The displacement of a body is given by r=√(a^2-t^2)+tcos t^2, where a is constant. Find velocity?

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The displacement of a body is given by r=√(a^2-t^2)+tcos t^2, where a is constant. Find velocity?

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Answer 1 · 2018-05-11T12:18:44

$- \frac{t}{{(a^{2} - t^{2})}^{\frac{1}{2}}} - 2 t^{2} sin (t^{2}) + cos (t^{2})$ $color(blue)(-t/((a^2-t^2)^(1/2))-2t^2sin(t^2)+cos(t^2)$

Explanation:

The change in displacement(distance) per unit of time is velocity.

Given:

$r = \sqrt{a^{2} - t^{2}} + t cos (t^{2})$ $r=sqrt(a^2-t^2)+tcos(t^2)$

$velocity = \frac{d r}{d t}$ $"velocity"=(dr)/(dt)$

Rewriting as:

$r = {(a^{2} - t^{2})}^{\frac{1}{2}} + t cos (t^{2})$ $r=(a^2-t^2)^(1/2)+tcos(t^2)$

$\frac{d r}{d t} ({(a^{2} - t^{2})}^{\frac{1}{2}} + t cos (t^{2})) = \frac{1}{2} {(a^{2} - t^{2})}^{- \frac{1}{2}} \cdot (- 2 t) +$ $(dr)/(dt)((a^2-t^2)^(1/2)+tcos(t^2))=1/2(a^2-t^2)^(-1/2)*(-2t)+$

$\to t (- sin (t^{2}) \cdot (2 t) + cos (t^{2})$ $->t(-sin(t^2)*(2t)+cos(t^2)$

$= \frac{- 2 t}{2 {(a^{2} - t^{2})}^{\frac{1}{2}}} - 2 t^{2} sin (t^{2}) + cos (t^{2})$ $=(-2t)/(2(a^2-t^2)^(1/2))-2t^2sin(t^2)+cos(t^2)$

$= - \frac{t}{{(a^{2} - t^{2})}^{\frac{1}{2}}} - 2 t^{2} sin (t^{2}) + cos (t^{2})$ $=-t/((a^2-t^2)^(1/2))-2t^2sin(t^2)+cos(t^2)$

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The displacement of a body is given by r=√(a^2-t^2)+tcos t^2, where a is constant. Find velocity?

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Explanation:

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