Answer to Question #248065 in Classical Mechanics for Lawrence

Question

40 units, 45o North of East

Answer 1

Given,

Magnitude of the vector = 40 units

Angle $= 45^\circ$ North of east East

Let the vector

Horizontal component $(\overrightarrow{r_x})=40 \cos (45^\circ) \hat{i}$

Vertical component $(\overrightarrow{r_y})=40\sin(45^\circ)\hat{j}$

Hence the required vector $\overrightarrow{r}=40\cos (45^\circ)\hat{i}+40\sin(45^\circ)\hat{j}$

$\Rightarrow |r|=\sqrt{(40\cos(45^\circ))^2+(40\sin (45^\circ))^2}$

$\Rightarrow |r|=\sqrt{20^2+20^2}=20\sqrt{2}$