Question

A dog running in an open field has components of velocity Vx = 2.6 m/s and Vy = -1.8 m/s at time t = 10 sec. The dog is running with constant acceleration of 0.45 m/s2 and direction 31.0º measured from +x-axis. At 20 sec, (a) what are the x- and y-components of dog’s velocity? (b) What are the magnitude and direction of dog’s velocity?

Answer 1

Let the initial velocity vector be:

$$\mathbf{v}_1 = (2.6, -1.8)\space m/s$$

and the time moments $t_1 = 10s, t_2 = 20s$.

The acceleration is:

$$\mathbf{a} = (0.45\cdot \cos31\degree, 0.45\cdot \sin31\degree)\space m/s^2$$

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

$$\mathbf{v}_2 = \mathbf{v}_1 + \mathbf{a}(t_2 - t_1)\\ \mathbf{v}_2 = (2.6, -1.8) + (0.45\cdot \cos31\degree, 0.45\cdot \sin31\degree)\cdot (20 -10) \approx\\ \approx (6.46, 0.518)\space m/s$$

The magnitude is:

$$v_2 = \sqrt{6.46^2 + 0.518^2} \approx 6.48 m/s$$

The direction is:

$$\theta = \arctan\left( \dfrac{0.518}{6.46} \right) \approx 4.58\degree$$

Answer. a) $v_{2x} = 6.46m/s, v_{2y} = 0.518m/s$. b) $v_2 = 6.48m/s, \theta = 4.58\degree.$