Vectors - Driving a car between time zones

  • #1
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Vectors -- Driving a car between time zones...

Homework Statement


A man is driving a car is 5000m due west from the line marking the eastern time zone. He travels at 30m/s along a straight road that runs 30 degrees north of east. How much time does it take the man to get to the eastern time zone

Homework Equations





The Attempt at a Solution


I drew it out on a cartesian plane and tried the cos 30 degrees= 5000/hypotenue
I know you have to find out the hypotenuse then use the 30m/s to get the time but somehow I'm not getting the right answer..
 
  • #2

Homework Statement


A man is driving a car is 5000m due west from the line marking the eastern time zone. He travels at 30m/s along a straight road that runs 30 degrees north of east. How much time does it take the man to get to the eastern time zone

Homework Equations





The Attempt at a Solution


I drew it out on a cartesian plane and tried the cos 30 degrees= 5000/hypotenue
I know you have to find out the hypotenuse then use the 30m/s to get the time but somehow I'm not getting the right answer..

(I'm adding some more words to your 1-word thread title. Please make your titles very descriptive of the problem you are asking about in your threads. Thanks)

Could you please look at your problem statement? I think there are some typos in it. It says the man is driving *from* the Eastern Time Zone, and then asks how long it takes to get *to* the Eastern Time Zone. Does it mean instead that the Eastern Time Zone is 5000m wide, and the man is driving from the eastern end to the western end? But the Eastern Time Zone is way wider than 5km...
 
  • #3

Homework Statement


A man is driving a car is 5000m due west from the line marking the eastern time zone. He travels at 30m/s along a straight road that runs 30 degrees north of east. How much time does it take the man to get to the eastern time zone

Homework Equations





The Attempt at a Solution


I drew it out on a cartesian plane and tried the cos 30 degrees= 5000/hypotenue
That's very good. :approve:

[itex] \cos \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}} [/itex]

You're off to the right start. :smile:

I know you have to find out the hypotenuse then use the 30m/s to get the time but somehow I'm not getting the right answer..

You are correct. The first thing you probably want to do is find the hypotenuse.

From there you can use of of your kinematics equations to find the time, Δt, that it takes to travel a known distance at a constant velocity. :wink:

But I can't help you more than that without seeing more work. If you show your work perhaps we can help point out any errors.

[Edit: berkeman, I interpreted the problem as the car is in the Central Time zone (UTC - 6:00), 5 km west of the Eastern Time zone's (UTC - 5:00) border.]
 
  • #4
[Edit: berkeman, I interpreted the problem as the car is in the Central Time zone (UTC - 6:00), 5 km west of the Eastern Time zone's (UTC - 5:00) border.]

Oh! That would start to make sense. Thanks :smile:
 

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