Simple Vector Calculations: Man Walking, Sign Lifting, and Jogging Displacement

  • Thread starter Thread starter FallenAngel
  • Start date Start date
  • Tags Tags
    Vectors
AI Thread Summary
The discussion revolves around vector calculations related to physics homework involving a man walking, a worker lifting a sign, and a jogger's displacement. Participants emphasize the importance of drawing a coordinate system with labeled axes to represent the vectors accurately. To calculate displacement, the magnitude can be found using the Pythagorean theorem, while direction is determined using trigonometric functions. There is frustration expressed over a lack of responses and the urgency of the homework deadline, along with reminders that effort must be shown to receive help. Overall, the thread highlights the need for foundational understanding in physics to tackle vector problems effectively.
FallenAngel
Messages
5
Reaction score
0
I am new to the US and I have taken physics and I need a little help.

For each vector given, draw a co-ordinate system and sketch the vector on a separate sheet of paper. be sure to label the angle that the vector makes with axis.

a: a man walks 800 meters in a direction of 20 degrees west of north

b: a worker lifts a sign by exerting a force along a rope of 250N at an angle of 50 degrees with respect to the ground.

----------------------------------------------------
a jogger runs 300 meters due east and then turns & runs 500 meters due south.

what is the total distance that she ran?

800m, yes?

so then it asks me to calculate the displacement [magnitude & direction]


how do i calculate magnitude and direction?
 
Physics news on Phys.org
If you have a vector (x,y), its magnitude is \sqrt{x^2 + y^2}, and the angle of orientation relative to the positive x-axis is \arctan{\frac{y}{x}}. (Use the signs of y and x to figure out what quadrant you are in)
 
Last edited:
In the first last example I got 583 meters due South East. It is easiest to break up the vectors into their X, and Y components. Essentially what you are doing is using the Distance formula. You usually want to add vectors when wanting to know a change in position, or change in velocity.
 
  • Like
Likes Schoolworl
...what?

I'm so lost. :(

X And Y? where?

What is a co - ordinate system.

Is it the same as drawing graphs?


>< I'm lost
 
Last edited by a moderator:
44 views and NO ONE has said anything?

This homework is for TOMORROW, FIRST PERIOD. :(
 
FallenAngel said:
44 views and NO ONE has said anything?

This homework is for TOMORROW, FIRST PERIOD. :(

How is it anyone's fault here that you've not done your homework? Firstly, you should've posted in the homework forums. Secondly, you have to show some effort before we can help.

By "draw a coordinate system" it means draw an x and y axis. Then draw the respective vectors on the diagrams, noting the angle to either the x or y axis.
 
Last edited:
I know nothing about physics I haven't done any of this before and I didn't know there were homework forums. - Now I do, but I didn't realize I had to post it there...
 
Last edited:
FallenAngel said:
I know nothing about physics I haven't done any of this before and I didn't know there were homework forums. - Now I do, but I didn't realize I had to post it there...

Well, your post has been moved now, so not to worry.

Have you made any attempts at the questions? For the first two questions have you tried setting up a coordinate system, like I mentioned above, and plotting the vectors? For the third question, yes, the total distance covered is 800m. To find the displacement, you will need to set up a coordinate system, and draw the two distance vectors. The resultant of these (the hypotenuse of the right angled triangle made) is the displacement. Its magnitude can be find by Pythagoras, and its direction by trigonometry.
 
A coordinate system is the x and y axis.
http://publib.boulder.ibm.com/infocenter/printer/v1r1/topic/com.ibm.printers.industrial/reference/hb9pr013.gif


for this part:

a: a man walks 800 meters in a direction of 20 degrees west of north

you can think as the positive y-axis as north, the positive x-axis as east, the negative y-axis as south and the west axis as west.
http://www.uscg.mil/diversity/COMPASS/Compass_pin.JPG

That was a quick pic I got from online.

So you travel 800 meters or units if you prefer, to the west of north. North is the positive y axis, and you go to the west at 20 degrees.

The rest are similar, and for the resultant, a previous member responsed to that part.

Sorry I did not read this thread in time.
 
  • #10
FallenAngel said:
I know nothing about physics I haven't done any of this before and I didn't know there were homework forums. - Now I do, but I didn't realize I had to post it there...
i've never taken physics either, work harddd, don't start to make excuses so early into the semester ... or ever!

if you need help, ask but show some work b4-hand.
 
Back
Top