Perplexed (Point particle in a Uniform Circular Motion

• CrimsonFirebird

  28 Feb 13, 23:38

  Good evening guys, CrimsonFirebird here. You do not need to know any of my background, only that I have just recently completed NS, A level 2 years ago and is currently reviewing what I have learnt during my JC days.

  I would need a second opinion on this question.
  
  
  A particle moves in the xy plane in a circle centered on the origin. At a certain instant the velocity and acceleration of the particle are 6.0 m/s and (3.0 + 4.0) m/s². What are the x and y coordinates of the particle at this moment?

  If you consider the vectors of both the accleration and the velocity, the accleration vector is not perpendicular to the velocity vector. 

  Hence, this question stumped me. Would need to see how other minds tackle this question.

  Thanks for letting me post and thanks in advance for the solutions.

  Regards

  CrimsonFirebird

   

  PS: For additional information, this is a multiple choice question and these are the following answers. Note that I do not believe in the philosophy of elimating answer and hence I did not consider it in the main section of the letter.
  

  	x = 0, y = -9.0 m
  		x = 0, y = +7.2 m
  		x = 0, y = +9.0 m
  		x = 0, y = -7.2 m
  		x = 6.0 m, y = -9.0 m

   

• CrimsonFirebird

  3 Mar 13, 00:32

  Good evening guys,

  I think I answered my question. The key word is the word "instant". They could have rephased the question better imo
  
  Thanks again for reading. Btw the answer is (0,9.0m)

• eagle

  6 Mar 13, 01:43

  Actually the question is ok. The acceleration (resultant) is not perpendicular to the velocity because this is a non-uniform circular motion.

  Since the velocity is 6i, hence we can conclude that x = 0 since that's the only location where velocity is perpendicular to the circle.

  Since velocity is at 6i, we can conclude that 4j is the centripetal acceleration, and 3i is the constant acceleration towards the right. 4j points upwards, so logically, y should be at -r

  Using a = v^2/r, r  = v^2 / a = 36 / 4 = 9
  Hence y should be at -9

  My answer is thus (0, -9) and not (0, 9).

• CrimsonFirebird

  8 Mar 13, 02:11

  Hey eagle,

  Thanks for replying. Thanks for the correction; oversight of mine.

  CrimsonFirebird.