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(B) The acceleration of the particle is given by $a = \frac{dv}{dt} = \frac{d}{dt}(t^2 - t) = 2t - 1$.$A$ particle retards when its acceleration and v

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$1/2 < t < 1$

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(B) The acceleration of the particle is given by $a = \frac{dv}{dt} = \frac{d}{dt}(t^2 - t) = 2t - 1$.$A$ particle retards when its acceleration and velocity are in opposite directions,i.e.,$a \cdot v Substituting the expressions for $v$ and $a$:$(2t - 1)(t^2 - t) $(2t - 1)t(t - 1) Since $t \ge 0$,we analyze the sign of the expression $t(2t - 1)(t - 1) 0$:$1$. For $0 0$,$(2t - 1) 0$.$2$. For $1/2 0$,$(2t - 1) > 0$,and $(t - 1) $3$. For $t > 1$: $t > 0$,$(2t - 1) > 0$,and $(t - 1) > 0$. The product is $(+)(+)(+) = (+)$,which is $> 0$.Therefore,the particle retards in the interval $1/2 < t < 1$.

$A$ 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

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