Particles moving along the axes.

  • Thread starter Thread starter niyati
  • Start date Start date
  • Tags Tags
    Axes Particles
Click For Summary
SUMMARY

Particle 1 accelerates at 5.45 m/s² along the x-axis, while Particle 2 accelerates at 7.56 m/s² along the negative y-axis, both starting from rest at the origin. After 6.15 seconds, the velocity of Particle 1 is calculated to be 33.5175 m/s in the positive x-direction, and Particle 2's velocity is 46.494 m/s in the negative y-direction. The relative speed of Particle 1 with respect to Particle 2 is determined using the Pythagorean theorem, yielding a speed of approximately 55.0 m/s.

PREREQUISITES
  • Understanding of kinematics and acceleration
  • Knowledge of vector addition and subtraction
  • Familiarity with the Pythagorean theorem
  • Basic concepts of relative motion
NEXT STEPS
  • Study vector addition in two dimensions
  • Learn about relative velocity in physics
  • Explore kinematic equations for uniformly accelerated motion
  • Investigate graphical methods for vector representation
USEFUL FOR

Physics students, educators, and anyone interested in understanding motion dynamics and relative velocity in two-dimensional space.

niyati
Messages
63
Reaction score
0
Particle 1 is moving on the x-axis with an acceleration of 5.45 m/s^2 in the positive x-direction. Particle 2 is moving on the y-axis with an acceleration of 7.56 m/s^2 in the negative y-direction. Both particles were at rest at the origin at t = 0 s. Find the speed of particle 1 with respect to particle 2 at 6.15 s. Answer in units of m/s.

I'm not sure what the question is asking for. The whole "with respect to" slightly confuses me. In the x-direction, the velocity after 6.15 is 33.5175. In the negative y-direction, it is 46.494. ...am I just suppose to draw the hypotenuse to the triangle formed and find that length?
 
Physics news on Phys.org
Yes. The velocity of #1 with respect to #2 is [tex]\vec{v}_1-\vec{v}_2[/tex]. Since [tex]\vec{v}_1[/tex] is perpendicular to [tex]\vec{v}_2[/tex], the speed (which is the magnitude of [tex]\vec{v}_1-\vec{v}_2[/tex]) is [tex](v_1^2+v_2^2)^{\scriptstyle{1/2}}[/tex].
 
Thank you!
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
Replies
2
Views
1K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
3
Views
1K
Replies
13
Views
6K
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K