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

To show that the swimmer is on the surface of the water, set the function equal to zero:

$h(12) = -\frac{1}{60}(12)^{3} + \frac{1}{4}(12) + \frac{3}{5}$

Calculating this, we have:

$h(12) = -\frac{1}{60}(1728) + 3 + 0.6 = 0$

Thus, the swimmer is on the surface at ( x = 12 ) m.

To find where the swimmer enters the water, set $h(x) = 0$:

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

Solving this equation, we can use numerical methods or graphing to find that the horizontal distance when she first enters is approximately ( x = 3 ) metres.

To find the derivative, apply the power rule:

$h'(x) = -\frac{1}{20}x^{2} + \frac{1}{4}$

Thus, the derivative is ( h'(x) = -\frac{1}{20}x^{2} + \frac{1}{4} ).

To find the horizontal distance at which the swimmer reaches the greatest depth, set the derivative to zero:

$-\frac{1}{20}x^{2} + \frac{1}{4} = 0$

Solving for $x$, we get:

$x = 4 \text{ m}$

To find the greatest depth, substitute $x = 4$ back into the original function:

$h(4) = -\frac{1}{60}(4)^{3} + \frac{1}{4}(4) + \frac{3}{5} = -\frac{64}{60} + 1 + \frac{3}{5} = \frac{11}{5} \text{ m}$

The percentage increases are calculated using the formula:

$\text{Percentage Increase} = \frac{(New - Old)}{Old} \times 100$

So for the sections:

• From 0-50m to 50-100m: (\frac{28.5 - 24.85}{24.85} \times 100 = 11.34%)
• From 50-100m to 100-150m: (\frac{29.33 - 28.5}{28.5} \times 100 \approx 2.91%)
• From 100-150m to 150-200m: (\frac{30.68 - 29.33}{29.33} \times 100 \approx 4.6%)

Firstly calculate the total time for the second swimmer:

• For the first 50m: 25.01
• For the next 50m (14.7% increase): (28.5 = 25.01 \times 1.147)
• Further to 150m: (29.51 = 28.5 \times 1.0291)
• Last 50m: (30.87 = 29.51 \times 1.046)

Total time for second swimmer is approximately 1:54.07. The difference with Michael Phelps (1:53.36) is: (1:54.07 - 1:53.36 = 0.71\text{ seconds}).