91Ó°ÊÓ

Since there is no acceleration in the east direction, the velocity in the east direction remains constant at \(40.0 \mathrm{m} / \mathrm{s}\).

We are given the constant acceleration of the particle in the south direction as \(2.50\,\mathrm{m}/\mathrm{s}^2\). We can use the final velocity formula to find the final velocity in the south direction: $$v_f = v_i + at$$, where \(v_f\) is the final velocity, \(v_i\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time duration. Since the particle's initial velocity in the south direction is \(0\,\mathrm{m}/\mathrm{s}\), the formula becomes: $$v_f = 0 + (2.50\,\mathrm{m}/\mathrm{s}^2)(8.00\,\mathrm{s})$$ Now, calculate the final velocity in the south direction: $$v_f = (2.50\,\mathrm{m}/\mathrm{s}^2)(8.00\,\mathrm{s}) = 20.0\,\mathrm{m}/\mathrm{s}$$ south.

We now have the velocities in both the east and the south direction. To find the resultant velocity, we will use the Pythagorean theorem: $$v = \sqrt{v_e^2 + v_s^2}$$ where \(v\) is the final velocity, \(v_e\) is the velocity in the east direction, and \(v_s\) is the velocity in the south direction. Substituting the values, we get: $$v = \sqrt{(40.0\,\mathrm{m}/\mathrm{s})^2 + (20.0\,\mathrm{m}/\mathrm{s})^2}$$ $$v = \sqrt{1600\,\mathrm{m^2}/\mathrm{s^2} + 400\,\mathrm{m^2}/\mathrm{s^2}}$$ $$v = \sqrt{2000\,\mathrm{m^2}/\mathrm{s^2}}$$ $$v = 44.7\,\mathrm{m}/\mathrm{s}$$ The final velocity of the particle at \(t=8.00\,\mathrm{s}\) is \(44.7\,\mathrm{m}/\mathrm{s}\) at an angle \(\theta\) south of east given by \(\tan^{-1}\left(\frac{v_s}{v_e}\right)\), which when calculated is approximately \(26.6^{\circ}\).