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Motion Under Variable Acceleration

The position ...

Question

The position of a particle moving along the $x$ -axis is expressed as $x = a t^{3} + b t^{2} + c t + d$ . The initial acceleration of the particle is

6 a

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2 b

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(a + b)

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(a + c)

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Solution

The correct option is B $2 b$
We know that
$v = \frac{d x}{d t} = 3 a t^{2} + 2 b t + c$
Now for acceleration,
Let the acceleration be A,
$A = \frac{d v}{d t} = 6 a t + 2 b$
To find initial acceleration, we put $t = 0$ in above expression
At $t = 0$
$A = 2 b$

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