To determine the magnitude of the acceleration (\(a\)) of the sailboat, you need to consider the change in velocity (\(\Delta \vec{v}\)) over the given time interval (\(\Delta t\)).
Given:
- Initial velocity (v_i): \(13.6 \, \text{m/s}\) at \(74.0^\circ\)
- Final velocity (v_f): \(17.1 \, \text{m/s}\) at \(79.8^\circ\)
- Time interval (\(\Delta t\)): \(2.62 \, \text{s}\)
First, break down the initial and final velocities into their components:
For the initial velocity (\(v_i\)):
\[
v_{ix} = 13.6 \cos(74.0^\circ)
\]
\[
v_{iy} = 13.6 \sin(74.0^\circ)
\]
For the final velocity (\(v_f\)):
\[
v_{fx} = 17.1 \cos(79.8^\circ)
\]
\[
v_{fy} = 17.1 \sin(79.8^\circ)
\]
Next, calculate these components:
\[
v_{ix} = 13.6 \cos(74.0^\circ) \approx 3.74 \, \text{m/s}
\]
\[
v_{iy} = 13.6 \sin(74.0^\circ) \approx 13.1 \, \text{m/s}
\]
\[
v_{fx} = 17.1 \cos(79.8^\circ) \approx 2.94 \, \text{m/s}
\]
\[
v_{fy} = 17.1 \sin(79.8^\circ) \approx 16.7 \, \text{m/s}
\]
Next, determine the change in velocity components:
\[
\Delta v_x = v_{fx} - v_{ix} = 2.94 - 3.74 \approx -0.8 \, \text{m/s}
\]
\[
\Delta v_y = v_{fy} - v_{iy} = 16.7 - 13.1 \approx 3.6 \, \text{m/s}
\]
Now calculate the magnitude of the change in velocity:
\[
\Delta \vec{v} = \sqrt{(\Delta v_x)^2 + (\Delta v_y)^2}
\]
\[
\Delta \vec{v} = \sqrt{(-0.8)^2 + (3.6)^2}
\]
\[
\Delta \vec{v} = \sqrt{0.64 + 12.96}
\]
\[
\Delta \vec{v} = \sqrt{13.6} \approx 3.69 \, \text{m/s}
\]
Now, using the formula for acceleration:
\[
a = \frac{\Delta \vec{v}}{\Delta t}
\]
\[
a = \frac{3.69 \, \text{m/s}}{2.62 \, \text{s}}
\]
\[
a \approx 1.41 \, \text{m/s}^2
\]
Therefore, the magnitude of the acceleration of the boat is approximately \[1.41 \, \text{m/s}^2\].
a sailboat is moving at 13.6 m/s at an angle of 74.0 degrees when a wind blows it for 2.62s. after, the boat moves with a velocity of 17.1 m/s in an 79.8 degrees angle. What is the magnitude of the acceleration of the boat?
a=[?] m/s^2
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