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Acceleration

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Question

The position of a particle along a straight line is given by $s = (t^{3} - 9 t^{2} - 15 t)$ m, here t is in seconds. Determine its maximum acceleration during the time interval $0 \leq t \leq 10 s .$

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Solution

$s = t^{3} - 9 t^{2} - 15 t v = 3 t^{2} - 18 t - 15 a = 6 t - 18 t h u s, a c c c e l e r a t i o n i s a l i n e a r l y i n c r e a s i n g f u n c t i o n a_{m a x} = 6 \times 10 - 18 = 42 m / s^{2}$

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