Question

A particle with velocity Vo = -3.8 i + 5.3 j (in m/s) at t = 0 undergoes a constant acceleration of magnitude 3.0 m/s2 at an angle of 130º from the +x-axis. What is the particle’s velocity at t = 5 sec?

Answer 1

Let the initial velocity vector be:

$$\mathbf{v}_1 = (-3.8, 5.3)\space m/s$$

and the time moments $t_1 = 0s, t_2 = 5s$.

The acceleration is:

$$\mathbf{a} = (3\cdot \cos130\degree, 3\cdot \sin130\degree)\space m/s^2$$

Then, by defintion, the velocity at $t_2$ is:

$$\mathbf{v}_2 = \mathbf{v}_1 + \mathbf{a}(t_2 - t_1)\\ \mathbf{v}_2 = (-3.8, 5.3) + (3\cdot \cos130\degree, 3\cdot \sin130\degree)\cdot (5 -0) \approx\\ \approx (-5.7, 7.6)\space m/s$$

Answer. $\mathbf{v}_2 = -5.7\mathbf{i} + 7.6\mathbf{j} \space (m/s)$.