Displacement is given by $X = 1 + 2 t + 3 t^{2}$ . Find the value of instantaneous acceleration.

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Solution

Given that,

$X = 1 + 2 t + 3 t^{2} \dots . (I)$

Now, the acceleration is

Differentiate of equation (I)

$x = 1 + 2 t + 3 t^{2}$

$v = \frac{d x}{d t} = 0 + 2 + 6 t$

Now, again differentiate

$a = \frac{d v}{d t} = \frac{d^{2} x}{d t^{2}} = 6 m / s^{2}$

Hence, the instantaneous acceleration is $6$ m/s²

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