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2.Motion in Straight Line
hard
A particle starts moving rectilinearly at time $t = 0$ such that its velocity $'v'$ changes with time $'t'$ according to the equation $v = t^2 - t$ where $t$ is in seconds and $v$ is in $m/s.$ The time interval for which the particle retards is
A$t < 1/2$
B$1/2 < t < 1$
C$t > 1$
D$t < 1/2$ and $t > 1$
Solution
Acceleration of the particle, $\vec{a}=\frac{d v}{d t}=2 t-1$
The particle retards when acceleration is oppostie to velocity. Hence, acceleration vector and velocity vector should be opposite to each other or the dot product of $\vec{a}$ and $\vec{v}$ should be negative. $\Rightarrow \vec{a} \cdot \vec{v}<0$
$\Rightarrow(2 t-1)\left(t^{2}-t\right)<0$
$\Rightarrow t(2 t-1)(t-1)<0$
$t$ is always positive. $\because(2 t-1)(t-1)<0$
$\Rightarrow$ Either $2 t-1<0$ or $t-1<0$
$\Rightarrow t<\frac{1}{2} s$ and $t<1 s .$ This is not possible. or $2 t-1<0$ and $t-1<0 \Rightarrow \operatorname{tg} t \frac{1}{2} s$ and $t<1 s .$ Hencec, the required time interval is $\frac{1}{2} < t < 1 $
The particle retards when acceleration is oppostie to velocity. Hence, acceleration vector and velocity vector should be opposite to each other or the dot product of $\vec{a}$ and $\vec{v}$ should be negative. $\Rightarrow \vec{a} \cdot \vec{v}<0$
$\Rightarrow(2 t-1)\left(t^{2}-t\right)<0$
$\Rightarrow t(2 t-1)(t-1)<0$
$t$ is always positive. $\because(2 t-1)(t-1)<0$
$\Rightarrow$ Either $2 t-1<0$ or $t-1<0$
$\Rightarrow t<\frac{1}{2} s$ and $t<1 s .$ This is not possible. or $2 t-1<0$ and $t-1<0 \Rightarrow \operatorname{tg} t \frac{1}{2} s$ and $t<1 s .$ Hencec, the required time interval is $\frac{1}{2} < t < 1 $
Standard 11
Physics
