1. **Problem statement:** A cyclist's velocity is given by the function $f(t) = (-t + 6) \cdot e^{t-3}$ for $t$ in seconds, describing acceleration and braking until stopping. The cyclist has already traveled 5 meters before observation. We need to: a) Show that $F(t) = (-t + 7) \cdot e^{t-3}$ is an antiderivative of $f(t)$ and find the function describing the distance traveled. b) Find the distance traveled at the time when the cyclist is fastest. c) Find the total distance traveled until the cyclist stops and the average velocity in the first 6 seconds. 2. **a) Show $F$ is an antiderivative of $f$ and find distance function:** Recall the derivative product rule: $\frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t)$. Let $F(t) = (-t + 7) e^{t-3}$. Calculate $F'(t)$: $$ F'(t) = \frac{d}{dt}(-t + 7) \cdot e^{t-3} + (-t + 7) \cdot \frac{d}{dt} e^{t-3} = (-1) e^{t-3} + (-t + 7) e^{t-3} $$ Simplify: $$ F'(t) = (-1) e^{t-3} + (-t + 7) e^{t-3} = (-1 - t + 7) e^{t-3} = (-t + 6) e^{t-3} = f(t) $$ Thus, $F$ is an antiderivative of $f$. The distance function $s(t)$ is the integral of velocity plus initial distance: $$ s(t) = F(t) - F(0) + 5 $$ Calculate $F(0)$: $$ F(0) = (-0 + 7) e^{0-3} = 7 e^{-3} $$ So, $$ s(t) = (-t + 7) e^{t-3} - 7 e^{-3} + 5 $$ 3. **b) Find distance at maximum velocity:** Find $t$ where $f(t)$ is maximum by solving $f'(t) = 0$. Calculate $f'(t)$: $$ f(t) = (-t + 6) e^{t-3} $$ Using product rule: $$ f'(t) = (-1) e^{t-3} + (-t + 6) e^{t-3} = (-1 - t + 6) e^{t-3} = (-t + 5) e^{t-3} $$ Set $f'(t) = 0$: $$ (-t + 5) e^{t-3} = 0 \implies -t + 5 = 0 \implies t = 5 $$ At $t=5$, velocity is maximum. Calculate distance at $t=5$: $$ s(5) = (-5 + 7) e^{5-3} - 7 e^{-3} + 5 = 2 e^{2} - 7 e^{-3} + 5 $$ 4. **c) Distance when cyclist stops and average velocity:** Cyclist stops when $f(t) = 0$: $$ (-t + 6) e^{t-3} = 0 \implies -t + 6 = 0 \implies t = 6 $$ Distance at $t=6$: $$ s(6) = (-6 + 7) e^{6-3} - 7 e^{-3} + 5 = 1 \cdot e^{3} - 7 e^{-3} + 5 $$ Average velocity over 6 seconds: $$ \text{average velocity} = \frac{\text{distance traveled in 6 s}}{6} = \frac{s(6) - 5}{6} = \frac{e^{3} - 7 e^{-3} + 5 - 5}{6} = \frac{e^{3} - 7 e^{-3}}{6} $$