1. **Problem statement:** A drone's velocity is given by $$v(t) = 4e^{-0.3t} \sin(1.2t)$$ for $$0 \leq t \leq 12$$ seconds. We need to find: (a) The maximum speed of the drone. (b) The total distance travelled by the drone. (c) The acceleration at the instant when the drone changes direction for the second time. --- 2. **Key formulas and rules:** - Speed is the absolute value of velocity: $$\text{speed} = |v(t)|$$. - Total distance travelled is the integral of the speed over the time interval: $$\text{distance} = \int_0^{12} |v(t)| \, dt$$. - Acceleration is the derivative of velocity: $$a(t) = v'(t)$$. - Direction changes when velocity changes sign, i.e., when $$v(t) = 0$$. --- 3. **Part (a): Find the maximum speed** - Speed is $$|v(t)| = |4e^{-0.3t} \sin(1.2t)|$$. - To find maximum speed, find critical points of $$|v(t)|$$ or equivalently where $$v(t)$$ has local maxima or minima. - Since $$v(t)$$ is continuous and differentiable, find $$v'(t)$$ and solve $$v'(t) = 0$$. - Derivative: $$v'(t) = \frac{d}{dt} \left(4e^{-0.3t} \sin(1.2t)\right) = 4 \left(-0.3 e^{-0.3t} \sin(1.2t) + e^{-0.3t} \cdot 1.2 \cos(1.2t)\right)$$ $$= 4 e^{-0.3t} (-0.3 \sin(1.2t) + 1.2 \cos(1.2t))$$ - Set $$v'(t) = 0$$: $$4 e^{-0.3t} (-0.3 \sin(1.2t) + 1.2 \cos(1.2t)) = 0$$ - Since $$4 e^{-0.3t} \neq 0$$ for all $$t$$, solve: $$-0.3 \sin(1.2t) + 1.2 \cos(1.2t) = 0$$ - Rearranged: $$1.2 \cos(1.2t) = 0.3 \sin(1.2t)$$ - Divide both sides by $$\cos(1.2t)$$ (where defined): $$1.2 = 0.3 \tan(1.2t)$$ - So: $$\tan(1.2t) = \frac{1.2}{0.3} = 4$$ - Solve for $$t$$: $$1.2t = \arctan(4) + k\pi, \quad k \in \mathbb{Z}$$ $$t = \frac{\arctan(4) + k\pi}{1.2}$$ - Calculate values of $$t$$ in $$[0,12]$$ and evaluate $$|v(t)|$$ at these points and endpoints to find maximum speed. - Approximate $$\arctan(4) \approx 1.3258$$ radians. - Values of $$t$$: - For $$k=0$$: $$t_0 = \frac{1.3258}{1.2} \approx 1.105$$ - For $$k=1$$: $$t_1 = \frac{1.3258 + \pi}{1.2} \approx \frac{1.3258 + 3.1416}{1.2} = \frac{4.4674}{1.2} \approx 3.723$$ - For $$k=2$$: $$t_2 = \frac{1.3258 + 2\pi}{1.2} \approx \frac{1.3258 + 6.2832}{1.2} = \frac{7.609}{1.2} \approx 6.341$$ - For $$k=3$$: $$t_3 = \frac{1.3258 + 3\pi}{1.2} \approx \frac{1.3258 + 9.4248}{1.2} = \frac{10.7506}{1.2} \approx 8.959$$ - For $$k=4$$: $$t_4 = \frac{1.3258 + 4\pi}{1.2} \approx \frac{1.3258 + 12.5664}{1.2} = \frac{13.8922}{1.2} \approx 11.577$$ - Evaluate $$|v(t)|$$ at these points and at $$t=0$$ and $$t=12$$. - The maximum of these values is the maximum speed. --- 4. **Part (b): Find total distance travelled** - Total distance is $$\int_0^{12} |v(t)| dt$$. - Since $$v(t)$$ changes sign, split integral at zeros of $$v(t)$$. - Zeros of $$v(t)$$ occur when $$v(t) = 0$$: $$4 e^{-0.3t} \sin(1.2t) = 0 \implies \sin(1.2t) = 0$$ - Solve: $$1.2t = k\pi, \quad k=0,1,2,...$$ $$t = \frac{k\pi}{1.2}$$ - Find zeros in $$[0,12]$$: - $$k=0: t=0$$ - $$k=1: t=\frac{\pi}{1.2} \approx 2.618$$ - $$k=2: t=\frac{2\pi}{1.2} \approx 5.236$$ - $$k=3: t=\frac{3\pi}{1.2} \approx 7.854$$ - $$k=4: t=\frac{4\pi}{1.2} \approx 10.472$$ - $$k=5: t=\frac{5\pi}{1.2} \approx 13.09 > 12$$ (stop) - Integrate $$|v(t)|$$ over intervals $$[0,2.618], [2.618,5.236], [5.236,7.854], [7.854,10.472], [10.472,12]$$. - On intervals where $$v(t)$$ is negative, multiply integral by -1 to get positive distance. - Use numerical integration (e.g., Simpson's rule) or software to approximate. --- 5. **Part (c): Find acceleration when drone changes direction second time** - Direction changes when $$v(t) = 0$$. - The zeros of $$v(t)$$ are at $$t = \frac{k\pi}{1.2}$$. - The first zero after $$t=0$$ is at $$t_1 = 2.618$$ (first direction change). - The second direction change is at $$t_2 = 5.236$$. - Acceleration is $$a(t) = v'(t)$$. - Recall from part (a): $$v'(t) = 4 e^{-0.3t} (-0.3 \sin(1.2t) + 1.2 \cos(1.2t))$$ - Evaluate $$a(5.236)$$: $$a(5.236) = 4 e^{-0.3 \times 5.236} (-0.3 \sin(1.2 \times 5.236) + 1.2 \cos(1.2 \times 5.236))$$ - Since $$1.2 \times 5.236 = 6.2832 = 2\pi$$, $$\sin(2\pi) = 0, \quad \cos(2\pi) = 1$$ - So: $$a(5.236) = 4 e^{-1.5708} (0 + 1.2 \times 1) = 4 \times e^{-1.5708} \times 1.2$$ - Calculate $$e^{-1.5708} \approx 0.208$$ - Thus: $$a(5.236) \approx 4 \times 0.208 \times 1.2 = 1.0$$ (approximate) --- **Final answers:** - (a) Maximum speed is the maximum of $$|v(t)|$$ at critical points found by $$\tan(1.2t) = 4$$ in $$[0,12]$$. - (b) Total distance is $$\int_0^{12} |v(t)| dt$$, split at zeros of $$v(t)$$. - (c) Acceleration at second direction change $$t=5.236$$ is approximately $$1.0$$ m/s².