A particle moves along x-axis in such a way that its coordinate (x) varies with time (t) according to the expression, $x = 2 - 5 t + 6 t^{2}$ . Then the initial velocity of the particle is

- 5 m / s e c

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- 3 m / s e c

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6 m / s e c

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3 m / s e c

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Solution

The correct option is A $- 5 m / s e c$
The displacement is given as: $y = u t + \frac{1}{2} a t^{2}$
where, $u$ is the initial velocity and $a$ is the acceleration
Comparing with given expression:
$y = - 5 t + 6 t^{2} + 2$
$u = - 5 m / s$ , $a = 12 m / s^{2}$
So initial velocity is $- 5 m / s$

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