Vectors - Driving a car between time zones

[Coco12]

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..

 

[berkeman]

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...

 

[collinsmark]

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.

$\cos \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}$

You're off to the right start.

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.

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.]

 

[berkeman]

[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

 

[Nickie]

When dealing with vectors, it is important to consider both magnitude and direction. In this scenario, the man is traveling at a speed of 30m/s, but the direction of his travel is 30 degrees north of east. This means that he is not traveling directly east, but rather at an angle.

To find the time it takes the man to get to the eastern time zone, we need to use the formula for distance, which is d = rt, where d is distance, r is rate (or speed), and t is time. However, since the man is not traveling directly east, we need to use the component of his speed that is in the eastward direction. This can be found using trigonometry.

Using the given information, we can find the eastward component of his speed using the cosine function: cos 30 = adjacent/hypotenuse. The adjacent side is the eastward component of his speed, which we can call x. The hypotenuse is the total speed of 30m/s. Therefore, we have:

cos 30 = x/30
x = 30(cos 30)
x = 30(0.866)
x = 25.98 m/s

Now we can plug this value into our distance formula: d = rt, where d is the distance to the eastern time zone (5000m), r is the eastward component of his speed (25.98m/s), and t is the time we are trying to find.

5000 = 25.98t
t = 5000/25.98
t = 192.41 seconds

Therefore, it will take the man approximately 192.41 seconds to reach the eastern time zone.