Question 1166452
This is a problem in one-dimensional kinematics that can be solved using the time-independent equation relating velocity, acceleration, and displacement.

## 📐 Kinematics Calculation

1.  **Define Direction and Variables:**
    Let **West be the positive direction** ($+$).

    * Initial Velocity ($\vec{v}_i$): $6.7 \text{ m/s}$ West $\implies v_i = +6.7 \text{ m/s}$
    * Final Velocity ($\vec{v}_f$): $1.4 \text{ m/s}$ West $\implies v_f = +1.4 \text{ m/s}$
    * Acceleration ($\vec{a}$): $3.4 \text{ m/s}^2$ East $\implies a = -3.4 \text{ m/s}^2$

2.  **Select the Kinematic Equation:**
    Since the time is not given, we use the equation that relates the final velocity, initial velocity, acceleration, and displacement ($\Delta x$):
    $$v_f^2 = v_i^2 + 2a \Delta x$$

3.  **Solve for Displacement ($\Delta x$):**
    $$\Delta x = \frac{v_f^2 - v_i^2}{2a}$$

4.  **Substitute Values and Compute:**
    $$\Delta x = \frac{(1.4 \text{ m/s})^2 - (6.7 \text{ m/s})^2}{2(-3.4 \text{ m/s}^2)}$$
    $$\Delta x = \frac{1.96 \text{ m}^2/\text{s}^2 - 44.89 \text{ m}^2/\text{s}^2}{-6.8 \text{ m/s}^2}$$
    $$\Delta x = \frac{-42.93 \text{ m}^2/\text{s}^2}{-6.8 \text{ m/s}^2}$$
    $$\Delta x \approx 6.3132 \text{ m}$$

The distance traveled is the magnitude of the displacement.

The distance the motorboat traveled during the acceleration is **$6.31$ m**.