MCQ
The position of a particle along $x$-axis at time $t$ is given by $x=2+t-3 t^2$. The displacement and the distance travelled in the interval $t=0$ to $t=1$ are respectively
- A$2,2$
- B$-2,2.5$
- C$0,2$
- ✓$-2,2.1$
✓
Answer
Correct option: D.
$-2,2.1$
d
(d)
(d)
$x=2+t-3 t^2, v=\frac{d x}{d t}=1-6 t$, velocity will become
zero at time, $0=1-6 t_0$ or $t_0=\frac{1}{6}\,s$,
Since, the given time $t=1 s$ is greater than $t_0=\frac{1}{6}\,s$,
distance $ > \mid$ displacement $\mid$
$\text { Displacement } s=x_f-x_i$
$=(2+1-3)-(2+0-0)=-2\,m$
$\text { Distance } d=\left|s_{0-t_0}\right|+\left|s_{t-t_0}\right|$
$=\frac{u^2}{2|a|}+\frac{1}{2}|a|\left(t-t_0\right)^2$
$\text { Comparing } v=1-6 t \text { with } v=u+u t \text {, we have }$
$u=1\,m / s \text { and } a=-6\,m / s ^2$
$\therefore \quad \text { Distance }=\frac{(1)^2}{2 \times 6}+\frac{1}{2} \times 6 \times\left(1-\frac{1}{6}\right)^2$
$=2.1\,m$
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