Question

Give the components of the velocity vector for wind blowing at 12 km/hr toward the southeast. (Assume north is the positive y-direction.)

Answer 1

Answer:

$\overline{V_x} = 6\sqrt{2}\\\overline{V_y} = -6\sqrt{2}$

components can be written as:

$\overline{V} = 6\sqrt{2}\,\hat{i}-6\sqrt{2}\,\hat{j}$

Step-by-step explanation:

The components of a velocity vector at any direction can be thought of separate vectors in 'i' and 'j' directions respectively. These vectors can be added to make the resultant velocity vector.

In our case the magnitude is provided:

$|\overline{V}|= 12$

the direction is provided to be southeast:

$\theta = 45$

$\theta$  is measured from positive x-axis clockwise towards the 4th quadrant. it is important to know that quadrant since the signs of the components will be where x is positive and y is negative as the direction is south east (+x,-y).

the components of the velocity will be:

$\overline{V_x} = 12\cos{(45)} = 12\dfrac{\sqrt{2}}{2} = 6\sqrt{2}\\\overline{V_y} = -12\sin{(45)} = -12\dfrac{\sqrt{2}}{2} = -6\sqrt{2}$

So that:

$\overline{V} = \overline{V_x}\hat{i}+\overline{V_x}\hat{j}$

$\overline{V} = 6\sqrt{2}\,\hat{i}-6\sqrt{2}\,\hat{j}$

This can be checked using since we know that magnitude of the above vector should be 12.

$|\overline{V}| = \sqrt{(12\cos{\theta})^2+(12\sin{\theta})^2}$

$|\overline{V}| = \sqrt{(6\sqrt{2})^2+(-6\sqrt{2})^2}\\|\overline{V}| = \sqrt{36(2)+36(2)}\\|\overline{V}| = \sqrt{144} = 12$