A particle starts moving rectilinearly at time \(t=0\) such that its velocity \(v\) changes with time \(t\)…

A particle starts moving rectilinearly at time \(t=0\) such that its velocity \(v\) changes with time \(t\) according to equation \(v=t^{2}-t\) where \(t\) is in seconds and \(v\) in \(\mathrm{m} \mathrm{s}^{-1}\). Find the time interval for which the particle retards.
  1. 0.5s < t < 1 s
  2. 0.5s < t < 2 s
  3. 0.5s < t < 3 s
  4. 0.5s < t < 5 s

Solution

Acceleration of the particle \(a=\frac{d \vec{v}}{d t}-2 t-1\)
The particle retards when acceleration is opposite to velocity. Hence, acceleration vector and velocity vector should be opposite to each ofher or the dot product of \(\vec{a}\) and \(\vec{v}\) should be negative.
\(\Rightarrow\)\(\vec{a} \cdot \vec{v} < 0\)
\(\Rightarrow\)\((2 t-1)\left(t^{2}-t\right) < 0\)
\(\Rightarrow\)\(t(2 r-1)(t-1) < 0\)
is always positive. \((2 t-1)(t-1) < 0\)
\(\therefore\) Either \(2 t-1 < 0\) or \(t-1 > 0\)
\(\Rightarrow \quad, < \frac{1}{2} \mathrm{~s}\) and \(t > 1 \mathrm{~s}\). This is not possible.
or \(2 t-1 > 0\) and \(t-1 < 0 \Rightarrow t > \frac{1}{2} \mathrm{~s}\) and \(t < 1 \mathrm{~s}\).
Hence, the required time interval is \(\frac{1}{2} \mathrm{~s} < t < 1 \mathrm{~s}\). ~

Asked in: JEE Mains - Motion In One Dimension - Test 3

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