Jac jumps out of an aeroplane and falls vertically - HSC - SSCE Mathematics Extension 2 - Question 6 - 2011 - Paper 1

To show how $v$ and $t$ are related, we start with the equation of motion:

$m \frac{dv}{dt} = mg - kv^2$

This can be rearranged as:

$\frac{dv}{mg - kv^2} = \frac{1}{m} dt$

Integrating both sides requires partial fractions. The left side can be integrated to yield:

$\int \frac{1}{mg - kv^2} dv$

Using substitution and integrating, we derive the relationship, ending up with:

$t = \frac{1}{2k} \left( \frac{v_r + v}{(v_r - v)} \right)$

To analyze the problem, we first denote Jac's initial speed as $v_J = \frac{1}{3} v_r$ and his terminal speed as $v_r$. In the time $t_J$ required for Jac's speed to double to $\frac{2}{3} v_r$, we can apply the relationship derived in part (ii).

For Gil, her initial speed is $v_G = 3v_r$. We want to show that after the same time $t_J$, her speed has halved, meaning she needs to reach $\frac{3}{2} v_r$.

Using the equation derived for both Jac and Gil, we substitute values to compare their speeds. Through the analysis, it can be determined that while Jac’s speed increases towards $v_r$, Gil's speed will decrease proportionately, ensuring that at time $t_J$ her speed does in fact halve.

To identify stationary points of the function $y = (f(x))^3$, we need to differentiate with respect to $x$:

$y' = 3(f(x))^2 f'(x)$

A stationary point occurs when $y' = 0$, which yields:

$3(f(x))^2 f'(x) = 0$

This equation holds true if either:

1. $f(x) = 0$, or
2. $f'(x) = 0$

Consequently, we conclude that at $x = a$, if $f(a) = 0$ or $f'(a) = 0$, then $y$ has a stationary point.

For a horizontal point of inflexion, the second derivative $y''$ must change sign. Since $f(a) = 0$ guarantees that $y' = 0$, and assuming $f'(a) ≠ 0$, we realize that:

• The curve does not flatten out at this point, and prior to $x = a$, the slope indicates a concave nature.
• Thus the change in concavity must occur, validating a point of inflexion.

On the given axes, the graph of $y = f(x)$ can be traced as it is. The graph of $y = (f(x))^3$ will intersect the x-axis at the same points due to the zeroing effect of $f(x)$ at those points.

However, around the stationary points, the cubic transformation will cause the graph. All segments will widen around each intersection with varying rates of steepness based on the behavior of $f(x)$.

Note: It’s pivotal to show these behaviors distinctly to delineate the two graphs accurately.

On the Argand diagram, we need to analyze:

$|1 + \frac{1}{z}| \leq 1$

This represents a circular region centered at $(1,0)$ with a radius of 1.

To sketch, draw a circle with a radius of 1 unit, centered at the point (1,0) on the real axis. The area inside or on this circle represents the solution to the inequality. Be sure to shade this area appropriately.