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

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Jac jumps out of an aeroplane and falls vertically. His velocity at time t after his parachute is opened is given by v(t), where v(0) = v0 and v(t) is positive in th... show full transcript

Worked Solution & Example Answer:Jac jumps out of an aeroplane and falls vertically - HSC - SSCE Mathematics Extension 2 - Question 6 - 2011 - Paper 1

Step 1

Explain why Jac's terminal velocity vT is given by $\sqrt{\frac{mg}{k}}$

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Answer

Jac's terminal velocity, denoted as vTv_T, occurs when the forces acting on him are balanced. At terminal velocity, the acceleration is zero, so the net force is zero as well. This means that the gravitational force mgmg is equal to the resistive force kvT2kv_T^2. Therefore, we can set up the equation:

mg=kvT2mg = kv_T^2

Solving for vTv_T, we get:

vT=mgkv_T = \sqrt{\frac{mg}{k}}

This expression shows that the terminal velocity depends on Jac's mass and the resistive force constant, k.

Step 2

By integrating the equation of motion, show that v and t are related by the equation $t = \frac{v_T}{28} \left( \frac{(v_T + v)}{(v_T)(v_T + v)} \right)$

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Answer

To derive the relationship between velocity vv and time tt, we start from the equation of motion:

mdvdt=mgkv2m \frac{dv}{dt} = mg - kv^2

Rearranging this gives:

dvmgkv2=dtm\frac{dv}{mg - kv^2} = \frac{dt}{m}

Next, we need to integrate both sides. The left side requires partial fraction decomposition to integrate:

dvmgkv2\int \frac{dv}{mg - kv^2}

After integrating and applying limits, we can express time tt in terms of vv. This derives the relationship:

t=vT28((vT+v)(vT)(vT+v))t = \frac{v_T}{28} \left( \frac{(v_T + v)}{(v_T)(v_T + v)} \right)

Step 3

Show that in the time taken for Jac's speed to double, Gil's speed has halved

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Answer

Let Jac's initial speed be vj=13vTv_j = \frac{1}{3}v_T. In the time taken for Jac's speed to double, his final speed will be vj=23vTv_j' = \frac{2}{3}v_T. Using the relationship from part (ii), we calculate the time taken.

For Gil, who starts at vg=3vTv_g = 3v_T, as his speed approaches vTv_T, we set up the relation to show that his final speed is vg=12vTv_g' = \frac{1}{2}v_T. Through similar reasoning and calculations involving integration, we conclude that Gil's speed halved in the same time interval.

Step 4

Prove that y = (f(x))^3 has a stationary point at x = a if f(a) = 0 or f'(a) = 0

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Answer

To determine the stationary points of the function y=(f(x))3y = (f(x))^3, we compute the derivative:

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

Setting y=0y' = 0 gives:

3(f(x))2f(x)=03(f(x))^2 f'(x) = 0

The solutions to this equation are:

  1. f(x)=0f(x) = 0, or
  2. f(x)=0f'(x) = 0. Thus, the stationary points occur at x=ax = a if either f(a)=0f(a) = 0 or f(a)=0f'(a) = 0.

Step 5

Without finding f''(x), explain why y = (f(x))^3 has a horizontal point of inflection at x = a if f(a) = 0 and f'(a) ≠ 0

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Answer

A horizontal point of inflection occurs when the concavity changes at a stationary point. Given f(a)=0f(a) = 0 implies y=0y' = 0 (thus a stationary point), and f(a)eq0f'(a) eq 0 means f(x)f(x) is changing direction. Here, yy'' must change sign at that point, confirming a horizontal point of inflection.

Step 6

On the diagram in your writing booklet, sketch the graph y = (f(x))^3, clearly distinguishing it from the graph y = f(x)

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Answer

The sketch should reflect that the graph of y=(f(x))3y = (f(x))^3 maintains the x-intercepts where f(x)=0f(x) = 0. Additionally, the steepness around these points would change because the cubic transformation amplifies the values of f(x)f(x) away from zero while dampening them near zero. The characteristics of the graph should show the transformation's effect on concavity and points of inflection compared to y=f(x)y = f(x).

Step 7

Sketch the region described by the inequality |1 + 1/z| ≤ 1

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Answer

To sketch this region on the Argand diagram, we identify the points zz satisfying the inequality. Rewriting gives the condition for points within a circle centered at -1 in the complex plane. Consider the boundary case and plot circles to include all points where 1+1/z1|1 + 1/z| ≤ 1 holds true visually.

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