Q99 / 125 · BITSAT 2009 · Physics
A particle starts moving rectilinearly at time $\mathrm{t}=0$ such that its velocity $\mathrm{v}$ changes with time $\mathrm{t}$ according to the equation $v=t^{2}-t$ where $t$ is in seconds and $v$ is in $\mathrm{m} / \mathrm{s}$. Find the time interval for which the particle retards.
- Correct answerA$\frac{1}{2}<\mathrm{t}<1$
- B$\frac{1}{2}>t>1$
- C$\frac{1}{4}<\mathrm{t}<1$
- D$\frac{1}{2}<t<\frac{3}{4}$
Solution
Acceleration of the particle $a=2 t-1$ The particle retards when acceleration is opposite to velocity.
$$
\Rightarrow \text { a } \cdot v<0 \Rightarrow(2 t-1)\left(t^{2}-t\right)<0 \Rightarrow t(2 t-1)
$$
$(\mathrm{t}-1)<0$
Now t is always positive
$$
\therefore(2 t-1)(t-1)<0
$$
or $2 \mathrm{t}-1<0$ and $\mathrm{t}-1>0 \Rightarrow \mathrm{t}<\frac{1}{2}$ and $\mathrm{t}>1$.
This is not possible or $2 \mathrm{t}-1>0 \& \mathrm{t}-1<0 \Rightarrow 1 / 2<\mathrm{t}<1$
Hence, the correct answer is option: 'A'.