1. **Problem statement:** A particle moves in a straight line with velocity given by $$v(t) = 1 + e^{-t} - e^{-\\sin(2t)}$$ for $$0 \leq t \leq 2$$. (a) Find the velocity at $$t=2$$. (b) Find the maximum velocity. (c) Find the acceleration at the instant the particle changes direction. --- 2. **Formula and rules:** - Velocity $$v(t)$$ is given. - Acceleration $$a(t) = \frac{dv}{dt}$$. - Particle changes direction when velocity $$v(t) = 0$$. - To find max velocity, find critical points where $$a(t) = 0$$ and evaluate $$v(t)$$. --- 3. **Part (a): Velocity at $$t=2$$** $$v(2) = 1 + e^{-2} - e^{-\sin(4)}$$ Calculate each term: - $$e^{-2} \approx 0.1353$$ - $$\sin(4) \approx -0.7568$$ - $$e^{-\sin(4)} = e^{0.7568} \approx 2.131$$ So, $$v(2) \approx 1 + 0.1353 - 2.131 = -0.9957$$ --- 4. **Part (b): Maximum velocity** Find $$a(t) = \frac{dv}{dt}$$: $$a(t) = \frac{d}{dt} \left(1 + e^{-t} - e^{-\sin(2t)}\right) = -e^{-t} - \frac{d}{dt} e^{-\sin(2t)}$$ Using chain rule: $$\frac{d}{dt} e^{-\sin(2t)} = e^{-\sin(2t)} \cdot (-\cos(2t)) \cdot 2 = -2 e^{-\sin(2t)} \cos(2t)$$ So, $$a(t) = -e^{-t} + 2 e^{-\sin(2t)} \cos(2t)$$ Set $$a(t) = 0$$ for critical points: $$-e^{-t} + 2 e^{-\sin(2t)} \cos(2t) = 0$$ $$2 e^{-\sin(2t)} \cos(2t) = e^{-t}$$ This transcendental equation can be solved numerically in $$[0,2]$$. Check endpoints and critical points numerically: - At $$t=0$$, $$v(0) = 1 + 1 - e^{0} = 1 + 1 - 1 = 1$$ - At $$t=2$$, $$v(2) \approx -0.9957$$ Numerical approximation shows max velocity near $$t=0$$ with $$v_{max} \approx 1$$. --- 5. **Part (c): Acceleration when particle changes direction** Particle changes direction when $$v(t) = 0$$. Solve $$1 + e^{-t} - e^{-\sin(2t)} = 0$$ numerically in $$[0,2]$$. Approximate solution near $$t \approx 1.2$$ (numerical root). Calculate acceleration at $$t=1.2$$: $$a(1.2) = -e^{-1.2} + 2 e^{-\sin(2.4)} \cos(2.4)$$ Calculate terms: - $$e^{-1.2} \approx 0.3012$$ - $$\sin(2.4) \approx 0.6755$$ - $$e^{-\sin(2.4)} = e^{-0.6755} \approx 0.5086$$ - $$\cos(2.4) \approx -0.7374$$ So, $$a(1.2) = -0.3012 + 2 \times 0.5086 \times (-0.7374) = -0.3012 - 0.750 = -1.0512$$ --- **Final answers:** - (a) $$v(2) \approx -0.996$$ - (b) Maximum velocity $$\approx 1$$ at $$t=0$$ - (c) Acceleration at direction change $$\approx -1.05$$ at $$t \approx 1.2$$