Query related to Two-Dimensional Motion

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Homework Help Overview

The discussion revolves around a problem in two-dimensional motion involving a particle moving in the X-Y plane with a constant acceleration. The original poster attempts to find the final velocity and position of the particle after a specified time, using the equations of motion and component analysis of velocity and acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss breaking down the components of acceleration and velocity, applying the equations of motion, and finding resultant velocities. Questions arise regarding the handling of components in directions perpendicular to acceleration and the necessity of calculating angles and magnitudes.

Discussion Status

Some participants provide feedback on the original poster's approaches, suggesting that while the methods may be valid, issues may lie in their implementation. There is a recognition of the need to consider both components of velocity and the potential for simplifying the problem without complex calculations.

Contextual Notes

Participants note that the problem does not explicitly require the angle and magnitude of the resultant vectors, focusing instead on the x and y components. There is also mention of the specific angle given in the problem, which relates to a common geometric triangle.

Gargi
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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.
 

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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
 
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?
 
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Either approach is correct. The problem is probably with their implementation to which we have no access. As @erobz said, please post your work.
 
Sure, I'll share my working:
First picture (longer one), shows my initial redundant approach. The second picture shows my alternate approach.
 

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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.
 
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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?
 
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?
 
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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.
 
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  • #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.
 
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