What is the displacement vector needed to reach a research station on a safari?

In summary: I can't remember what it was, but I think it was multiplying 1.91 by the cosine of 14.3. That may have been it.Thanks for the help!
  • #1
Nyxious
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0
Hello all! I've been struggling with this problem for a couple of hours and I just can't seem to wrap my head around on how to do it. Here it is:

1. On a safari, a team of naturalists sets out toward a research station located 6.49 km away in a direction 38.5 ° north of east. After traveling in a straight line for 1.91 km, they stop and discover that they have been traveling 14.3 ° north of east, because their guide misread his compass. What are (a) the magnitude and (b) the direction (as a positive angle relative to due east) of the displacement vector now required to bring the team to the research station?I've attempted the question and found the magnitude to be 4.58 km. The angle, however, I can't seem to get. If someone can explain how to get both the magnitude and the angle I would appreciate it very much. The way I found the magnitude was just subtracting the final destination by the distance traveled. I'm not sure this is how it's done.
Attempt
I tried the standard way to get the resultant, but this didn't work for either the magnitude or the angle. I tried finding the x component and y component with the function
x comp = ( 6.49 x Cos 38.5 ) + ( 1.91 x Cos 14.3 )
y comp = ( 6.49 x Sin 38.5 ) + ( 1.91 x Sin 14.3 )

I hope I gave enough information to help someone help me. I really want to learn so a full explanation would be very appreciate. Thanks!
 
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  • #2
It may help to draw a picture. Pay attention to signs.

HowFarWe'veCome + HowFarWe'veYetToGo = HowFarWeWillHaveGone.
 
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  • #3
Hey there! Your approach with components is on the right track, but slightly off. And the distance left won't quite be 4.58 km -- that would be if they traveled in a straight line toward the destination, but since they were at an angle, it will actually be a greater distance for them to travel. Think of the problem in terms of three vectors: the vector from the start to the station (1), the vector from the start to where they stopped after going 1.91 km (2), and the vector from there to the station (3 -- the one you're solving for). Vectors 2 and 3 must add up to vector 1, and you know the components of vectors 1 and 2. Try solving for the components of vector 3 from there.
 
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  • #4
jbriggs444 said:
It may help to draw a picture. Pay attention to signs.

HowFarWe'veCome + HowFarWe'veYetToGo = HowFarWeWillHaveGone.
That was actually the first thing I did. That's actually how I kinda assumed the magnitude to be 4.58km. The angle however, is a mystery to me.
 
  • #5
jackarms said:
Hey there! Your approach with components is on the right track, but slightly off. And the distance left won't quite be 4.58 km -- that would be if they traveled in a straight line toward the destination, but since they were at an angle, it will actually be a greater distance for them to travel. Think of the problem in terms of three vectors: the vector from the start to the station (1), the vector from the start to where they stopped after going 1.91 km (2), and the vector from there to the station (3 -- the one you're solving for). Vectors 2 and 3 must add up to vector 1, and you know the components of vectors 1 and 2. Try solving for the components of vector 3 from there.
Thanks for the help! I seem to be having another problem arise with my original magnitude I found. I see why 4.58km wouldn't work in this situation, but the program we use for homework is counting this correct. Any ideas why? I feel like my magnitude should be larger.

Also, the angle to get from where they stopped after going 1.91km to the final destination is completely confusing me. At this point, I'm not sure how to even begin :confused:
 
  • #6
Nyxious said:
That was actually the first thing I did. That's actually how I kinda assumed the magnitude to be 4.58km. The angle however, is a mystery to me.

The equation you wrote down does not match what you say you did though.
 
  • #7
jbriggs444 said:
The equation you wrote down does not match what you say you did though.
I know, i said the way I did it did not work with getting the magnitude. I went back and found a magnitude of 8.269km which was incorrect, and then i tried subtracting the components and got 4.8120km, which is close to what i got when I just subtracted the final destination (6.49km) and the distance already traveled (1.91km), which was 4.58km. Apparently the site is counting the 4.58km right, which I'm not sure why. The angle is still a problem also.
 
  • #8
Okay, I went back and did everything again and here is what I got:
C (or the distance remaining to get to the research lab) = 4.8120km
The angle = 47.9°

Can someone check to see if they got the same or if I'm still missing something?EDIT: Got it correct. Thanks for the help guys!
 
Last edited:

1. What is a displacement vector?

A displacement vector is a mathematical representation of the movement of an object from one point to another. It includes both the magnitude (size or length) of the displacement and the direction in which the object moved. It is typically denoted by an arrow pointing from the initial position to the final position of the object.

2. How do you calculate the displacement vector?

The displacement vector can be calculated by subtracting the initial position vector from the final position vector. This means finding the difference between the x-coordinates and y-coordinates of the two points. The resulting vector will have a magnitude and direction that represents the displacement of the object.

3. What is the difference between displacement vector and distance?

Displacement vector and distance are two different concepts. Distance is a scalar quantity that measures the total length of the path traveled by an object, while displacement vector is a vector quantity that represents the shortest distance between the initial and final positions of the object. Distance is always positive, but displacement can be positive, negative, or zero depending on the direction of movement.

4. How do you represent displacement vector graphically?

Displacement vectors can be represented graphically by drawing arrows on a coordinate plane. The length of the arrow represents the magnitude of the displacement, and the direction of the arrow represents the direction of movement. The initial point of the arrow is placed at the initial position of the object, and the final point of the arrow is placed at the final position of the object.

5. What are some real-life examples of displacement vector problems?

Displacement vector problems can be seen in many real-life situations, such as a car traveling from one location to another, a person walking to a destination, or a ball being thrown into the air. In each of these examples, the displacement vector would represent the movement of the object from its initial position to its final position.

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