To solve this problem, we'll use the relationship between tangential acceleration, centripetal acceleration, and the net acceleration of an object moving in a circular path.

The given details are:

Radius of the circular path, $$r = 5 \, \text{m}$$

Tangential acceleration, $$a_t = 10 \, \text{m/s}^2$$

Angle between net acceleration and centripetal acceleration, $$\theta = 30^\circ$$

First, let's denote the centripetal acceleration as $$a_c$$ and the net acceleration as $$a_{net}$$.

We know that the net acceleration is the vector sum of the centripetal acceleration and the tangential acceleration. Since the angle between the net acceleration and the centripetal acceleration is given as $$30^\circ$$, we can use trigonometric relationships to solve for the centripetal acceleration.

From the definition of net acceleration, we can establish:

$$a_{\text{net}} = \sqrt{a_c^2 + a_t^2}$$

Given that:

$$\cos \theta = \frac{a_c}{a_{\text{net}}}$$

Substituting the values we get:

$$\cos 30^\circ = \frac{a_c}{\sqrt{a_c^2 + a_t^2}}$$

Knowing $$\cos 30^\circ = \frac{\sqrt{3}}{2}$$, we substitute it into the equation:

$$\frac{\sqrt{3}}{2} = \frac{a_c}{\sqrt{a_c^2 + (10 \, \text{m/s}^2)^2}}$$

Rearranging and squaring both sides, we get:

$$(\frac{\sqrt{3}}{2})^2 = \frac{a_c^2}{a_c^2 + 100}$$

This simplifies to:

$$\frac{3}{4} = \frac{a_c^2}{a_c^2 + 100}$$

Cross-multiplying gives:

$$3(a_c^2 + 100) = 4a_c^2$$

$$3a_c^2 + 300 = 4a_c^2$$

$$300 = a_c^2$$

$$a_c = \sqrt{300} = 10\sqrt{3} \, \text{m/s}^2$$

Now, identifying that centripetal acceleration is given by:

$$a_c = \frac{v^2}{r}$$

Substituting values, we have:

$$10\sqrt{3} = \frac{v^2}{5}$$

Solving for $$v$$ gives:

$$10\sqrt{3} \times 5 = v^2$$

$$v^2 = 50\sqrt{3}$$

$$v = \sqrt{50\sqrt{3}}$$

Converting and simplifying:

$$v = (\sqrt{50}) (\sqrt[4]{3})$$

$$v \approx 7.07 \cdot 1.32 = 9.3 $$

Thus, the instantaneous speed is $$9.3 \, \text{m/s}$$.

The correct answer is:

Option D: $$9.3 \mathrm{~ms}^{-1}$$

Circular Motion

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