The distance $S$ described in time $t$ by aparticle moving in a straightline is given by $S = t^{5} - 40 t^{3} + 30 t^{2} + 80 t - 250$ then the maximum acceleration is

260 {units/sec}^{2}

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- 260 {units/sec}^{2}

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130 {units/sec}^{2}

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380 {units/sec}^{2}

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Solution

The correct option is B $- 260 {units/sec}^{2}$
$v = \frac{d s}{d t} = s t^{4} - 120 t^{2} + 60 t + 80$
$a = \frac{d v}{d t} = 20 t^{3} - 240 t + 60$
$\frac{d a}{d t} = 60 t^{2} - 240$
$= 60 (t^{2} - 4)$
$\frac{d a}{d t} = 0$ at $t = \pm 2$
a is maximum at $x = 2$
$a (2) = - 260$

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