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A particle travels in a straight line, such that for a short time $2 s \leq t \leq 6 s$ , its motion is described by $v = (4 / a) m / s$ , where $a$ is in $m / s^{2}$ . If $v = 6 m / s$ when $t = 2 s$ , determine the particle's acceleration when $t = 3 s$ .

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$1.603 m / s^{2}$

$2.603 m / s^{2}$

$0.903 m / s^{2}$

$0.603 m / s^{2}$

answer is D.

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Detailed Solution

$a = \frac{4}{v} or \frac{d v}{d t} = \frac{4}{v} ∴ \int_{6}^{v} v d v = \int_{2}^{t} 4 d t ∴ \frac{v^{2}}{2} - 18 = 4 t - 8 v = \sqrt{8 t + 20} ∴ a = \frac{d v}{d t} = \frac{4}{\sqrt{8 t + 20}} At t = 3 s, a = 0.603 m / s^{2}$

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