1. **Problem statement:** We are given the velocity function $v(t) = 2 \sin\left(e^{t/4}\right) + 1$ and acceleration $a(t) = \frac{1}{2} e^{t/4} \cos\left(e^{t/4}\right)$ for a particle moving along the x-axis with initial position $x(0) = 2$. We need to solve parts (c) and (d): (c) Find the total distance traveled by the particle from $t=0$ to $t=6$. (d) Find the position of the particle at the time it changes direction exactly once in $0 \leq t \leq 6$. --- 2. **Key formulas and concepts:** - Total distance traveled is the integral of the speed (absolute value of velocity): $$\text{Distance} = \int_0^6 |v(t)| \, dt$$ - The particle changes direction when velocity changes sign, i.e., when $v(t) = 0$. - Position at time $t$ is given by: $$x(t) = x(0) + \int_0^t v(s) \, ds$$ --- 3. **Step (c): Total distance traveled** - Since $v(t)$ is not explicitly simple, we find zeros of $v(t)$ in $[0,6]$ to split the integral where velocity changes sign. - Solve $v(t) = 0$: $$2 \sin\left(e^{t/4}\right) + 1 = 0 \implies \sin\left(e^{t/4}\right) = -\frac{1}{2}$$ - The general solution for $\sin y = -\frac{1}{2}$ is: $$y = \frac{7\pi}{6} + 2k\pi \quad \text{or} \quad y = \frac{11\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}$$ - Let $y = e^{t/4}$, so: $$t = 4 \ln y$$ - Find $t$ values in $[0,6]$: - For $k=0$, $y_1 = \frac{7\pi}{6} \approx 3.665$, $t_1 = 4 \ln(3.665) \approx 4.98$ - For $k=0$, $y_2 = \frac{11\pi}{6} \approx 5.76$, $t_2 = 4 \ln(5.76) \approx 7.5$ (outside interval) - Only $t_1 \approx 4.98$ is in $[0,6]$. - Split integral: $$\text{Distance} = \int_0^{4.98} |v(t)| dt + \int_{4.98}^6 |v(t)| dt$$ - Check sign of $v(t)$ in each interval: - At $t=0$, $v(0) = 2 \sin(e^0) + 1 = 2 \sin(1) + 1 > 0$ (positive) - At $t=5.5$, $v(5.5) = 2 \sin(e^{5.5/4}) + 1$; since $t=5.5 > 4.98$ and velocity crosses zero at $4.98$, $v(5.5)$ is negative. - So: $$\text{Distance} = \int_0^{4.98} v(t) dt - \int_{4.98}^6 v(t) dt$$ - Use position function: $$x(t) = 2 + \int_0^t v(s) ds$$ - Calculate: $$\int_0^{4.98} v(t) dt = x(4.98) - 2$$ $$\int_0^6 v(t) dt = x(6) - 2$$ - Therefore: $$\text{Distance} = (x(4.98) - 2) - (x(6) - x(4.98)) = 2x(4.98) - x(6) - 2$$ - Numerically approximate $x(4.98)$ and $x(6)$ by integrating $v(t)$ (using numerical methods or software). --- 4. **Step (d): Position at direction change** - The particle changes direction at $t \approx 4.98$ where $v(t) = 0$. - Position at this time: $$x(4.98) = 2 + \int_0^{4.98} v(t) dt$$ - Numerically approximate this integral to find $x(4.98)$. --- **Summary:** - Find $t$ where $v(t)=0$ in $[0,6]$: $t \approx 4.98$. - Total distance: $$\int_0^{4.98} v(t) dt - \int_{4.98}^6 v(t) dt = 2x(4.98) - x(6) - 2$$ - Position at direction change: $$x(4.98) = 2 + \int_0^{4.98} v(t) dt$$ Numerical integration is required for exact values.