Question:medium

The relation between the displacement 'x' (in metre) and the time 't' (in second) of a particle is $ t = 2x^2+3x $. If the displacement of the particle is 25 cm from the origin (x=0), then the acceleration of the particle is

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When given $t=f(x)$ instead of $x=f(t)$, you can find acceleration using the formula $ a = -f''(x) / [f'(x)]^3 $, where $f'(x) = dt/dx$. In this problem, $f'(x)=4x+3$ and $f''(x)=4$, so $a = -4/(4x+3)^3$.

Updated On: Mar 30, 2026

$ +\frac{1}{16} \text{ms}^{-2} $
$ -\frac{1}{16} \text{ms}^{-2} $
$ +\frac{1}{8} \text{ms}^{-2} $
$ -\frac{1}{8} \text{ms}^{-2} $

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The Correct Option is B

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The relation between the displacement 'x' (in metre) and the time 't' (in second) of a particle is \( t = 2x^2+3x \). If the displacement of the particle is 25 cm from the origin (x=0), then the acceleration of the particle is

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The Correct Option is B

Solution and Explanation

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