A particle moves according to the equation $s = 1 - \frac{1}{t^{2}}$ $s=1-1/t^2$ , how do you find its acceleration?

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A particle moves according to the equation $s = 1 - \frac{1}{t^{2}}$ $s=1-1/t^2$ , how do you find its acceleration?

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Answer 1 · 2017-01-11T05:49:57

Assuming $s$ $s$ is the position function, recall that acceleration is the second derivative of position. Therefore, take two derivatives in a row to obtain the acceleration.

The first derivative gives you velocity, and the second gives you acceleration.

$\frac{d s}{d t} = \to v = \frac{2}{t^{3}}$ $(ds)/(dt) = vecv = 2/t^3$

$\to a = \frac{d^{2} s}{{d t}^{2}} = \frac{d}{d t} [\frac{d s}{d t}] = \frac{d}{d t} [\frac{2}{t^{3}}] = - \frac{6}{t^{4}}$ $color(blue)(veca) = (d^2s)/(dt^2) = d/(dt)[(ds)/(dt)] = d/(dt)[2/t^3] = color(blue)(-6/t^4)$

What acceleration does the particle have after $3 s$ $"3 s"$ have passed? Is it slowing down or speeding up?

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A particle moves according to the equation $s = 1 - \frac{1}{t^{2}}$ $s=1-1/t^2$ , how do you find its acceleration?

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A particle moves according to the equation $s = 1 - \frac{1}{t^{2}}$ $s=1-1/t^2$ , how do you find its acceleration?