#224 - {Official} Star Trek Trivia

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{Official} Star Trek Trivia #224

Random Trivia - Basic Physics

Know Your Physics Applications!

 

 

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A shuttle (on Earth) is traveling at a fixed altitude with a negligible wind velocity. The shuttle is headed N 30^o W at a speed of 500 miles per hour. As the shuttle reaches a certain point it encounters a wind with a velocity of 70 miles per hour in the direction N 45^o E. What are the resultant speed and direction of the shuttle (the angle)?

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??????????????????????

 

Answering: Send myself a private message (do not answer here!)

 

Points: 7

 

 

Master Q

StarTrek_Master_Q@yahoo.com

Edited September 24, 2003 by master_q

[master_q 0]

Answer:

 

 

It seems that only one person got this question correct. However, it seems that many were confused on the notation of degrees. So I completely ignored the degrees.

The previous question was pretty easy, but this one it seems that I should show my work on solving my problem.

 

First Make Vector Graph! (Show all the forces acting on shuttle)

v1 = 500 <cos 120, sin 120>

= <-250, 250(3)^(1/2)>

Velocity of wind is:

v1 = 70<cos 45, sin 45>

= <35(2)^(1/2), 35 (2)^(1/2)>

So the velocity of the shuttle is . . . .

v = v1 + v2

<-250+35(2)^(1/2), 250(3)^(1/2), 35(2)^(1/2)>

= <-200.5, 482.5>

speed of shuttle . . .

||v|| = {(-200.5)^2 + (482.5)^2}^(1/2)

= 522.5 miles per hour

 

If theta is the direction angle of the flight path, then . . .

tan theta = (482.5)/(-200.5)

= -2.4065

So

theta = 180 + arctan(-2.4065)

= 180 - 67.4

= 112.6

 

(Yes, there are many other ways to solve this problem)

 

 

Master Q

StarTrek_Master_Q@yahoo.com

Edited September 26, 2003 by master_q