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  • In some appropriate units, time $(t)$ and position $(x)$ relation of a moving particle is given by $t=x^2+x$. The acceleration of the particle is

Q.16) In some appropriate units, time $(t)$ and position $(x)$ relation of a moving particle is given by $t=x^2+x$. The acceleration of the particle is

A) $+\frac{2}{2 x+1}$

B) $-\frac{2}{(x+2)^3}$

C) $-\frac{2}{(2 x+1)^3}$

D) $+\frac{2}{(x+1)^3}$

Answers (1)

best_answer


Solution:
Given: $t=x^2+x \Rightarrow x(t)$
Differentiate to get velocity:

$$
\frac{d t}{d x}=2 x+1 \Rightarrow \frac{d x}{d t}=\frac{1}{2 x+1}
$$
Differentiate again for acceleration:

$$
a=\frac{d^2 x}{d t^2}=\frac{d}{d t}\left(\frac{1}{2 x+1}\right)
$$
Using chain rule:

$$
\frac{d}{d t}=\frac{d}{d x} \cdot \frac{d x}{d t}
$$
So,

$$
a=\frac{d}{d x}\left(\frac{1}{2 x+1}\right) \cdot \frac{1}{2 x+1}=\left(\frac{-2}{(2 x+1)^2}\right) \cdot \frac{1}{2 x+1}=\frac{-2}{(2 x+1)^3}
$$
Hence, the answer is option (3) $-2 /(2 x+1)^3$.

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Dimpy

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