To find the final velocity of the boat after it accelerates, we can use one of the equations of motion. The equation that relates initial velocity, final velocity, acceleration, and distance is:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity,
- \( a \) is the acceleration,
- \( s \) is the distance traveled.
Given:
- \( u = 15 , \text{m/s} \) (initial velocity),
- \( a = 2.0 , \text{m/s}^2 \) (acceleration),
- \( s = 430 , \text{m} \) (distance).
We need to calculate \( v \). Plugging in the values we have:
\[ v^2 = (15 , \text{m/s})^2 + 2 \times (2.0 , \text{m/s}^2) \times (430 , \text{m}) \]
Calculating \( (15 , \text{m/s})^2 \):
\[ (15 , \text{m/s})^2 = 225 , \text{m}^2/\text{s}^2 \]
Calculating \( 2 \times (2.0 , \text{m/s}^2) \times (430 , \text{m}) \):
\[ 2 \times 2.0 \times 430 = 4.0 \times 430 = 1720 , \text{m}^2/\text{s}^2 \]
Now, substituting these values back into the equation:
\[ v^2 = 225 , \text{m}^2/\text{s}^2 + 1720 , \text{m}^2/\text{s}^2 \]
\[ v^2 = 1945 , \text{m}^2/\text{s}^2 \]
Taking the square root of both sides to find \( v \):
\[ v = \sqrt{1945} \approx 44.15 , \text{m/s} \]
Therefore, the final velocity of the boat after the acceleration is approximately 44.15 m/s.