TS EAMCET · Physics · Motion In One Dimension
The relation between the displacement ' \(x\) ' (in metre) and the time ' \(t\) ' (in second) of a particle is \(\mathrm{t}=2 x^2+3 x\). If the displacement of the particle is 25 cm from the origin \((x=0)\), then the acceleration of the particle is
- A \(+\frac{1}{16} \mathrm{~ms}^{-2}\)
- B \(-\frac{1}{16} \mathrm{~ms}^{-2}\)
- C \(+\frac{1}{8} \mathrm{~ms}^{-2}\)
- D \(-\frac{1}{8} \mathrm{~ms}^{-2}\)
Answer & Solution
Correct Answer
(B) \(-\frac{1}{16} \mathrm{~ms}^{-2}\)
Step-by-step Solution
Detailed explanation
\(\frac{dt}{dx} = 4x+3\) \(v = \frac{dx}{dt} = \frac{1}{4x+3}\) \(a = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{1}{4x+3}\right) = -\frac{1}{(4x+3)^2} \cdot 4 \frac{dx}{dt}\) \(a = -\frac{4}{(4x+3)^2} \cdot \frac{1}{4x+3} = -\frac{4}{(4x+3)^3}\)…
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