A body starts from rest on a long inclined plane of slope 45°. The coefficient of friction between the body and the plane varies as $$\mu = 0.3x$$, where x is distance travelled down the plane. The body will have maximum speed (for $$g = 10$$ m/s$$^2$$) when $$x$$ =

A body starts from rest on an inclined plane with a slope of 45°. The coefficient of friction is given by $$\mu = 0.3x$$, where $$x$$ is the distance traveled down the plane. We need to find the distance $$x$$ at which the body achieves maximum speed, with $$g = 10 \text{m/s}^2$$.

First, consider the forces acting on the body. The gravitational force component pulling the body down the plane is $$mg \sin 45^\circ$$. Since $$\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$$, this component is $$mg \cdot \frac{1}{\sqrt{2}} = \frac{mg}{\sqrt{2}}$$.

The frictional force opposes the motion and is given by $$\mu \times \text{normal reaction}$$. The normal reaction is $$mg \cos 45^\circ = mg \cdot \frac{1}{\sqrt{2}} = \frac{mg}{\sqrt{2}}$$. Substituting $$\mu = 0.3x$$, the frictional force is $$0.3x \cdot \frac{mg}{\sqrt{2}}$$.

The net force down the plane is the gravitational component minus the frictional force:

$$F_{\text{net}} = \frac{mg}{\sqrt{2}} - 0.3x \cdot \frac{mg}{\sqrt{2}} = \frac{mg}{\sqrt{2}} (1 - 0.3x)$$

By Newton's second law, $$F_{\text{net}} = ma$$, so:

$$ma = \frac{mg}{\sqrt{2}} (1 - 0.3x)$$

Dividing both sides by $$m$$:

$$a = \frac{g}{\sqrt{2}} (1 - 0.3x)$$

Substituting $$g = 10 \text{m/s}^2$$:

$$a = \frac{10}{\sqrt{2}} (1 - 0.3x)$$

Simplifying $$\frac{10}{\sqrt{2}}$$ by multiplying numerator and denominator by $$\sqrt{2}$$:

$$\frac{10}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$$

So:

$$a = 5\sqrt{2} (1 - 0.3x) = 5\sqrt{2} - 1.5\sqrt{2} x$$

Acceleration $$a$$ is the derivative of velocity with respect to time, but we need velocity as a function of distance $$x$$. Using the chain rule:

$$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$$

So:

$$v \frac{dv}{dx} = 5\sqrt{2} - 1.5\sqrt{2} x$$

Factor out $$\sqrt{2}$$:

$$v \frac{dv}{dx} = \sqrt{2} (5 - 1.5x)$$

To avoid decimals, note that $$1.5 = \frac{3}{2}$$, so:

$$v \frac{dv}{dx} = \sqrt{2} \left(5 - \frac{3}{2}x\right)$$

Separate variables:

$$v dv = \sqrt{2} \left(5 - \frac{3}{2}x\right) dx$$

Integrate both sides. The body starts from rest at $$x = 0$$, so initial conditions are $$v = 0$$ when $$x = 0$$:

$$\int_0^v v dv = \sqrt{2} \int_0^x \left(5 - \frac{3}{2}x\right) dx$$

Left side:

$$\left[ \frac{v^2}{2} \right]_0^v = \frac{v^2}{2} - 0 = \frac{v^2}{2}$$

Right side:

$$\sqrt{2} \left[ 5x - \frac{3}{2} \cdot \frac{x^2}{2} \right]_0^x = \sqrt{2} \left[ 5x - \frac{3}{4} x^2 \right] - 0$$

So:

$$\frac{v^2}{2} = \sqrt{2} \left(5x - \frac{3}{4} x^2\right)$$

Multiply both sides by 2:

$$v^2 = 2\sqrt{2} \left(5x - \frac{3}{4} x^2\right) = 2\sqrt{2} \cdot 5x - 2\sqrt{2} \cdot \frac{3}{4} x^2 = 10\sqrt{2} x - \frac{6\sqrt{2}}{4} x^2 = 10\sqrt{2} x - \frac{3\sqrt{2}}{2} x^2$$

Factor out $$\sqrt{2}$$:

$$v^2 = \sqrt{2} \left(10x - \frac{3}{2} x^2\right)$$

Since $$v$$ is non-negative, $$v^2$$ is maximum when $$v$$ is maximum. Let $$s = v^2 = \sqrt{2} \left(10x - \frac{3}{2} x^2\right)$$. To find the maximum, take the derivative of $$s$$ with respect to $$x$$ and set it to zero:

$$\frac{ds}{dx} = \sqrt{2} \cdot \frac{d}{dx}\left(10x - \frac{3}{2} x^2\right) = \sqrt{2} \left(10 - \frac{3}{2} \cdot 2x\right) = \sqrt{2} (10 - 3x)$$

Set $$\frac{ds}{dx} = 0$$:

$$\sqrt{2} (10 - 3x) = 0 \implies 10 - 3x = 0 \implies 3x = 10 \implies x = \frac{10}{3} \approx 3.33 \text{m}$$

To confirm it is a maximum, check the second derivative:

$$\frac{d^2s}{dx^2} = \sqrt{2} \cdot (-3) = -3\sqrt{2} < 0$$

Since the second derivative is negative, $$s$$ (and thus $$v$$) is maximum at $$x = \frac{10}{3} \text{m}$$.

Comparing with the options:

A. 9.8 m

B. 27 m

C. 12 m

D. 3.33 m

Hence, the correct answer is Option D.