# displacement problem

• Jan 24th 2010, 09:25 AM
displacement problem
an airplane travels 200 miles due west then 150 miles 60 degrees north of west What is the resultant displacement
• Jan 24th 2010, 10:27 AM
1005
Quote:

an airplane travels 200 miles due west then 150 miles 60 degrees north of west What is the resultant displacement

You need to assign a reference for your displacement vectors and then sum them. Most intuitive for me would be to think of the directions as if they were on a map: west being negative x-axis, east being positive x-axis, north being positive y-axis, and south being negative y-axis. From there, you see the first vector is $\displaystyle -200$ and the second vector is $\displaystyle (150 \angle 180^\circ - 60^\circ)$ or $\displaystyle (150 \angle 120^\circ)$

Drawing pictures helps with vector problems such as this one. Do do vector addition, you need to break each vector into rectangular components, and add them.
x: $\displaystyle -200 + 150 \cos (120^\circ)$
y: $\displaystyle 150\sin (120^\circ)$

$\displaystyle (-200 + 150 \cos (120^\circ))\hat i \ + (150\sin (120^\circ))\hat j \$
Where $\displaystyle \hat i \$ and $\displaystyle \hat j \$ are unit vectors designating only direction and not magnitude.
If you wanted the answer expressed in phasor form, then do this:
$\displaystyle (\sqrt{(-200 + 150 \cos (120^\circ))^2 + (150\sin (120^\circ))^2}\angle \arctan (\frac{150\sin (120^\circ)}{-200 + 150 \cos (120^\circ)}))$
• Jan 24th 2010, 11:02 AM
skeeter
Quote:

an airplane travels 200 miles due west then 150 miles 60 degrees north of west What is the resultant displacement

you can use the law of cosines to find the magnitude of the displacement ...

$\displaystyle d = \sqrt{200^2 + 150^2 - 2(200)(150)\cos(120^\circ)}$

then the law of sines to find the direction of the displacement relative to west ...

$\displaystyle \frac{\sin{\theta}}{150} = \frac{\sin(120^\circ)}{d}$

$\displaystyle \theta = \arcsin\left[\frac{150\sin(120^\circ)}{d}\right]$
• Jan 24th 2010, 02:40 PM
for some reason the answer key says the displacement is facing north of east i thought it would be north of west
• Jan 24th 2010, 02:42 PM
skeeter
Quote:

for some reason the answer key says the displacement is facing north of east i thought it would be north of west

• Jan 24th 2010, 06:32 PM
Quote:

Originally Posted by 1005
You need to assign a reference for your displacement vectors and then sum them. Most intuitive for me would be to think of the directions as if they were on a map: west being negative x-axis, east being positive x-axis, north being positive y-axis, and south being negative y-axis. From there, you see the first vector is $\displaystyle -200$ and the second vector is $\displaystyle (150 \angle 180^\circ - 60^\circ)$ or $\displaystyle (150 \angle 120^\circ)$

Drawing pictures helps with vector problems such as this one. Do do vector addition, you need to break each vector into rectangular components, and add them.
x: $\displaystyle -200 + 150 \cos (120^\circ)$
y: $\displaystyle 150\sin (120^\circ)$

$\displaystyle (-200 + 150 \cos (120^\circ))\hat i \ + (150\sin (120^\circ))\hat j \$
Where $\displaystyle \hat i \$ and $\displaystyle \hat j \$ are unit vectors designating only direction and not magnitude.
$\displaystyle (\sqrt{(-200 + 150 \cos (120^\circ))^2 + (150\sin (120^\circ))^2}\angle \arctan (\frac{150\sin (120^\circ)}{-200 + 150 \cos (120^\circ)}))$