The position coordinate of a particle moving along a straight line is given by $x = 4 t^{3}$ - $3 t^{2}$ + $4 t$ + $5$ . Then what is acceleration at $t = 1$ ?

18

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Solution

The correct option is A $18$
$x (t) = 4 t^{3} - 3 t^{2} + 4 t + 5$
On differentiating with respect to time we get the velocity of the particle as a function of time,
$v (t) = \frac{d x}{d t} = 12 t^{2} - 6 t + 4$
Again differentiating with respect to time to get the acceleration of the particle as a function of time,
$a (t) = \frac{d v}{d t} = 24 t - 6$
$∴$ at $t = 1$ ,
$a = 24 (1) - 6 = 18 m / s^{2}$

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