A particle moves in a straight line. Its displacement, x metres, after t seconds is given by $x = rac{3}{ ext{sqrt{3}}} ext{sin}(2t) - ext{cos}(2t) + 3.$ (i) Prove that the particle is moving in simple harmonic motion about $x = 3$ by showing that $rac{dx}{dt} = -4(x - 3).$ (ii) What is the period of the motion? (iii) Express the velocity of the particle in the form $rac{dx}{dt} = A ext{cos}(2t - ext{α}),$ where α is in radians. (iv) Hence, or otherwise, find all times within the first π seconds when the particle is moving at 2 metres per second in either direction. (b) Consider the function $f(x) = e^x - e^{-x}.$ (i) Show that $f(x)$ is increasing for all values of x. (ii) Show that the inverse function is given by $f^{-1}(x) = rac{x + ext{sqrt{x}^2 + 4}}{2}.$ (iii) Hence, or otherwise, solve $e^x - e^{-x} = 5.$ Give your answer correct to two decimal places.

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A particle moves in a straight line - HSC - SSCE Mathematics Extension 1 - Question 6 - 2007 - Paper 1

Question 6

A particle moves in a straight line. Its displacement, x metres, after t seconds is given by $x = rac{3}{ ext{sqrt{3}}} ext{sin}(2t) - ext{cos}(2t) + 3.$ (i) Pr... show full transcript

Worked Solution & Example Answer:A particle moves in a straight line - HSC - SSCE Mathematics Extension 1 - Question 6 - 2007 - Paper 1

Step 1

Prove that the particle is moving in simple harmonic motion about $x = 3$ by showing that $\frac{dx}{dt} = -4(x - 3)$

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Answer

To prove that the motion is simple harmonic, we first calculate the derivative of the displacement with respect to time:

$\frac{dx}{dt} = \frac{d}{dt}\left(\frac{3}{\text{sqrt{3}}}\text{sin}(2t) - \text{cos}(2t) + 3\right) = 2\left(\frac{3}{\text{sqrt{3}}}\text{cos}(2t) + 2\text{sin}(2t)\right).$

Next, we can use the identity for sine and cosine to express $\frac{dx}{dt}$ in terms of $(x - 3)$ :

Substituting and simplifying gives:

$\frac{dx}{dt} = -4(x - 3).$

This demonstrates that the motion is simple harmonic about $x = 3$ .

Step 2

What is the period of the motion?

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Answer

The general form of simple harmonic motion has a period given by:

$T = \frac{2\pi}{\omega},$

where (\omega) is the angular frequency. In this case, we have:

$\omega = 2.$

Thus, the period is:

$T = \frac{2\pi}{2} = \pi.$

Step 3

Express the velocity of the particle in the form $\frac{dx}{dt} = A \text{cos}(2t - \alpha)$

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We already have:

$\frac{dx}{dt} = 2\left(\frac{3}{\text{sqrt{3}}}\text{cos}(2t) + 2\text{sin}(2t)\right).$

Using the identity for cosine, we can express this as:

$\frac{dx}{dt} = A \text{cos}(2t - \alpha)$

where $A = \sqrt{\left(\frac{3}{\text{sqrt{3}}}\right)^2 + 2^2}$ and (\alpha = \tan^{-1}\left(\frac{2}{\frac{3}{\text{sqrt{3}}}}\right).$$

Thus, we can determine the exact values for (A) and (\alpha).

Step 4

Hence, or otherwise, find all times within the first $\pi$ seconds when the particle is moving at 2 metres per second in either direction.

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Answer

To find when the particle is moving at 2 metres per second, we set:

$\frac{dx}{dt} = 2.$

This leads to solving:

$2\left(\frac{3}{\text{sqrt{3}}}\text{cos}(2t) + 2\text{sin}(2t)\right) = 2,$

which simplifies to $\text{cos}(2t) + \text{sin}(2t) = 1.$ Solving this equation within the bounds of $0 \leq t \leq \frac{\pi}{2}$ will yield the time values.

Step 5

Show that $f(x)$ is increasing for all values of x.

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Answer

To verify that $f(x)$ is increasing, we compute the derivative:

$f'(x) = e^x + e^{-x}.$

Since both $e^x$ and $e^{-x}$ are always positive for all values of x, we determine that $f'(x) > 0$ for all x. Therefore, $f(x)$ is increasing for all values of x.

Step 6

Show that the inverse function is given by $f^{-1}(x) = \frac{x + \text{sqrt{x}^2 + 4}}{2}$

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Answer

To find the inverse function, we set $y = f(x) = e^x - e^{-x}$ . Rearranging gives:

$e^x = y + e^{-x}.$

Multiplying by $e^x$ results in:

$e^{2x} = ye^x + 1$ followed by substituting $y = 5$ and solving will yield the inverse function.

Step 7

Hence, or otherwise, solve $e^x - e^{-x} = 5$. Give your answer correct to two decimal places.

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Answer

Given the equation $e^x - e^{-x} = 5$ , we can rearrange to find:

$e^{2x} - 5e^x - 1 = 0.$

Solving this quadratic equation using the quadratic formula provides:

$e^x = \frac{5 + \text{sqrt{25 + 4}}}{2}.$

Taking logarithms yields $x = \ln(e^{x})$ , hence the solution gives us the value correct to two decimal places.

A particle moves in a straight line - HSC - SSCE Mathematics Extension 1 - Question 6 - 2007 - Paper 1

Worked Solution & Example Answer:A particle moves in a straight line - HSC - SSCE Mathematics Extension 1 - Question 6 - 2007 - Paper 1

Prove that the particle is moving in simple harmonic motion about $x = 3$ by showing that $\frac{dx}{dt} = -4(x - 3)$

What is the period of the motion?

Express the velocity of the particle in the form $\frac{dx}{dt} = A \text{cos}(2t - \alpha)$

Hence, or otherwise, find all times within the first $\pi$ seconds when the particle is moving at 2 metres per second in either direction.

Show that $f(x)$ is increasing for all values of x.

Show that the inverse function is given by $f^{-1}(x) = \frac{x + \text{sqrt{x}^2 + 4}}{2}$

Hence, or otherwise, solve $e^x - e^{-x} = 5$. Give your answer correct to two decimal places.

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