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polet [3.4K]
3 years ago
6

Vector Calculus

Mathematics
1 answer:
djyliett [7]3 years ago
4 0

It seems like <em>σ</em><em>(t)</em> is the positive function of the particle. Compute its acceleration by differentiating this function twice.

<em>σ</em><em>(t)</em> = {<em>t </em>², sin(<em>t </em>), cos(<em>t</em> )}

<em>σ</em><em>'(t)</em> = {2<em>t</em>, cos(<em>t</em> ), -sin(<em>t</em> )}

<em>σ</em><em>''(t)</em> = {2, -sin(<em>t</em> ), -cos(<em>t</em> )}

When <em>t</em> = 0, the acceleration on the particle is

<em>σ</em><em>''</em> (0) = {2, -sin(0), -cos(0)} = {2, 0, -1}

Then the force acting on the particle at time <em>t</em> = 0 is

<em>F</em> = {2<em>m</em>, 0, -<em>m</em>}

by Newton's second law.

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