Question 128: Motion in 1D

Question

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

Accepted Answer

Given the relationship between time $ t $ and position $ x $ of a moving particle:
$ t = x^2 + x $
First, find the derivative of $ t $ with respect to $ x $:
$ \frac{d t}{d x} = 2x + 1  $
The velocity $ v $ is the inverse of this derivative, as it's given by the derivative of position $ x $ with respect to time $ t $:
$ v = \frac{d x}{d t} = \frac{1}{2x + 1} $
Next, calculate the derivative of $ v $ with respect to $ x $:
$ \frac{d v}{d x} = \frac{-2}{(2x + 1)^2} $
The acceleration $ a $ is then the product of velocity $ v $ and the derivative of velocity with respect to $ x $:
$ a = v \cdot \frac{d v}{d x} = \frac{1}{2x + 1} \cdot \left(\frac{-2}{(2x + 1)^2}\right) $
Simplifying this expression, we find:
$ a = -\frac{2}{(2x + 1)^3} $

Answer

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

Answer

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

Answer

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

Answer

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

Motion in 1D

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

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