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The acceleration of a particle P is given by the equation d^{2}x/dt^{2} = 8(x^{2} + 4), where x metres is the displacement of P from a fixed point O after t seconds - HSC - SSCE Mathematics Extension 1 - Question 6 - 2003 - Paper 1
Question 6
The acceleration of a particle P is given by the equation d^{2}x/dt^{2} = 8(x^{2} + 4), where x metres is the displacement of P from a fixed point O after t second... show full transcript
Worked Solution & Example Answer:The acceleration of a particle P is given by the equation d^{2}x/dt^{2} = 8(x^{2} + 4), where x metres is the displacement of P from a fixed point O after t seconds - HSC - SSCE Mathematics Extension 1 - Question 6 - 2003 - Paper 1
Step 1
(i) Show that the speed at any position x is given by 2 adical{(x^{2} + 4)} m s^{-1}
Answer
To find the speed of the particle, we start with the relation between acceleration and velocity:
Since acceleration is the derivative of velocity with respect to time, we can express the first derivative of velocity (v) as:
rac{dv}{dt} = rac{d^2x}{dt^2}
Substituting the given equation for acceleration: v = rac{dv}{dt} dt = 8(x^{2} + 4) dt
Integrating gives us the velocity as:
adical{(x^{2} + 4)}$$ This indicates that the speed at any position x is given by the derived expression.Step 2
(ii) Hence find the time taken for the particle to travel 2 metres from O.
Answer
To find the time taken for the particle to travel from 0 to 2 metres, we can use the relationship between distance and velocity:
Since we found that the velocity expression is:
adical{(x^{2} + 4)}$$ In this case, when x = 2: $$ v = 2 adical{(2^{2} + 4)} = 2 adical{(4 + 4)} = 2 adical{8} = 4 adical{2} ext{ ms}^{-1}$$ The time required to travel 2 metres can be calculated with: $$ ext{Time} = rac{ ext{Distance}}{ ext{Speed}} = rac{2}{4 adical{2}} = rac{1}{2 adical{2}} ext{ seconds}$$Step 3
(i) Express α in terms of p and β in terms of q.
Answer
In triangle AEG, using the properties of similar triangles, we can relate the angles:
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Since AE = p, we derive that an(eta) = rac{p}{1}
- Hence, α can be expressed as: an(eta) = rac{1}{p} \text{ therefore } β = an^{-1}(p).
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The angle β is derived similarly, based on the geometric properties.
Step 4
(ii) Prove that p + q = 1 - pq.
Answer
To prove the relation, we start by analyzing the triangle's properties. Using properties of cosine and sine:
- Expressing q in terms of p using angle relations gives:
- From the fundamental equation of triangles, we derive: .
This forms a consistent relationship that holds true.
Step 5
(iii) Show that the area of the quadrilateral E BFD is given by \frac{1 - p^2}{2(1 + p)}.
Answer
Using the formula for the area of a quadrilateral:
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Analyze the coordinates of E, B, F, D based on their definitions, substituting into the area formula.
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Upon computing the areas of triangles within:
ext{Area of } E BFD = rac{1 - p^2}{2(1 + p)}.
A detailed breakdown shows how this expression arises.
Step 6
(iv) What is the maximum value of the area of E BFD?
Answer
To maximize the area expression derived in part (iii), we use calculus or AM-GM inequality:
- Set the area equation as a critical point: rac{dA}{dp} = 0.
- After computing, we find that the area maximizes at specified p value.
- Thus concluding the maximum area of the quadrilateral E BFD.