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: [2025]

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Q.
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: [2025]

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

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

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

4 $- \frac{2}{{(2 x + 1)}^{3}}$

Ans.
(4)

Given, $t = x^{2} + x$

Differentiating with respect to 't' on both sides,

$\Rightarrow \frac{d t}{d t} = 2 x \frac{d x}{d t} + \frac{d x}{d t}$

$\Rightarrow 1 = 2 x v + v$ ...(i)

$\Rightarrow 1 = v (2 x + 1)$

$v = \frac{1}{2 x + 1}$ ...(ii)

Again, differentiating equation (i) with respect to 't',

$0 = 2 (\frac{d x}{d t} . v + x \frac{d v}{d t}) + \frac{d v}{d t}$

$0 = 2 (v^{2} + x . a) + a$

$0 = 2 v^{2} + 2 x a + a$

$0 = [2 v^{2} + a (2 x + 1)]$

$\Rightarrow 2 v^{2} = - a (2 x + 1)$

Using equation (ii), we get

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

$∴$ $a = \frac{- 2}{{(2 x + 1)}^{3}}$

Solution

Q.
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: [2025]

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

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

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

4 $- \frac{2}{{(2 x + 1)}^{3}}$

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Solution

Q. In some appropriate units, time (t) and position (x) relation of a moving particle is given by t=x2+x. The acceleration of the particle is: [2025]

1 +2(x+1)3

2 +22x+1

3 -2(x+2)3

4 -2(2x+1)3

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Q.
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: [2025]

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

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

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

4 $- \frac{2}{{(2 x + 1)}^{3}}$