If $S = t^{3} - 4 t^{2} - 5 t$
Find velocity & acceleration at $t = 2$

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Solution

$s = t^{3} - 4 t^{2} - 5 t$

$v = \frac{d s}{d t} = 3 t^{2} - 8 t - 5$

$v |_{t = 2} = 3 (2)^{2} - 8 (2) - 5$

$= 12 - 16 - 5 = - 9$ m/s

$a = \frac{d^{2} s}{d t^{2}} = 6 t - 8$

$9 |_{t = 2} = 6 (2) - 8 = 4 m / s^{2}$

$∴$ Velocity at $t = 2 s = 9$ m/s (opposite to direction motion)
acceleration at $t = 2 s = 4 m / s^{2}$ (in direction of motion).

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a = 3 t^{2} - 2 t + 5

(m / s^{2})

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Express $x^{\frac{3}{4}} ({log}_{3} x)^{2} + {log}_{3} x - \frac{5}{4} = \sqrt{3}$ in terms of t where t = ${log}_{3}$ x.

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