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Standard XII

Physics

1st Equation of Motion

The position ...

Question

The position of a particle moving on a straight line path is given by $x = 12 + 18 t + 9 t^{2} m$ and $t$ in $s$ . Its acceleration at any instant is

18 m s^{- 2}

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45 m s^{- 2}

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9 m s^{- 2}

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12 m s^{- 2}

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Solution

The correct option is A $18 m s^{- 2}$
$a = \frac{d^{2} x}{d t^{2}}$
$= \frac{d^{2}}{d t^{2}} (12 + 18 t + 9 t^{2})$
$= 18 m / s^{2}$

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The position of a particle moving on a straight line path is given by x=12+18t+9t2m and t in s. Its acceleration at any instant is

The correct option is A 18ms−2a=d2xdt2=d2dt2(12+18t+9t2)=18m/s2

The position of a particle moving on a straight line path is given by $x = 12 + 18 t + 9 t^{2} m$ and $t$ in $s$ . Its acceleration at any instant is

The correct option is A $18 m s^{- 2}$
$a = \frac{d^{2} x}{d t^{2}}$
$= \frac{d^{2}}{d t^{2}} (12 + 18 t + 9 t^{2})$
$= 18 m / s^{2}$