anonymous
  • anonymous
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."
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
Is anyone there?

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ParthKohli
  • ParthKohli
Yes, hello. I suppose that the subscripts "n" and "t" stand for normal and tangential respectively.
anonymous
  • anonymous
That is correct!
anonymous
  • anonymous
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.
anonymous
  • anonymous
Are you still there?
ParthKohli
  • ParthKohli
Yes, I'm reading through the passage you wrote above.
anonymous
  • anonymous
Oh. Sorry. Carry on.
anonymous
  • anonymous
ParthKohli
  • ParthKohli
Sorry, my connection has been shabby.
ParthKohli
  • ParthKohli
If I recall correctly,\[a_T = \frac{d}{dt}v(t)\]
anonymous
  • anonymous
Right.
ParthKohli
  • ParthKohli
\[v(t) = (2 + 2t) \hat i + (2t) \hat j \]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.
anonymous
  • anonymous
How'd you come up with a= 2i + 2j as total acceleration? Or that value along the velocity vector for that matter?
ParthKohli
  • ParthKohli
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 \(((2+2t)\hat i + (2t)\hat j) _{t=2} = 6 \hat i + 4 \hat j\).
ParthKohli
  • ParthKohli
I hope that makes sense... do comment if there's anything I left out :)
anonymous
  • anonymous
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?
anonymous
  • anonymous
No, wait...that made no sense. Also, I'm wrong.
anonymous
  • anonymous
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.
ParthKohli
  • ParthKohli
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.
anonymous
  • anonymous
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?
ParthKohli
  • ParthKohli
Ah, no. |dw:1442908978829:dw|
ParthKohli
  • ParthKohli
\(a_T \) is along the velocity, and \(a_N\) is perpendicular to that.
ParthKohli
  • ParthKohli
That dotted line from \(a\) to \(a_T\) should be perpendicular, so I drew it a little wrongly.
ParthKohli
  • ParthKohli
OK, so what's the definition of tangential and normal acceleration that you know?
anonymous
  • anonymous
Tangential acceleration is related to the rate of change of the object's speed; \[a = \frac{ dv }{ dt} u _{t}\] And normal acceleration is related to the change in the object's direction; \[a _{n} = v\frac{ d \theta }{ dt } u _{n} = \frac{ v^2 }{ \rho } u _{n}\], where \[\rho = \frac{ dr }{ dtheta }\], as rho is the radius of curvature. Thus, the total acceleration is \[a = a _{t} u _{t} + a _{{n}} u _{n} = \frac{ dv }{ dt } u _{t} + \frac{ v^2 }{ \rho } u _{n}\]
anonymous
  • anonymous
BTW, that took forever to write using that equation button function.
ParthKohli
  • ParthKohli
Wow... Instead of separately applying the definition of normal acceleration, you should just find \(\vec a_T\) and then use \(\vec a_T + \vec a_N = \vec a\) which applies because they're, well, components of acceleration.
anonymous
  • anonymous
And that would be doable; if I could find out what \[a _{t}\] and \[a\] actually ARE. So now we're just right back where I started.
ParthKohli
  • ParthKohli
\(a_T\) is the component of acceleration in the direction of \( \vec v\) as I drew in that diagram. Here, \(\vec v = 6 \hat i + 4 \hat j\) at \(t =2 \) and \(\vec a = 2\hat i + 2 \hat j\) as we calculated. To find the component of \(\vec a = 2 \hat i + 2 \hat j\) along \( \vec v = 6 \hat i + 6 \hat j\), we just find \(|\vec a| \cos \theta = | \vec a | \frac{\vec a \cdot \vec v}{|\vec a||\vec v|} = \frac{\vec a \cdot \vec v}{| \vec v | }\).
ParthKohli
  • ParthKohli
So the magnitude of the component of \(\vec a\) along \( \vec v\) is\[\frac{(2 \hat i + 2 \hat j)\cdot (6 \hat i + 4 \hat j)}{|6\hat i + 4 \hat j|}\]\[= \frac{20}{\sqrt{52}} = \frac{ 10}{\sqrt {13}}\]If we multiply that by \(\hat v \), that is the unit vector in the direction of \(\vec v\) as well as \(\vec a_T\), then we get \(\vec a_T\) in vector-form.
anonymous
  • anonymous
Okay, so multiply 2.774 by the unit vector in direction of v, and then take the magnitude of THAT.
anonymous
  • anonymous
Make that multiplying 1.387 by the unit vector. I didn't see how you reduced. Whoops.
ParthKohli
  • ParthKohli
\[\hat v = \frac{3}{\sqrt{13}} \hat i + \frac{2}{\sqrt{13}}\hat j\]So\[\vec a_T = a_T \cdot \hat v = \frac{10}{\sqrt{13}} \left(\frac{3}{\sqrt{13}}\hat i + \frac{2}{\sqrt{13}}\hat j \right)\]
ParthKohli
  • ParthKohli
That \(10/\sqrt{13}\) I found earlier is in the magnitude of the tangential acceleration. Now I know that tangential acceleration is always along the velocity vector, so I multiplied it by the direction vector of velocity to get \(a_T\) as a vector.
ParthKohli
  • ParthKohli
Now \(\hat a = 2 \hat i + 2 \hat j \) and \(\vec a_T = \frac{30}{13}\hat i + \frac{20}{13}\hat j \) so we now use \(\vec a_N + \vec a_T = \vec a\)
anonymous
  • anonymous
Okay, so now I just plug in the tangential acceleration and the total acceleration and solve for normal.
anonymous
  • anonymous
Well, I want magnitude of \[a _{n}\] and \[a _{t}\] for my answers, so I suppose if I use the magnitude of a an plug in my two valuse I could solve for a\[a _{n}\]. My answers keep getting in the range of 0.572-0.579. Mastering is telling me that I'm really close, but I don't know where I'm off and it won't give me credit due to some rounding error.
anonymous
  • anonymous
Apparently, my answer WAS pretty close; the real answer for a_t was 0.555 m/s^2. Don't know where my rounding went wrong though.
anonymous
  • anonymous
Anyway, thanks for your help! I wouldn't have gotten through this problem without you! I'm going to go ahead and close this thread now.
ParthKohli
  • ParthKohli
Wow... I totally forgot about this. We did it!

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