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Mechanics - Motion with constant Acceleration - Immediate help?

Two particles, X and Y, are moving on parallel horizontal tracks. At time t=0 both particles pass O (a fixed point). Both particles moves to the left of O. Velocity of X = 3 m/s and Acceleration of X = 5 m/s^2.

Velocity of Y = 7 m/s and Acceleration of Y = 2 m/s^2.

FIND;-

(1) the velocity of Y at the instant when X changes direction of motion.

My answer is 8.2 m/s. I found that X changes direction when t=(3/5) seconds.

(2) the time that elapses, from t=0, before both particles have the same velocity.

My thoughts:

I tried to sketch a velocity time graph for both particles. I guess for the two particles to have the same velocity, they should intersect. But then, particle Y is moving to the left and particle X to the right(changes direction of motion).

I solved for t by equating 2 equations. 3+5t = 7+2t , t = (4/3) but answers says (10/3) seconds.

I can't figure it out.

(3) the time that elapses from t=0, before X overtakes Y ???????????????????????

Please share all your valuable inputs/ideas.

THANKS IN ADVANCE.

1 Answer

Relevance
  • 7 years ago

    Sorry, but I don't think this one is going to fly!

    a) It looks like the acceleration of X must be negative - but that still doesn't give the right answer for (2)

    b) As I think you've seen, X never does overtake Y, whether one or both accelerations are negative.

    All I can suggest is that you go back to the original question and check for negative signs and decimal points.

    Sorry that's not the answer you were looking for.

    Source(s): Retired graduate engineer.
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