Question

The position of a particle along $x-$ axis at time $t$ is given by $x=1+t-t^2\text{m}$. The distance travelled by the particle in first $2\text{sec}$ is

• A. $1\text{m}$
• B. $2\text{m}$
• C. $2.5\text{m}$
• D. $3\text{m}$

Answer 1

The correct option is C $2.5\text{m}$

$x=1+t-t^2$
At $t=0\text{sec},x_0=1\text{m}$
We know that, $v=\frac{dx}{dt}=1-2t$
So, at $t=\frac{1}{2}\text{sec},v=0$
And displacement, $x_{1/2}=1+\frac{1}{2}-\frac{1}{4}=\frac{5}{4}=1.25\text{m}$
Distance travelled by the particle from $t=0$ to $t=\frac{1}{2}\text{sec}$
$=|x_{1/2}-x_0|=1.25\text{m}-1\text{m}=0.25\text{m}$
Position of particle at the end of $2\text{sec}$ is given by
$x_2=1+2-2^2=-1\text{m}$
Distance travelled by the particle $t=\frac{1}{2}\text{sec}$ to $t=2\text{sec}$
$=|x_2-x_{\frac{1}{2}}|=|-1\text{m}-1.25\text{m}|=2.25\text{m}$
Total distance travelled by the particle from $t=0\text{sec}$ to $t=2\text{sec}$ will be
$\therefore x=0.25\text{m}+2.25\text{m}=2.5\text{m}$