"Describe the motion..." mechanics question (1 Viewer)

[catha230]

Hi all, I am a bit confused by the last part of the following question. I wonder whether the velocity will ever each 0 or not. The thing is v=cos^2(x) which means when x=pi/2 v=0. However, the expression for x in terms of t is x=tan^-1(t) which means that each will never reach pi/2? If that is the case then v will never reach zero? However the restriction for x is between 0 and pi/2 which means x can be pi/2 but if so then what will t be? infinity? but t can never physically be infinity,,,
Sorry if the paragraph above does not make much sense. I actually don't know what I am trying to say lol.

Thank you so much for the help!

 

[fan96]

Your reasoning is correct.

Another way to look at it is the following:

If is any point that is not , then the particle will start off with a positive velocity, which means its displacement will increase.
But when its displacement increases, its velocity () gets smaller (because decreases from to (monotonically) on the interval ).

So the particle will always be moving forward and slowing down but should never actually come to a stop.

(The case where is is different and I leave that to you to consider...)

 

[Drongoski]

To elaborate:

When t=0 (start), x = 0 and v = cos^2(0) = 1 ms^(-1)
The acceleration -2cos^3(x)sinx = -cos^2(x)sin2x is always negative for the range of x
So the particle is continually slowing down(decelerating); the negative acceleration is also tending to 0, but does not attain 0.
So in the long run, the velocity tends to 0, but never quite attains 0, i.e. doesn't come to a stop. But in reality, after a few minutes, the particle for all intent and purposes has come to a (virtual) stop.

 

[catha230]

Your reasoning is correct.

Another way to look at it is the following:

If is any point that is not , then the particle will start off with a positive velocity, which means its displacement will increase.
But when its displacement increases, its velocity () gets smaller (because decreases from to (monotonically) on the interval ).

So the particle will always be moving forward and slowing down but should never actually come to a stop.

(The case where is is different and I leave that to you to consider...)

Oh I understood it now. Thanks for the help!!!