TS EAMCET · Physics · Motion In One Dimension
A particle is moving along the \(Y\)-axis. The position of the particle from the origin as a function of time \((t)\) is given as \(y(t)=10 t e^{-2 t}\). How far is the particle from the origin when it stops momentarily? ( \(y\) is given in units of metre and \(t\) is in units of second)
- A \(5 \mathrm{~m}\)
- B \(5 \mathrm{e} \mathrm{m}\)
- C \(\frac{5}{e} m\)
- D \(10 \mathrm{~m}\)
Answer & Solution
Correct Answer
(C) \(\frac{5}{e} m\)
Step-by-step Solution
Detailed explanation
The position of the particle is given as \(y(t)=10 t e^{-2 t}\) \(\ldots\) (i) Velocity of the particle is given as \(v=\frac{d}{d t} y(t)=\frac{d}{d t}\left(10 t e^{-2 t}\right)\) \(=10\left[t e^{-2 t}(-2)+e^{-2 t} \cdot 1\right]\)…
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