A motorboat accelerates uniformly from a velocity of 6.6 m/s to the west to a velocity of 1.3 m/s to the west. If its acceleration was 2.1 m/s2 to the east, how far did it travel during the acceleration? Answer in units of m.

To find the distance traveled during the acceleration of the motorboat, we can use the kinematic equation:

\(v_f^2 = v_i^2 + 2a \cdot d\)

where:
\(v_f\) is the final velocity,
\(v_i\) is the initial velocity,
\(a\) is the acceleration, and
\(d\) is the distance traveled.

In this case, the initial velocity (\(v_i\)) is 6.6 m/s to the west, the final velocity (\(v_f\)) is 1.3 m/s to the west, and the acceleration (\(a\)) is 2.1 m/s² to the east.

First, let's convert the velocities to positive values to make calculations easier. We know that the direction is west, so we'll assign negative values to them. Thus, the initial velocity (\(v_i\)) becomes -6.6 m/s, and the final velocity (\(v_f\)) becomes -1.3 m/s.

Now, let's plug the known values into the equation:

\((-1.3 \, \text{m/s})^2 = (-6.6 \, \text{m/s})^2 + 2 \cdot (2.1 \, \text{m/s}^2) \cdot d\)

Simplifying:

\(1.69 \, \text{m/s}^2 = 43.56 \, \text{m/s}^2 - 4.2 \, \text{m/s}^2 \cdot d\)

Combining like terms:

\(4.2 \, \text{m/s}^2 \cdot d = 43.56 \, \text{m/s}^2 - 1.69 \, \text{m/s}^2\)

\(4.2 \, \text{m/s}^2 \cdot d = 41.87 \, \text{m/s}^2\)

Now divide both sides by \(4.2 \, \text{m/s}^2\):

\(d = \frac{41.87 \, \text{m/s}^2}{4.2 \, \text{m/s}^2}\)

\(d = 9.97 \, \text{m}\)

Therefore, the motorboat traveled approximately 9.97 meters during the acceleration.