1. **Problem statement:** A particle moves in a straight line with velocity given by $$v(t) = 1 + e^{-t} \sin t$$ for $$0 \leq t \leq 2$$. (a) Find the velocity at $$t=2$$. (b) Find the maximum velocity. (c) Find the acceleration when the particle changes direction. --- 2. **Formula and rules:** - Velocity $$v(t)$$ is given. - Acceleration $$a(t)$$ is the derivative of velocity: $$a(t) = v'(t)$$. - Particle changes direction when velocity $$v(t) = 0$$. - To find max velocity, find critical points where $$v'(t) = 0$$ and evaluate. --- 3. **Part (a): Velocity at $$t=2$$** $$v(2) = 1 + e^{-2} \sin 2$$ Calculate numeric value: $$e^{-2} \approx 0.1353, \sin 2 \approx 0.9093$$ $$v(2) \approx 1 + 0.1353 \times 0.9093 = 1 + 0.123 = 1.123$$ --- 4. **Part (b): Maximum velocity** Find $$v'(t)$$: $$v'(t) = \frac{d}{dt} \left(1 + e^{-t} \sin t\right) = 0 + \frac{d}{dt} (e^{-t} \sin t)$$ Use product rule: $$v'(t) = e^{-t} \cos t - e^{-t} \sin t = e^{-t} (\cos t - \sin t)$$ Set $$v'(t) = 0$$: $$e^{-t} (\cos t - \sin t) = 0$$ Since $$e^{-t} \neq 0$$, solve: $$\cos t - \sin t = 0 \implies \cos t = \sin t$$ This occurs at: $$t = \frac{\pi}{4} \approx 0.785$$ (within $$[0,2]$$) Evaluate $$v(t)$$ at critical point and endpoints: - $$v(0) = 1 + e^{0} \sin 0 = 1 + 1 \times 0 = 1$$ - $$v(2) \approx 1.123$$ (from part a) - $$v(\pi/4) = 1 + e^{-\pi/4} \sin(\pi/4)$$ Calculate: $$e^{-\pi/4} \approx e^{-0.785} \approx 0.456$$ $$\sin(\pi/4) = \frac{\sqrt{2}}{2} \approx 0.707$$ $$v(\pi/4) \approx 1 + 0.456 \times 0.707 = 1 + 0.322 = 1.322$$ Maximum velocity is approximately $$1.322$$ at $$t = \frac{\pi}{4}$$. --- 5. **Part (c): Acceleration when particle changes direction** Particle changes direction when $$v(t) = 0$$: $$1 + e^{-t} \sin t = 0 \implies e^{-t} \sin t = -1$$ Since $$e^{-t} > 0$$ and $$\sin t \in [-1,1]$$, check if $$\sin t = -e^{t}$$ is possible for $$t \in [0,2]$$. At $$t=0$$, $$v(0) = 1$$; at $$t=2$$, $$v(2) \approx 1.123$$, velocity is positive. Check if velocity crosses zero: Try $$t \approx 1.8$$: $$v(1.8) = 1 + e^{-1.8} \sin 1.8$$ $$e^{-1.8} \approx 0.165, \sin 1.8 \approx 0.973$$ $$v(1.8) \approx 1 + 0.165 \times 0.973 = 1 + 0.160 = 1.16 > 0$$ Try $$t \approx 0.1$$: $$v(0.1) = 1 + e^{-0.1} \sin 0.1 \approx 1 + 0.905 \times 0.0998 = 1 + 0.090 = 1.09 > 0$$ Velocity does not become zero in $$[0,2]$$, so particle does not change direction in this interval. Therefore, no acceleration at direction change in $$[0,2]$$. If we consider the theoretical instant where $$v(t) = 0$$, solve numerically: No solution in $$[0,2]$$. --- 6. **Acceleration formula:** $$a(t) = v'(t) = e^{-t} (\cos t - \sin t)$$ **Final answers:** - (a) $$v(2) \approx 1.123$$ - (b) Maximum velocity $$\approx 1.322$$ at $$t = \frac{\pi}{4}$$ - (c) Particle does not change direction in $$[0,2]$$, so no acceleration at direction change.