A particle moves along x-axis in such a way that its coordinate ( $x$ , in metres) 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

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- 3 m / s

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6 m / s

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3 m / s

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Solution

The correct option is B $- 5 m / s$
Given: $x = 6 t^{2} - 5 t + 2$
velocity $= \frac{d x}{d t} = 12 t - 5$
Initial velocity means velocity at t $=$ 0
Therefore, initial velocity, $v = (12 \times 0) - 5 = - 5 m / s$

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