91Ó°ÊÓ

Using Newton's second law, \(F = ma\), where \(F\) is the net force, \(m\) is the mass, and \(a\) is the acceleration. We are given the net force and acceleration, so we can solve for the mass of the sailboat: $$ m = \frac{F}{a} $$ We are given \(F = 0.375\,\text{N}\) and \(a = 0.30\,\text{m/s}^2\). Plugging these values into the formula: $$ m = \frac{0.375\,\text{N}}{0.30\,\text{m/s}^2} = 1.25\,\text{kg} $$ So, the mass of the sailboat is \(1.25\,\text{kg}\).

We are given that the sailboat is moving to the west at \(0.33\,\text{m/s}\). The southward component of the initial velocity is zero since the boat is initially not moving in the south direction. So, the initial velocity components are as follows: $$ v_{w0} = 0.33\,\text{m/s}, $$ and $$ v_{s0} = 0\,\text{m/s}. $$

We are given that the wind blows at \(28^{\circ}\) south of west and has a magnitude of \(0.30\,\text{m/s}^2\). We can determine the westward and southward components of acceleration using basic trigonometry as follows: $$ a_{w} = a \cos(28^{\circ}) $$ and $$ a_{s} = a \sin(28^{\circ}) $$ where \(a_w\) and \(a_s\) are the westward and southward components of acceleration, respectively. Plug in the given values: $$ a_{w} = 0.30\, \text{m/s}^2 \cos(28^{\circ}) \approx 0.26\, \text{m/s}^2 $$ and $$ a_{s} = 0.30\, \text{m/s}^2 \sin(28^{\circ}) \approx 0.14\, \text{m/s}^2 $$

To find the new velocity components, we use the formula: $$ v_f = v_0 + at $$ For the westward component: $$ v_{wf} = v_{w0} + a_{w}t $$ We know that \(v_{w0} = 0.33\,\text{m/s}\), \(a_{w} = 0.26\,\text{m/s}^2\), and \(t = 2.0\,\text{s}\). Plug in these values: $$ v_{wf} = 0.33\,\text{m/s} + (0.26\,\text{m/s}^2)(2.0\,\text{s}) = 0.85\,\text{m/s} $$ For the southward component: $$ v_{sf} = v_{s0} + a_{s}t $$ We know that \(v_{s0} = 0\,\text{m/s}\), \(a_{s} = 0.14\,\text{m/s}^2\), and \(t = 2.0\,\text{s}\). Plug in these values: $$ v_{sf} = 0\,\text{m/s} + (0.14\,\text{m/s}^2)(2.0\,\text{s}) = 0.28\,\text{m/s} $$

To find the magnitude of the new velocity, we use the Pythagorean theorem: $$ v = \sqrt{v_{wf}^2 + v_{sf}^2} $$ We know that \(v_{wf} = 0.85\,\text{m/s}\) and \(v_{sf} = 0.28\,\text{m/s}\). Plug in these values: $$ v = \sqrt{(0.85\,\text{m/s})^2 + (0.28\,\text{m/s})^2} \approx 0.90\,\text{m/s} $$ To find the direction, we use inverse tangent function: $$ \theta = \tan^{-1}\left(\frac{v_{sf}}{v_{wf}}\right) $$ We know that \(v_{wf} = 0.85\,\text{m/s}\) and \(v_{sf} = 0.28\,\text{m/s}\). Plug in these values: $$ \theta = \tan^{-1}\left(\frac{0.28\,\text{m/s}}{0.85\,\text{m/s}}\right) \approx 18.3^{\circ} $$ So, the new velocity of the boat is approximately \(0.90\,\text{m/s}\) in a direction \(18.3^{\circ}\) south of west.