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How do you find tangential and normal velocity from a curve?

I know how to find dy/dx, but I have no idea how to obtain ut and un and dv/dt.

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closed as off-topic by Gert, tpg2114 May 14 at 21:36

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  • $\begingroup$ Hi. Could you please show some workings and your thought process. We don't answer exercise type questions here but we're happy to nudge someone who is struggling in the right direction. $\endgroup$ – Ollie113 May 14 at 13:42
  • $\begingroup$ I don't understand how to find ut. I have no idea where to start. $\endgroup$ – Yolanda Hui May 14 at 13:52
  • $\begingroup$ By definition there is no normal velocity to a curve, velocity is tangent to a curve. Acceleration on the other hand has components along and normal to the curve. $\endgroup$ – ggcg May 14 at 14:25
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For a body moving on a curve, the tangential component is along the curve. The velocity is along the curve, so tangential component is same as the velocity of body. The normal component is perpendicular to ta tangential component and is zero.

However, it may have non--zero component of tangential and normal acceleration.

Here, $u_t$ and $u_n$ are unit vectors in tangential and normal directions(can be found from dy/dx)

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  • $\begingroup$ I know how to find dy/dx, but how do you get ut and un from dy/dx at x=1? Dy/dx only gives the slope. $\endgroup$ – Yolanda Hui May 14 at 13:58
  • $\begingroup$ A unit vector making angle $\theta$ with horizontal is $cos \theta \vec i + sin\theta \vec j $ .You get $\theta $ from dy/dx $\endgroup$ – Tojrah May 14 at 14:02
  • $\begingroup$ I did that exactly, but the answer is wrong. imgur.com/a/oLAGd2y $\endgroup$ – Yolanda Hui May 14 at 14:09
  • $\begingroup$ I can't see what is wrong in your answer. $\endgroup$ – Tojrah May 14 at 14:14
  • $\begingroup$ the answer is 0.5. my answer doesn't add up to 0.5 $\endgroup$ – Yolanda Hui May 14 at 14:15

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