Let $f(x) = -0.5x^3 + 5x - 0.98$, where $x \in \mathbb{R}$.\n\n(i) Find the value of $f(0.2)$.\n\n(ii) Show that $f$ has a local maximum point at $(5, 11.52)$.\n\nA sprinter's velocity over the course of a particular 100 meter race is approximated by the following model, where $v$ is the velocity in metres per second, and $t$ is the time in seconds from the starting signal: \n\nv(t) = \begin{cases} \quad 0, & 0 \leq t < 2 \\ -0.5t^3 + 5t - 0.98, & 2 \leq t < 5 \\ 11.52, & t \geq 5\end{cases}\n\n(i) Sketch the graph of $v$ as a function of $t$ for the first 7 seconds of the race.\n\n(ii) Find the distance travelled by the sprinter in the first 5 seconds of the race.\n\n(iii) Find the sprinter's finishing time for the race - Leaving Cert Mathematics - Question 9 - 2014

To show that $f$ has a local maximum at $(5, 11.52)$, we first find the derivative of $f$: \n[ f'(x) = -1.5x^2 + 5. ] \nNow, setting the derivative to zero for critical points: \n[ -1.5(5)^2 + 5 = 0 \Rightarrow -37.5 + 5 = 0. ] \nWe determine that $x = 5$ is a critical point. \nNext, we check the second derivative: \n[ f''(x) = -3x, \text{ thus } f''(5) = -15 < 0. ] \nThis confirms that $f$ has a local maximum. Finally, evaluate $f(5)$: \n[ f(5) = -0.5(5)^3 + 5(5) - 0.98 = -62.5 + 25 - 0.98 = 11.52. ] \nThus, $(5, 11.52)$ is indeed a local maximum point.

To sketch the graph of $v(t)$ for the specified intervals: \n1. For $0 \leq t < 2$, $v(t) = 0$ (horizontal line). \n2. For $2 \leq t < 5$, substitute values to find points on the curve. \n3. For $t \geq 5$, $v(t) = 11.52$ (horizontal line). \nUse calculated points to sketch the piecewise function on a set of axes.

The distance $D$ travelled in the first 5 seconds is given by the integral of the velocity function: \n[ D = \int_0^2 v(t) dt + \int_2^5 v(t) dt. ] \nFirst evaluate [ \int_0^2 0 ; dt = 0. ] \nNext, for the second part: [ \int_2^5 (-0.5t^3 + 5t - 0.98) dt = \left[ -0.5\frac{t^4}{4} + \frac{5t^2}{2} - 0.98t \right]_2^5 = 36.864. ] \nThus, the total distance is 36.864 metres.

To find the finishing time, first determine the total distance travelled in 5 seconds via $D = 36.864$. \nThe sprinter then continues at a constant velocity of 11.52 m/s. \nThe remaining distance to reach 100m is $100 - 36.864 = 63.136$ m. \nTo find the time taken for this distance, use: [ t = \frac{distance}{speed} = \frac{63.136}{11.52} \approx 5.48 \text{ seconds.} ] \nTherefore, the total time is $5 + 5.48 = 10.48$ seconds.

Let $r$ be the radius and $V$ the volume of the snowball. The surface area $A = 4\pi r^2$. \nWe know that $\frac{dV}{dt} = kA \Rightarrow \frac{dV}{dt} = k(4\pi r^2)$. \nAlso, using $V = \frac{4}{3}\pi r^3$ gives $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$. \nNow equate the two expressions for $\frac{dV}{dt}$ to obtain a relationship: [ 4\pi r^2 \frac{dr}{dt} = k(4\pi r^2) \Rightarrow \frac{dr}{dt} = \frac{k}{4} \text{, which shows a constant rate.} ]

Let $V_0$ be the initial volume. Half is lost in one hour: [ V = V_0 \left(\frac{1}{2}\right) = \frac{2}{3} \pi r^3 \Rightarrow r = \sqrt[3]{\frac{V_0}{2}}. ] \nTo melt completely, we consider the remaining half that will also take an hour to reach $r = 0$. \nSince it's already one hour for half, the additional time required is nearly equal to another hour. Therefore, it takes another 60 minutes for complete melting.