A particle is moving in a straight line so that after ‘ $t$ ’ seconds Its distance is ‘ $s$ ’ ( $i n c m s$ ) from a fixed point on the line is given by $s = f (t) = 8 t + t^{3}$ find
i) the velocity at time $t = 2 s e c$ .
ii) the initial velocity
iii) acceleration at $t = 2 s e c$ .

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Detailed Solution

Given $s = f (t) = 8 t + t^{3}$

Velocity $(V) = \frac{d s}{d t} = 8 + 3 t^{2}$

i) at $t = 2 s e c$

$V = 8 + 3 (2)^{2} = 20 cm / \sec$

ii) initial velocity $; t = 0$

$V = 8 + 3 (0)^{2} = 8 cm / \sec$

iii) acceleration $a = \frac{d v}{d t} = 6 t at t = 2$

$a = 6 (2) = 12 cm / \sec^{2}$

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