Question

A particle's constant acceleration is south at $2.50 \mathrm{m} / \mathrm{s}^{2}\( At \)t=0,\( its velocity is \)40.0 \mathrm{m} / \mathrm{s}$ east. What is its velocity at \(t=8.00 \mathrm{s} ?\)

Short answer

Answer: The final velocity of the particle at 8.00 seconds is 44.7 m/s at an angle of approximately 26.6° south of east.

Step-by-step solution

Find velocity in the east direction

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

Find the final velocity in the south direction

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.

Find the final velocity using the Pythagorean theorem

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 \(26.6^{\circ}\).