Query related to Two-Dimensional Motion

  • #1
Gargi
5
0
Homework Statement
A particle moves in the X-Y plane with a constant acceleration of 1.5 m/s^2 in the direction making an angle of 37° with the X-axis. At t = 0, the particle is at the origin and its velocity is 8.0 m/s along the X-axis. Find the velocity and the position of the particle at t = 4.0 s. The graph corresponding to the question is attached herewith.
Relevant Equations
v = u + at.
v² = u² + 2as.
s = ut + ½at²
My initial approach to this question was breaking the components of acceleration in the x and y axes and applying the three equations of motion to find the final velocity as well as the final position. As we were expected to find the net final velocity of the particle, I found the resultant of the two components of final velocity in both the axes. This led to some complex solving of roots.

Alternate approach which went wrong: I broke the component of the initial velocity in the direction of the acceleration and find the final velocity in the direction of acceleration. The final velocity achieved using this did not match the right answer.

Please tell me what is wrong with my approach. Please tell me in the comments regarding any further requirements.
 

Attachments

  • Capture1.PNG
    Capture1.PNG
    1.9 KB · Views: 6
Physics news on Phys.org
  • #2
Gargi said:
Homework Statement: A particle moves in the X-Y plane with a constant acceleration of 1.5 m/s^2 in the direction making an angle of 37° with the X-axis. At t = 0, the particle is at the origin and its velocity is 8.0 m/s along the X-axis. Find the velocity and the position of the particle at t = 4.0 s. The graph corresponding to the question is attached herewith.
Relevant Equations: v = u + at.
v² = u² + 2as.
s = ut + ½at²

My initial approach to this question was breaking the components of acceleration in the x and y axes and applying the three equations of motion to find the final velocity as well as the final position. As we were expected to find the net final velocity of the particle, I found the resultant of the two components of final velocity in both the axes. This led to some complex solving of roots.

Alternate approach which went wrong: I broke the component of the initial velocity in the direction of the acceleration and find the final velocity in the direction of acceleration. The final velocity achieved using this did not match the right answer.

Please tell me what is wrong with my approach. Please tell me in the comments regarding any further requirements.
Share your calculations, and please use latex for the formatted mathematics. See LaTeX Guide
 
  • #3
Gargi said:
I broke the component of the initial velocity in the direction of the acceleration and find the final velocity in the direction of acceleration. The final velocity achieved using this did not match the right answer.
What about the other component of the initial velocity?
 
  • Like
Likes erobz
  • #4
Either approach is correct. The problem is probably with their implementation to which we have no access. As @erobz said, please post your work.
 
  • #5
Sure, I'll share my working:
First picture (longer one), shows my initial redundant approach. The second picture shows my alternate approach.
 

Attachments

  • 20240403_003042.jpg
    20240403_003042.jpg
    40.9 KB · Views: 6
  • 20240403_003054.jpg
    20240403_003054.jpg
    47.6 KB · Views: 6
  • #6
Your attempt on the left is correct and I agree with your numbers.
Your attempt on the left correctly calculates the velocity change in the direction of the acceleration. However it does not add to that the unchanging velocity component in the direction perpendicular to the acceleration. @haruspex noted that in post #3.
 
  • Like
Likes erobz
  • #7
kuruman said:
Your attempt on the left is correct and I agree with your numbers.
Your attempt on the left correctly calculates the velocity change in the direction of the acceleration. However it does not add to that the unchanging velocity component in the direction perpendicular to the acceleration. @haruspex noted that in post #3.
So I need to find a resultant value in the latter approach as well considering the unchanged component of velocity too?
Can we avoid roots by any other means, any other method?
 
  • #8
Gargi said:
So I need to find a resultant value in the latter approach as well considering the unchanged component of velocity too?
Yes
Gargi said:
Can we avoid roots by any other means, any other method?
Do you mean avoid square roots?
 
  • Like
Likes Gargi
  • #9
Gargi said:
This led to some complex solving of roots.
There is no need for messing with square roots. The problem is asking for two vectors: velocity and position. Since the problem does not ask you specifically for an angle and a magnitude, you only need to find their ##x## and ##y## components. That's what you did in your first method, which is the quickest way to solve this problem. Finding the speed at ##t=4## s is unnecessary. Just write
##\mathbf v=(12.8~\mathbf{\hat x}+3.61~\mathbf{\hat y})~##m/s for the velocity at ##t=4~##s
##\mathbf r=(41.6~\mathbf{\hat x}+7.22~\mathbf{\hat y})~\text{m}## for the position at ##t=4~##s.

The second method is longer because you have to mess with square roots and you still have to add the components of the vectors in the direction perpendicular to the acceleration, a step that you forgot to include.
 
  • Like
Likes Gargi
  • #10
"an angle of 37°"
Handy fact: that angle is often specified because it very nearly the acute angle in a 3-4-5 right-angled triangle.
 
  • Like
Likes Gargi

Similar threads

Replies
5
Views
787
  • Introductory Physics Homework Help
Replies
6
Views
952
  • Introductory Physics Homework Help
Replies
8
Views
541
  • Introductory Physics Homework Help
2
Replies
56
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
954
  • Introductory Physics Homework Help
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
873
  • Introductory Physics Homework Help
Replies
6
Views
947
Back
Top