Question

A particle is moving along $x-$axis. Its $x-$coordinate versus time graph is shown.


Its $v-t$ graph will be:

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

If $x-t$ graph is given then its slope $\left(m\right)$ represent the velocity of the particle in motion.

$v=\frac{dx}{dt}=\tan \theta$
[$\theta \to$ angle with $+x-$axis in anti-clockwise direction]


For $AB$,
Time interval is $0-4\text{s}$
Slope of $AB=m$
$m=\tan \theta =\tan 0^{\circ }$
$m=0$ [$\theta =0^{\circ }$ as $AB||\text{x-axis}$]
$\Rightarrow v=0$

For $BD$,
Time interval is $4-10\text{s}$
Slope of $BD=m$
$m=\tan \theta =v$

From $\Delta BFD,\tan \alpha =\frac{12}{6}=2$
$[\tan \alpha =\frac{BF}{FD}]$
$\tan (180-\theta )=2[\because \alpha =180-\theta ]$
$\Rightarrow -\tan \theta =2$
Or, $\tan \theta =-2$
$\therefore v=-2\text{m/s}$

For $DE$,
Time interval is $10-14\text{s}$
Slope of $DE$ is,
$m=\tan \beta =v$
From triangle,
$\Delta DEG,\tan \beta =\frac{8}{4}=2$
$[\tan \beta =\frac{EG}{DG}]$
$\Rightarrow v=2\text{m/s}$
Now, $v-t$ graph will be


$\therefore$Option $\left(b\right)$ is correct.