Question

I tried getting help in the engineering section, but no one's around to help, apparently, so I'll just copy/paste my question here.

I've gotten about halfway through this one question and I'm a little confused as to what to do next. Is there anyone out there who can help me? The question is thus; "The motion of a particle is defined by the equations and x=(2t+t2)m and y=(t2)m, where t is in seconds. Find v_n, v_t, a_n and a_t where (t=2) seconds."

Answer 1

I've already found v_n and v_t, I want to know how to obtain a_n and a_t. I know the general equation for polar acceleration, but it doesn't seem to be all that applicable here. So I've been working on this problem for awhile. Just to prove that I'm not trying to be lazy, I'll explain how I found v_n and v_t. The definition of velocity is that it is tangent to the position curve. Theregore, there can BE no normal velocity vector. Thus, v_n = 0. As for v_t, that is just velocity times the unit tangent vector. To find this, I derive the x and y components of the position vector, and then plug in the value t= 2, from which I then calculate the magnitude of the velocity. This gets me v_t = 7.21 m/s.

Answer 2

I figured math could be a good place to look for help since a lot of peple are online, and a lot of this involves differential equations anyway.

Answer 3

Is anyone there?

Answer 4

Yes, hello. I suppose that the subscripts "n" and "t" stand for normal and tangential respectively.

Answer 5

That is correct!

Answer 6

Also, thanks for coming to help. I've kind of been lurking here for a couple of hours waiting for someone to come. Figured it was time I started actively hunting people down.

Answer 7

Are you still there?

Answer 8

Yes, I'm reading through the passage you wrote above.

Answer 9

Oh. Sorry. Carry on.

Answer 10

@ParthKohli ?

Answer 11

Sorry, my connection has been shabby.

Answer 12

If I recall correctly,\[a_T = \frac{d}{dt}v(t)\]

Answer 13

Right.

Answer 14

\[v(t) = (2 + 2t) \hat i + (2t) \hat j \]Now the total acceleration is \(\vec a = 2\hat i + 2 \hat j\) but we have to find its component along the velocity vector at \(t=2\), which is \(6\hat i + 4 \hat j\), to find the tangential acceleration.

I'm not quite sure if I've got my facts right, so do check.

Answer 15

How'd you come up with a= 2i + 2j as total acceleration? Or that value along the velocity vector for that matter?

Answer 16

A better way would be to find the speed function because we know what the velocity is, and then differentiate that to obtain the expression for tangential acc.\[v(t) = \sqrt{(2+2t)^2 + (2t)^2} = \sqrt{8t^2 + 8t + 4}\]BTW, the total acceleration was obtained by differentiating the velocity vector wrt time. And the velocity vector at \(t=2\) is \(((2+2t)\hat i + (2t)\hat j) _{t=2} = 6 \hat i + 4 \hat j\).

Answer 17

I hope that makes sense... do comment if there's anything I left out :)

Answer 18

So if that's the magnitude of acceleration, I'm guessing that means a_t = 2 and a_n = 2 m/s^2? Have I been overthinking it this whole time?

Answer 19

No, wait...that made no sense. Also, I'm wrong.

Answer 20

Okay, yeah. I get how you got total acceleration. Now I'm just trying to figure out how you go from there to the tangential and normal components.

Answer 21

OK, here's a good way to think about it:

We know that tangential acceleration is along the tangent of the path. And normal acceleration is the component that is perpendicular to that.

Now convince yourself that the velocity vector is always tangent to the path.

And that means that tangential acceleration is along the velocity vector, and the normal acceleration is perpendicular to that.

Answer 22

Okay, so...tangential acceleration is the velocity along the unit vector (so basically just the value for velocity) and normal acceleration is, what...cross product?

Answer 23

Ah, no.