1. **State the problem:** We are given the displacement function of an object: $$s(t) = t - \sin 2t$$ for $$0 \leq t \leq \pi$$. We need to find the derivative $$s'(t)$$, analyze where the object is moving or stationary, and find the distance travelled between two stationary points. 2. **Find the derivative $$s'(t)$$:** Using the derivative rules: - The derivative of $$t$$ is 1. - The derivative of $$\sin 2t$$ is $$2 \cos 2t$$ by the chain rule. So, $$ s'(t) = 1 - 2 \cos 2t $$ 3. **Stationary points:** The object is stationary when $$s'(t) = 0$$. Given one stationary point is $$t = \frac{\pi}{6}$$. 4. **Show $$s'(t) > 0$$ between the two stationary points:** The two stationary points satisfy: $$ 1 - 2 \cos 2t = 0 \implies \cos 2t = \frac{1}{2} $$ From the unit circle, $$\cos \theta = \frac{1}{2}$$ at $$\theta = \frac{\pi}{3}$$ and $$\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$$. Since $$\theta = 2t$$, the two stationary points are: $$ 2t = \frac{\pi}{3} \implies t = \frac{\pi}{6} $$ $$ 2t = \frac{5\pi}{3} \implies t = \frac{5\pi}{6} $$ To check $$s'(t) > 0$$ between $$t = \frac{\pi}{6}$$ and $$t = \frac{5\pi}{6}$$, pick a test value, e.g., $$t = \frac{\pi}{2}$$: $$ s'(\frac{\pi}{2}) = 1 - 2 \cos (2 \times \frac{\pi}{2}) = 1 - 2 \cos \pi = 1 - 2(-1) = 1 + 2 = 3 > 0 $$ Thus, $$s'(t) > 0$$ between the two stationary points. 5. **Find the distance travelled between $$t = \frac{\pi}{6}$$ and $$t = \frac{5\pi}{6}$$:** Distance travelled is the absolute change in displacement: $$ \Delta s = s\left(\frac{5\pi}{6}\right) - s\left(\frac{\pi}{6}\right) $$ Calculate each: $$ s\left(\frac{5\pi}{6}\right) = \frac{5\pi}{6} - \sin \left(2 \times \frac{5\pi}{6}\right) = \frac{5\pi}{6} - \sin \frac{5\pi}{3} $$ Since $$\sin \frac{5\pi}{3} = -\frac{\sqrt{3}}{2}$$, $$ s\left(\frac{5\pi}{6}\right) = \frac{5\pi}{6} - \left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6} + \frac{\sqrt{3}}{2} $$ Similarly, $$ s\left(\frac{\pi}{6}\right) = \frac{\pi}{6} - \sin \left(2 \times \frac{\pi}{6}\right) = \frac{\pi}{6} - \sin \frac{\pi}{3} $$ Given $$\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$$, $$ s\left(\frac{\pi}{6}\right) = \frac{\pi}{6} - \frac{\sqrt{3}}{2} $$ Therefore, $$ \Delta s = \left(\frac{5\pi}{6} + \frac{\sqrt{3}}{2}\right) - \left(\frac{\pi}{6} - \frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6} + \frac{\sqrt{3}}{2} - \frac{\pi}{6} + \frac{\sqrt{3}}{2} = \frac{4\pi}{6} + \sqrt{3} = \frac{2\pi}{3} + \sqrt{3} $$ **Final answers:** - $$s'(t) = 1 - 2 \cos 2t$$ - $$s'(t) > 0$$ between $$t = \frac{\pi}{6}$$ and $$t = \frac{5\pi}{6}$$ - Distance travelled between these points is $$\frac{2\pi}{3} + \sqrt{3}$$