Vectors and finding relative speed

Click For Summary
SUMMARY

The discussion focuses on calculating the speed of Particle A with respect to Particle B using vector components. Particle A has a velocity of (15i - 10j) or 18.03 m/s, while Particle B has a velocity of (5i + 15j) or 15.81 m/s. The correct approach involves subtracting the vector components directly, ignoring the angles, as the vectors are in the x-y plane. The initial attempt to calculate a k component was incorrect, leading to confusion about the problem's requirements.

PREREQUISITES
  • Understanding of vector components in two dimensions
  • Familiarity with unit vectors i and j
  • Knowledge of basic vector subtraction
  • Ability to apply the Pythagorean theorem in vector calculations
NEXT STEPS
  • Learn how to perform vector subtraction in two dimensions
  • Study the concept of relative velocity in physics
  • Explore examples of vector addition and subtraction with angles
  • Practice solving problems involving unit vectors and their applications
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and vector analysis, as well as educators looking for examples of vector calculations in two dimensions.

bluetriangle
Messages
7
Reaction score
0

Homework Statement


Find Particle A's speed with respect to Particle B.

I am given:
Particle A: (15i - 10j) or 18.03 m/s
angle A: -33.69 degrees

Particle B: (5i + 15j) or 15.81 m/s
angle B: 71.57 degrees

(Both angles with respect to x axis)

Homework Equations


Had there not been any angles, I know I would have to subtract the two vector quantities but I don't know how to solve it with angles.

The Attempt at a Solution


I know the k components, found by using Pythagorean Thereom, do I simply subtract k(A) - k(B) ?

that would be 37.88 k.
Is this correct?
 
Physics news on Phys.org
Hi Blue, and welcome to PF :-)

"Had there not been any angles": well, simply ignore them and use the vector components...

Under 2. Homework Equations , I don't see even a single equation.

Under 3. Attempt at solution, you demonstrate a severe lack of understanding the problem statement.

The ##\bf \hat\imath## and ##\bf \hat\jmath## are unit vectors in the x- and y direction, respectively. That should tell you the velocity vectors are in the x-y plane and the z component is zero. I have no idea what you think you are doing when you calculate k(A) or k(B). You don't show your work either, so it is very difficult for me to guess how you manage to come up with 37.88 k. Where does it come from ? What does it mean ? Particle A's speed with respect to particle B is 37.88 k ?

There is one way I can find 37.875, but i shudder to type it out.

If the exercise as given is too complicated, try an easier one to start with: same question for Particle A: (15i - 10j) Paricle B: (0i + 0j)
 

Similar threads

Replies
14
Views
2K
Replies
20
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
1K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
23
Views
9K
Replies
28
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K