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Direction when the magnitude of the instantaneous velocity is 0 
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#1
Jul111, 10:46 PM

P: 4

Hi guys, sorry to ask such a basic question, but I'm studying for the MCAT and need to get my fundamentals down!
Anyway, my question is this: Consider an object which has been thrown straight up into the air. It will rise, then at the very top of its ascent, it changes direction and comes back down. Now, I'm pretty sure that at the moment when the object reaches its maximum height, the velocity becomes zero, although there must still be an acceleration (due to gravity), or else the object wouldn't fall back down again. So, on a velocitytime graph, we would see a point where the velocity drops to zero, corresponding to the moment the object reaches its maximum height, right? Let's call this point q. But I am confused, because, since velocity is a vector, it has a direction. In this case, the direction would be up as we approach q from the left, and down as we approach q from the right. So what would the direction of the velocity vector be, when we are actually at that point? I am tempted to say that, since the magnitude of the velocity vector is 0m/s, the object isn't actually moving in *any* direction. But if that is the case, then *when* does the direction of the object officially change from up to down? Again, sorry to ask such a newbie question. If someone could elaborate on the behavior of the object at this point I would seriously appreciate it! 


#2
Jul111, 10:52 PM

P: 351




#3
Jul111, 11:16 PM

P: 351

Keep in mind that the velocity changes magnitude, but the acceleration is constant. The only thing it does is change its direction. The velocity magnitude gets smaller and smaller, but not the acceleration.



#4
Jul211, 06:21 AM

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Direction when the magnitude of the instantaneous velocity is 0
The acceleration never changes direction. It always points down.
A vector of zero length can point in any direction. 


#5
Jul211, 09:45 AM

P: 18

Vanadium is absolutely correct.A zero vector has indefinite direction



#6
Jul211, 09:50 AM

P: 351

Yes, I've got it backwards. I think this is a discontinuity question that doesn't have a satisfying answer. I guess Newton would have said when you move infinitesimally to the right in your graph. But infinitesimal is not zero.



#7
Jul211, 11:34 AM

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There is no discontinuity here. Position is continuous. Velocity is continuous. Acceleration is not only continuous, it's constant.



#8
Jul211, 12:02 PM

P: 351

The velocity's function is continuous, but not its vector.
Crudely : velocity: [itex]\cup[/itex] vector: [itex]\uparrow[/itex] ∅ [itex]\downarrow[/itex] None of this is getting at the questioner's problem. There is at least a psychological discontinuity. The velocity vector was up, then it was indefinite, then it was down. 


#9
Jul211, 04:59 PM

P: 1,403

It will of course remain continuous if you express it in spherical coordinates. [tex] (r(t),\phi(t),\theta(t)) [/tex] but [itex] \phi(t) [/itex] and [itex] \theta(t) [/itex], are not continous functions of t. I don't see there is any problem with that. 


#10
Jul211, 05:45 PM

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P: 1,503

All functions in question here are continuous and differentiable on an open interval. There is no discontinuity. You could easily plot all three and they would all be smooth curves with no discontinuities or infinities or any sort (assuming the ball hits the ground at some point). The direction of a zero vector is indefinite but not undefined. It simply doesn't have a direction because it isn't moving.



#11
Jul211, 08:40 PM

P: 4

Thanks for the help everyone! So from reading everyone's comments, I'm thinking that it works like this: No matter how close you get to the left of the point in time where the velocity becomes 0, the velocity vector points up. Then at this point, the direction changes to indefinite. Then infinitely close to this point, but on the right side, the direction changes to down. I was confused for the reason that danR said, this is kind of a confusing problem because the direction of the velocity vector seems to be a discrete quality of the vector, which abruptly changes at the point where the velocity becomes 0. Think about the limits on either side of this point. From the left, the limiting values would approach 0, but with a direction of up. Whereas from the right, the limiting values approach 0, but with a direction of down. This is why I was confused!



#12
Jul211, 09:12 PM

P: 454

Hi.
At time t=0 the object is at the top q where where position x = 0. Let tau be a small positive amount of time. t=tau; velocity v = g tau, x= 1/2 g tau^2. t=0; v = 0, x=0 t= tau; v = g tau, x= 1/2 g tau^2. I find here nothing mysterious. Regards. 


#13
Jul311, 12:08 AM

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#14
Jul311, 12:28 AM

P: 4

Haha true enough! Thanks guys. I totally get it. Like I said, its been a while since I took a physics course!



#15
Jul311, 10:53 AM

P: 351

I will concede all that. The terms 'no direction', 'any direction', 'null direction' are rather confusing to me, but if it's called '0 direction', I have something numerical to specify in a vector operation. But computer programs have failed on 'discontinuous' shifts in wind direction gusts, for example, when vectors are involved.



#16
Jul311, 04:17 PM

HW Helper
P: 7,169

You could choose to define the vector's direction as always upwards, and then the vectors velocity would transition from positive to zero to negative.



#17
Jul311, 04:58 PM

Sci Advisor
P: 2,950

I think a better way to visualize this is to visualize the vector itself. It gets smaller and smaller and smaller and eventually disappears and then gets bigger again but going down. You will see that the vector smoothly transitions from pointing up to pointing down.



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