The position of a moving particle in space is vgiven by r(t) = (- 76- 21) i + (108t° - 2890) j + (181 + 11) k t20. (i) Determine position , velocity , acceleration and speed of the particle at t=2. (ii) Find the vector equation of the straight line tangent to the curve at the point Q(- 82,26,173). () r(2)=(D.0.J) v (2) = (O.O.0) a (2) = (O.0.J) Speed = || v (2)|| = (ii) Equation of thangent line r (s) = (- 82,26,173)+( SE R

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A: We have to find x

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A:

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A: NOTE: Refresh your page if you can't see any equations. . here we have

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Q: refer  to  image  below.

A:

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A: Solution