while surveying a cave, a spelunker follows a passge 180m straight west, then 210 m in a direction 45 degrees east of south and then 280m at 30.0 degrees east of north. After a fourth unmeasured displacement, she finds herself back where she started. Use vector components to find the magnitude and direction of the fourth displacement. Then check the reasonablemess of your answer with a graphical sum?
I made the drawing out for this problem but Im having a little trouble going from there
8 answers
In your drawing, did you use squared paper , and did you draw the displacements to scale? If you did, you can measure the distance from the last known movement to the starting point and let me know what it is.
i did not do it to scale
Sarah, since I am new here, I need to know a little of what you have already done at school.
What have you done in vectors?
Does P(4,5) mean anything to you?
Are you familiar with sin() and cos()?
What have you done in vectors?
Does P(4,5) mean anything to you?
Are you familiar with sin() and cos()?
kinda... i don't have a very good teacher right now so i am completely lost and need about the basics of everything ive seen the sin and cos before and know what a vector is but i don't know how to solve them
That's all I need to know. As long as I am covering something you have done at school, then it will benefit you. That's what counts.
By the way, teachers don't like to be judged "good" or "bad", you could very well say instead "I don't understand my teacher very well" without hurting anyone.
Can you start by decomposing each displacement vectors (total three of them) into the components along the x- and y-directions? I will give you an example.
A vector has to have magnitude and direction. The magnitudes of the first three have been given, namely 180,210 and 280.
Now the directions.
Generally, the cardinal directions (North, South, East and West) fit very well into the Cartesian coordinates system by placing East along the positive x-axis, and North along the y-axis.
This way, North-east will be at +45 degrees with respect to the x-axis. South-east will be at -45 degrees with respect to the x-axis, since angles are measured counter-clockwise in the Cartesian coordinates system.
Finally, since North is 90 degrees, so 30 degrees east of north will be 90-30=60 degrees.
So far so good?
It would be nice if you could use squared paper to do the following to give a better visualization of the displacements and for verification purposes. Put the origin, O(0,0) at mid-height of the paper and about 1/3 of the width from the right.
Let the starting point be at the origin, namely O(0,0).
The first displacement vector is vector OA, at 180 m. towards West (180 degrees with the positive x-axis). You can draw that point A at 180 m. to the left of the origin.
Now to move from A to B, you need to go 210 m. at -45 degrees with the x-axis. You can do that if you have a set-square, or use the squares of the squared paper to get the 45 degrees.
Can you post to let me know if you can follow all the steps so far?
By the way, teachers don't like to be judged "good" or "bad", you could very well say instead "I don't understand my teacher very well" without hurting anyone.
Can you start by decomposing each displacement vectors (total three of them) into the components along the x- and y-directions? I will give you an example.
A vector has to have magnitude and direction. The magnitudes of the first three have been given, namely 180,210 and 280.
Now the directions.
Generally, the cardinal directions (North, South, East and West) fit very well into the Cartesian coordinates system by placing East along the positive x-axis, and North along the y-axis.
This way, North-east will be at +45 degrees with respect to the x-axis. South-east will be at -45 degrees with respect to the x-axis, since angles are measured counter-clockwise in the Cartesian coordinates system.
Finally, since North is 90 degrees, so 30 degrees east of north will be 90-30=60 degrees.
So far so good?
It would be nice if you could use squared paper to do the following to give a better visualization of the displacements and for verification purposes. Put the origin, O(0,0) at mid-height of the paper and about 1/3 of the width from the right.
Let the starting point be at the origin, namely O(0,0).
The first displacement vector is vector OA, at 180 m. towards West (180 degrees with the positive x-axis). You can draw that point A at 180 m. to the left of the origin.
Now to move from A to B, you need to go 210 m. at -45 degrees with the x-axis. You can do that if you have a set-square, or use the squares of the squared paper to get the 45 degrees.
Can you post to let me know if you can follow all the steps so far?
yeah i can follow i am good so far
Sorry, I was preoccupied with a long post which I just sent out.
Good!
Continue with the third vector BC at 60 degrees with the x-axis. Measure with a protractor correctly the angle. The magnitude is 280 m.
After that, you only have to measure the vector CO, which is the final but yet unknown vector. You need to measure the length and the angle with the x-axis.
Tell me what you find.
Good!
Continue with the third vector BC at 60 degrees with the x-axis. Measure with a protractor correctly the angle. The magnitude is 280 m.
After that, you only have to measure the vector CO, which is the final but yet unknown vector. You need to measure the length and the angle with the x-axis.
Tell me what you find.
You should get a vector length of 143.55 and at an angle of 220.91 degrees with the positive x-axis.
Here's how you can do it by decomposing the vectors into the x- and y- components.
What you have done on the squared paper can be accurately calculated by making a table of calculations. We calculate the x- and y-components of each vector, and add independently the x- and the y-components. When we reach the point C, we will know how far point C is compared with the origin, thus the vector CO.
Here's the table that I have started for you:
Vector L(len.) angle dx dy
OA 180 180 -180 0
AB 210 -45 +148.49 -148.49
BC 280 60 +140 +242.49
Note:
dx = L cos(angle)
dy = L sin(angle)
From this table, you can calculate
the coordinates starting from O by adding the corresponding dx, dy.
Pt coordinates
O 0, 0
A 0-180=-180, 0+0 = 0
B -180+148.49=-31.51, 0-148.49=-148.49
C -31.51+140=108.49, -148.49+242.49=94
Can you now find the final unknown vector?
Here's how you can do it by decomposing the vectors into the x- and y- components.
What you have done on the squared paper can be accurately calculated by making a table of calculations. We calculate the x- and y-components of each vector, and add independently the x- and the y-components. When we reach the point C, we will know how far point C is compared with the origin, thus the vector CO.
Here's the table that I have started for you:
Vector L(len.) angle dx dy
OA 180 180 -180 0
AB 210 -45 +148.49 -148.49
BC 280 60 +140 +242.49
Note:
dx = L cos(angle)
dy = L sin(angle)
From this table, you can calculate
the coordinates starting from O by adding the corresponding dx, dy.
Pt coordinates
O 0, 0
A 0-180=-180, 0+0 = 0
B -180+148.49=-31.51, 0-148.49=-148.49
C -31.51+140=108.49, -148.49+242.49=94
Can you now find the final unknown vector?
1 answer