NEET · Physics · STD 11 - 2. motion in straight line
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 \(-\frac{2}{(x+2)^3}\)
- B \(-\frac{2}{(2 x+1)^3}\)
- C \(+\frac{2}{(x+1)^3}\)
- D \(+\frac{2}{2 x+1}\)
Answer & Solution
Correct Answer
(B) \(-\frac{2}{(2 x+1)^3}\)
Step-by-step Solution
Detailed explanation
\(t=x^2+x \)
\( \frac{ dt }{ dx }=\frac{ d }{ dx }\left[ x ^2+ x \right] \)
\( \frac{1}{ v }=2 x +1 \)
\( v=(2 x+1)^{-1} \ldots\text{(i)} \)
\( a=\frac{v d v}{d x}=(2 x+1)^{-1}\left[-1(2 x+1)^{-2} \times 2\right] \)
\( a=-1(2 x+1)^{-3} \times 2 \)
\( a=-2(2 x+1)^{-3} \)
\(=-\frac{2}{(2 x+1)^3}\)
\( \frac{ dt }{ dx }=\frac{ d }{ dx }\left[ x ^2+ x \right] \)
\( \frac{1}{ v }=2 x +1 \)
\( v=(2 x+1)^{-1} \ldots\text{(i)} \)
\( a=\frac{v d v}{d x}=(2 x+1)^{-1}\left[-1(2 x+1)^{-2} \times 2\right] \)
\( a=-1(2 x+1)^{-3} \times 2 \)
\( a=-2(2 x+1)^{-3} \)
\(=-\frac{2}{(2 x+1)^3}\)
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