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anonymous
 one year ago
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."
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 all that applicable. Either that, or I'm just an idiot and I'm missing something obvious. Could use some advice here.
anonymous
 one year ago
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." 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 all that applicable. Either that, or I'm just an idiot and I'm missing something obvious. Could use some advice here.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.0Now, I know that acceleration is \[a = \frac{ dv }{ dt } u _{t} + v \frac{ d \theta }{ dt} u _{n}\], and that \[\frac{ d \theta }{ dt } = \frac{v}{\rho} \], leading to the second term becoming \[\frac{ v^2 }{ \rho } = \frac{ v^2 }{ \frac{ ds }{ d theta} }\]. Problem is, I don't have theta, nor do I have any way of finding it. I thought maybe drawing the curve might help, but I've got nothing. Maybe if I set up x and y as a unit triangle and use a geometric property?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is anyone going to reply to this thread? I could really use a sounding board here.

Koikkara
 one year ago
Best ResponseYou've already chosen the best response.0@iambatman Could help you !

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, awesome! I didn't think anyone was reading this thread.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How do I get in contact with @iambatman ?
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