Solve $$\frac{3x}{x-2} \leq 1.$$ --- An aircraft flying horizontally at $V$ m s$^{-1}$ releases a bomb that hits the ground 4000 m away, measured horizontally - HSC - SSCE Mathematics Extension 1 - Question 4 - 2001 - Paper 1

We know the distance the bomb must travel horizontally before hitting the ground is 4000 m. The horizontal position after $t$ seconds is given by:

$x = Vt.$

Setting $x = 4000$, we have:

$4000 = Vt \implies t = \frac{4000}{V}.$

Now considering the vertical motion: The vertical position after $t$ seconds is given by:

$y = -5t^2.$

Using the relationship for the angle of descent, where the bomb strikes the ground at an angle of 45° to the vertical, we establish:

$\tan(45°) = 1 = \frac{y}{x} = \frac{-5t^2}{4000} \implies -5t^2 = 4000.$

Substituting for $t$:

$-5\left(\frac{4000}{V}\right)^2 = 4000 \implies -\frac{5 \cdot 4000^2}{V^2} = 4000.$

Solving for $V$ gives:

$-5 \cdot 4000 = 4000V^2 \implies -5 = V^2 \implies V = 20 \text{ m/s}.$

This is a first-order linear ordinary differential equation. We separate variables:

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

Integrating both sides:

$\ln|x| = -4t + C,$

where $C$ is the integration constant. Exponentiating gives:

$x = e^{-4t + C} = e^{C} e^{-4t}.$

Letting $A = e^{C}$ results in:

$x(t) = Ae^{-4t}.$

Now, we apply initial conditions: When $t = 0$, $x = 3$:

$3 = Ae^{0} \implies A = 3.$

At $t = 0$, if we also know $x = -2$ when it satisfies the condition from the other statement, we will have the complete function. Thus the solution is:

$x(t) = 3e^{-4t}.$