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How much time for this person to drive across time zones? (vectors)

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Homework Statement


A man drives a car starting 5.00 km due West from the line marking the Eastern time zone. He travels at 30 m/s along a straight road that runs in a direction E 30° N. How much time does it take the man to get to the Eastern time zone? (The man must travel along the road: no off-roading!)

Homework Equations


N/A

The Attempt at a Solution


I tried drawing and getting an understanding but that didn't work either. I would just like someone to show me how to get to the answer which is 3.2 minutes, it comes with the question for some reason. To be more exact, like what steps do I need to take in order for me to go into the right direction. What I've done so far is that I've assumed that the 30m/s is the hyptoneuse and the angle right below that is 30 degrees.
 

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  • #2
gneill
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I tried drawing and getting an understanding but that didn't work either.
Can you show us your drawing?
I would just like someone to show me how to get to the answer which is 3.2 minutes, it comes with the question for some reason.
Helpers can't show you step by step how to do your homework (it's against the forum rules). But they can give guidance by pointing out where you've gone right or wrong in the work that you've shown. They may also give informative hints where it doesn't just give away the solution altogether; the idea is for the questioner to learn how to do the work themselves.
What I've done so far is that I've assumed that the 30m/s is the hyptoneuse and the angle right below that is 30 degrees.
Sounds reasonable. Can you think of a way to determine how long (in kilometers) that hypotenuse is? Again, it would be good to see your diagram.
 
  • #3
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Can you show us your drawing?

Helpers can't show you step by step how to do your homework (it's against the forum rules). But they can give guidance by pointing out where you've gone right or wrong in the work that you've shown. They may also give informative hints where it doesn't just give away the solution altogether; the idea is for the questioner to learn how to do the work themselves.

Sounds reasonable. Can you think of a way to determine how long (in kilometers) that hypotenuse is? Again, it would be good to see your diagram.
Here's my diagram I drew and the picture that came with the question. Also, to get from m/s to km I would assume you should multiply by 3.6?
 

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  • #4
gneill
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Here's my diagram I drew and the picture that came with the question. Also, to get from m/s to km I would assume you should multiply by 3.6?
upload_2018-10-6_15-57-2.png

Okay, a couple of things. Speeds and distances have different units so there's no direct conversion between them. There has to be a "conversion constant" of some sort. In this case speed and distance are related by time: d = vt.

In your diagram you should label the hypotenuse with some variable representing the unknown distance, say D. The angle and horizontal distance labels are fine. Now, what can you do to find the value of D? What do you know about the lengths of the sides of a 30°-60°-90° triangle? Or perhaps you know some trig relationship?
 

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  • #5
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View attachment 231819
Okay, a couple of things. Speeds and distances have different units so there's no direct conversion between them. There has to be a "conversion constant" of some sort. In this case speed and distance are related by time: d = vt.

In your diagram you should label the hypotenuse with some variable representing the unknown distance, say D. The angle and horizontal distance labels are fine. Now, what can you do to find the value of D? What do you know about the lengths of the sides of a 30°-60°-90° triangle? Or perhaps you know some trig relationship?
Yea, this is my last question of the booklet. So, throughout the booklet we were doing trig relationships such as cos, sign, tan, etc. So I would assume this would have something to do with that. So I'm assuming it doesn't have anything to do with the sides of a triangle, well it does, but not like with the angles.
 
  • #6
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View attachment 231819
Okay, a couple of things. Speeds and distances have different units so there's no direct conversion between them. There has to be a "conversion constant" of some sort. In this case speed and distance are related by time: d = vt.

In your diagram you should label the hypotenuse with some variable representing the unknown distance, say D. The angle and horizontal distance labels are fine. Now, what can you do to find the value of D? What do you know about the lengths of the sides of a 30°-60°-90° triangle? Or perhaps you know some trig relationship?
So, what possibly could be done with the trig relationships?
 
  • #7
gneill
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Yea, this is my last question of the booklet. So, throughout the booklet we were doing trig relationships such as cos, sign, tan, etc. So I would assume this would have something to do with that. So I'm assuming it doesn't have anything to do with the sides of a triangle, well it does, but not like with the angles.
Well, the trig functions are all based upon the ratios of side lengths for right-angled triangles. If you remember your high school geometry you might recall that there are a few triangles with certain angles that have well-known side length relationships. The 30°-60°-90° is one of them. But feel free to use the trig functions, too.
 
  • #8
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Well, the trig functions are all based upon the ratios of side lengths for right-angled triangles. If you remember your high school geometry you might recall that there are a few triangles with certain angles that have well-known side length relationships. The 30°-60°-90° is one of them. But feel free to use the trig functions, too.
I'm confused on how to use the trig functions. So is the hypotenuse a unknown variable and I would use a trig function to find it? Or is the 30m/s the hypotenuse? And once I have the hypotenuse, how would I convert to it to time or is it the time?
 
  • #9
gneill
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I'm confused on how to use the trig functions. So is the hypotenuse a unknown variable and I would use a trig function to find it? Or is the 30m/s the hypotenuse? And once I have the hypotenuse, how would I convert to it to time or is it the time?
The unknown distance D is the hypotenuse. 30 m/s is the speed at which the car drives along that distance. How long does it take to drive a distance D at speed V?

You should review your trig function definitions to find one that involves your known (given) side length and the unknown side length.
 
  • #10
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The unknown distance D is the hypotenuse. 30 m/s is the speed at which the car drives along that distance. How long does it take to drive a distance D at speed V?

You should review your trig function definitions to find one that involves your known (given) side length and the unknown side length.
Oh so instead of just sin cosin and tan it would be either sin law or co-sign law??
 
  • #11
gneill
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Oh so instead of just sin cosin and tan it would be either sin law or co-sign law??
Since you have a right-triangle you can stick to the basic trig functions: sin, cos, tan.
 
  • #12
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Since you have a right-triangle you can stick to the basic trig functions: sin, cos, tan.
So how do I get the answer in minutes?
 
  • #13
gneill
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So how do I get the answer in minutes?
Have you found the distance yet?
 
  • #14
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Have you found the distance yet?
Sort of but I'm not sure if it is right. I did 5 cos 30 which would get me 4.33, which im assuming is the hyptoneuse? I'm not sure if this is right tho.
 
  • #15
gneill
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Sort of but I'm not sure if it is right. I did 5 cos 30 which would get me 4.33, which im assuming is the hyptoneuse? I'm not sure if this is right tho.
Nope. Look up the definition of cosine. What's the ratio?
 
  • #16
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Nope. Look up the definition of cosine. What's the ratio?
I got the answer. I did sin law, which got me the answer in seconds which I then converted to minutes which got me 3.2. Thanks everyone for the help
 
  • #17
gneill
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I got the answer. I did sin law, which got me the answer in seconds which I then converted to minutes which got me 3.2. Thanks everyone for the help
Can you show us the calculations that you did?
 
  • #18
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Can you show us the calculations that you did?
I already handed in the assignment, but firstly I did a/sin a then c/sin c which got me a number, I divided that number by 30 which is how many m/s, then that got me that one side, then I divided that number by 60 to get me the answer in minutes.
 
  • #19
gneill
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I already handed in the assignment, but firstly I did a/sin a then c/sin c which got me a number, I divided that number by 30 which is how many m/s, then that got me that one side, then I divided that number by 60 to get me the answer in minutes.
That's not exactly a clear description of the mathematical operations you carried out. Better to type out the math and allow it to be evaluated by our helpers. Best of luck to you on the marking of your assignment. :smile:
 

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