Stumped on this physics problem.

CraKaJaX

Lifer
Dec 26, 2004
11,905
148
101
Snow is falling vertically at a constant speed of 8.0 m/s. At what angle from the vertical do the snowflakes appear to be falling as viewed by the driver of a car traveling on a straight, level road with a speed of 50 km/h?

I really have no idea where to even start.
 

spidey07

No Lifer
Aug 4, 2000
65,469
5
76
Calculate distance traveled in one second for the car and the snow. Then use trigonometry to solve for the angle. Probably not the "right" way to do it, but I like brute force. ;)
 

homercles337

Diamond Member
Dec 29, 2004
6,345
3
71
Originally posted by: spidey07
Calculate distance traveled in one second for the car and the snow. Then use trigonometry to solve for the angle. Probably not the "right" way to do it, but I like brute force. ;)

Actually thats correct. The car and snowflake are both vectors. Do the dot product and take the arcsine. Thats your angle.
 

RallyMaster

Diamond Member
Dec 28, 2004
5,582
0
0
spidey07's method could actually work...

in one second, the snow travels 8 meters downwards while the car moves 50km/h * 1000m/1km * 1 hour/3600 s = 13.8888889 meters.

since the snow is traveling vertical and the car moving horizontal, you set tan theta = 8 m / 13.8888889 m and solve for theta.

dot product:
<0, -8, 0> dot <13.88889, 0, 0> = 0 + 0 + 0?

A dot B/ (magA * magB) = cos theta is the only equation that i know using dot product that finds you an angle between the two vectors. since the dot product is 0, then the two are falling perpendicular which only means the two vectors are at 90º.
 

Born2bwire

Diamond Member
Oct 28, 2005
9,840
6
71
Originally posted by: homercles337
Originally posted by: spidey07
Calculate distance traveled in one second for the car and the snow. Then use trigonometry to solve for the angle. Probably not the "right" way to do it, but I like brute force. ;)

Actually thats correct. The car and snowflake are both vectors. Do the dot product and take the arcsine. Thats your angle.

The dot product provides a projection of one vector onto the other. Since the vectors are normal to each other, the projection is zero. Here you would want the cross product of the vectors and take the inverse sine. In addition, you want to take the cross product of the directional vectors since the norm of the cross product is the product of the norms of the two vectors and the sine of the angle between them.
 

RallyMaster

Diamond Member
Dec 28, 2004
5,582
0
0
In the use of the cross product, you have A cross B / (magA * magB) = sin theta

Set up a 3x3 matrix:
| i j k|
| 0 -8 0|
|13.9 0 0|

This thread is helping me review for my multivariable test coming up on October 4th. I feel like I'm actually being productive...
 

Born2bwire

Diamond Member
Oct 28, 2005
9,840
6
71
Originally posted by: RallyMaster
In the use of the cross product, you have A cross B / (magA * magB) = sin theta

Set up a 3x3 matrix:
| i j k|
| 0 -8 0|
|13.9 0 0|

This thread is helping me review for my multivariable test coming up on October 4th. I feel like I'm actually being productive...

It's all an illusion.

 

Vegitto

Diamond Member
May 3, 2005
5,234
1
0
Could you solve this problem by using the solution to the Doppler effect?

EDIT: No wait, that won't work..

EDIT2: Especially since with this solution, when the source is actually straight above the beholder, there is no angle between them, just distance.. But Pythagoras works nicely here, right?