Question

A pilot flies her route in two straight-line segments. The displacement vector A for the first segment has a magnitude of 243 km and a direction 30.0o north of east. The displacement vector for the second segment has a magnitude of 178 km and a direction due west. The resultant displacement vector is R = A + B and makes an angle ? with the direction due east. Using the component method, find (a) the magnitude of R and (b) the directional angle ?.
(a) R = km
(b) ? = degrees

Answer 1

Answer:

a) $R=126Km$

b) $\theta=74.6\textdegree$

Explanation:

From the question we are told that:

1st segment

243km at Angle=30

2nd segment

178km West

Resolving to the X axis

$F_x=243cos30+178$

$F_x=33.44Km$

Resolving to the Y axis

$F_y=243sin30+178sin0$

$R=\sqrt{F_y^2+F_x^2}$

$F_y=121.5Km$

Therefore

Generally the equation for Directional Angle is mathematically given by

$\theta=tan^{-1}\frac{F_y}{F_x}$

$\theta=tan^{-1}\frac{121.5}{33.44}$

$\theta=74.6\textdegree$

Generally the equation for Magnitude is mathematically given by

$R=\sqrt{F_y^2+F_x^2}$

$R=\sqrt{33.44^2+121.5^2}$

$R=126Km$