 + \frac{3}{5} = 0$

Rearranging, we can estimate using numerical or graphical methods to find $x = 3$ m, where the swimmer first enters.

Differentiating, we get:

$h'(x) = \frac{1}{20}x^2 - \frac{1}{2}x$

This represents the rate of change of the swimmer's height at any horizontal distance $x$.

Setting $h'(x) = 0$:

$\frac{1}{20}x^2 - \frac{1}{2}x = 0$

Factoring gives:

$x(x - 10) = 0$

Thus, $x = 0$ or x = 10 m. The swimmer reaches her greatest depth at $x = 10$ m.

Substituting $x = 10$ back into $h(x)$:

$h(10) = \frac{1}{60}(10^3) - \frac{1}{4}(10^2) + \frac{3}{5} = \frac{1000}{60} - 25 + \frac{3}{5}$

Calculating yields:

$= 16.67 - 25 + 0.6 = -7.73 \, m\text{ (greatest depth)}$

Calculating the percentage increase:

• From 0-50 m to 50-100 m: $\frac{28.5 - 24.85}{24.85} \times 100 = 2.9\%$

• From 50-100 m to 100-150 m: $\frac{29.33 - 28.5}{28.5} \times 100 = 2.9\%$

• From 100-150 m to 150-200 m: $\frac{30.68 - 29.33}{29.33} \times 100 = 4.6\%$

Calculating the other swimmer's total time:

• For the first 50 m: $25.01\text{ s}$
• And applying the percentage increases accordingly:

1. For 0-50 m, finishing at 25.01s
2. Adding relevant percentages, we see: $1:54:07 - 1:53:36 = 0.71\text{ seconds difference.}$