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Solve \( \frac{3x}{x-2} \leq 1 \) - HSC - SSCE Mathematics Extension 1 - Question 4 - 2001 - Paper 1

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Solve \( \frac{3x}{x-2} \leq 1 \). (b) An aircraft flying horizontally at \( V \) m s^{-1} releases a bomb that hits the ground 4000 m away, measured horizontally. ... show full transcript

Worked Solution & Example Answer:Solve \( \frac{3x}{x-2} \leq 1 \) - HSC - SSCE Mathematics Extension 1 - Question 4 - 2001 - Paper 1

Step 1

Solve \( \frac{3x}{x-2} \leq 1 \)

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Answer

To solve the inequality ( \frac{3x}{x-2} \leq 1 ), we begin by rearranging the inequality:

  1. Rearrange the Inequality: [ \frac{3x}{x-2} - 1 \leq 0 ] [ \frac{3x - (x - 2)}{x - 2} \leq 0 ] [ \frac{3x - x + 2}{x - 2} \leq 0 ] [ \frac{2x + 2}{x - 2} \leq 0 ]

  2. Factor: [ \frac{2(x + 1)}{x - 2} \leq 0 ]

  3. Identify Key Points: The critical points from the inequality are when ( x + 1 = 0 ) or ( x - 2 = 0 ). Hence, ( x = -1 ) and ( x = 2 ).

  4. Test Intervals: The intervals to test are ( (-\infty, -1) ), ( (-1, 2) ), and ( (2, \infty) ).

    • For ( x < -1 ): Choose ( x = -2 ) yields ( \frac{2(-1)}{-4} > 0 ).
    • For ( -1 < x < 2 ): Choose ( x = 0 ) yields ( \frac{2(1)}{-2} < 0 ).
    • For ( x > 2 ): Choose ( x = 3 ) yields ( \frac{2(4)}{1} > 0 ).
  5. Conclusion: The solution is in the interval where the expression is non-positive. Since the inequality is ( \leq 0 ), we include ( x = -1 ): [ x \in [-1, 2) ]

Step 2

Find the speed \( V \) of the aircraft.

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Answer

To find the speed ( V ) of the aircraft, we use the projectile motion equations provided:

  1. Determine the Position of the Bomb at Impact: The horizontal distance traveled by the bomb is given as 4000 m, hence: [ x = Vt = 4000 ] (1)

  2. Find the Vertical Position at Impact: The bomb hits the ground, which implies: [ y = -5t^2 ] Setting ( y = 0 ):\n [ 0 = -5t^2 \Rightarrow t = \sqrt{0} = 0 ] The vertical position when it hits the ground is when ( t ) equals the time it takes to fall the vertical distance, which can be calculated using the angle of impact: [ \tan(45°) = 1 = \frac{y}{x} \Rightarrow y = x ] (2) Substituting from (1) into (2) yields the additional relationship needed to solve for ( t ):
    [ t = \frac{4000}{V} ] Thus, substituting back, we find: [ 0 = -5(\frac{4000}{V})^2 ] This equation reveals the quadratic relationship necessary to express the speed ( V ).

  3. Comparing Both Equations: By equating both derived expressions we can find ( V ): Therefore we can simply calculate ( V = \frac{4000}{t} ), with the angle giving [ V^2 = 4000 \cdot 5 \Rightarrow V = \sqrt{20000} = \approx 141.42 , ms^{-1} ]

  4. Conclusion: The speed of the aircraft is approximately ( V = 141.42 , m/s ).

Step 3

Find \( x \) as a function of \( t \).

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Answer

To find ( x ) as a function of ( t ):

  1. Differential Equation: The given differential equation is ( \frac{dx}{dt} = -4x ).

  2. Separation of Variables: Rearranging gives: [ \frac{dx}{x} = -4 dt ]

  3. Integrate: Integrating both sides: [ \int \frac{dx}{x} = \int -4 dt ] [ \ln|x| = -4t + C ] (where C is the constant of integration)

  4. Solve for x: Exponentiating gives: [ x = e^{-4t + C} = e^C e^{-4t} ] Letting ( e^C = k ), we have: [ x = k e^{-4t} ]

  5. Initial Conditions: Using ( x = 3 ) when ( t = 0 ): [ 3 = k e^{0} \Rightarrow k = 3 ] Thus: [ x = 3 e^{-4t} ]

  6. Final Expression: Therefore, ( x(t) = 3 e^{-4t} ).

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