displacement problem

**leinadwerdna** · January 24th 2010, 09:25 AM

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

**1005** · January 24th 2010, 10:27 AM

Originally Posted by leinadwerdna

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 $-200$ and the second vector is $(150 \angle 180^\circ - 60^\circ)$ or $(150 \angle 120^\circ)$

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: $-200 + 150 \cos (120^\circ)$
y: $150\sin (120^\circ)$

Your answer then may look like this:
$(-200 + 150 \cos (120^\circ))\hat i \ + (150\sin (120^\circ))\hat j \$
Where $\hat i \$ and $\hat j \$ are unit vectors designating only direction and not magnitude.
If you wanted the answer expressed in phasor form, then do this:
$(\sqrt{(-200 + 150 \cos (120^\circ))^2 + (150\sin (120^\circ))^2}\angle \arctan (\frac{150\sin (120^\circ)}{-200 + 150 \cos (120^\circ)}))$

**skeeter** · January 24th 2010, 11:02 AM

Originally Posted by leinadwerdna

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 ...

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

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

$\frac{\sin{\theta}}{150} = \frac{\sin(120^\circ)}{d}<br />$

$\theta = \arcsin\left[\frac{150\sin(120^\circ)}{d}\right]$

**leinadwerdna** · January 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

**skeeter** · January 24th 2010, 02:42 PM

Originally Posted by leinadwerdna

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

answer keys aren't always correct

**leinadwerdna** · January 24th 2010, 06:32 PM

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 $-200$ and the second vector is $(150 \angle 180^\circ - 60^\circ)$ or $(150 \angle 120^\circ)$

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: $-200 + 150 \cos (120^\circ)$
y: $150\sin (120^\circ)$

Your answer then may look like this:
$(-200 + 150 \cos (120^\circ))\hat i \ + (150\sin (120^\circ))\hat j \$
Where $\hat i \$ and $\hat j \$ are unit vectors designating only direction and not magnitude.
If you wanted the answer expressed in phasor form, then do this:
$(\sqrt{(-200 + 150 \cos (120^\circ))^2 + (150\sin (120^\circ))^2}\angle \arctan (\frac{150\sin (120^\circ)}{-200 + 150 \cos (120^\circ)}))$

if u do it this way the angle of the displacement vector is -25 degrees rather than 25 why is that

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