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Question

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


A
6a
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B
2b
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C
a+b
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D
a+c
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Solution

The correct option is B $$2b$$
The relationship between position and time is given by the equation,
$$x=at^3+bt^2+ct+d$$
Velocity, $$v=\dfrac{dx}{dt}$$
$$v=3at^2+2bt+c$$
Acceleration, $$a=\dfrac{dv}{dt}$$
$$a=6at+2b$$
The initial acceleration at time $$t=0sec$$
$$a|_{t=0sec}=6a(0)+2b$$
$$a|_{t=0sec}=2b$$
The correct option is B.

Physics

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