Question

The velocity ($v$) versus displacement ($x$) graph of a particle moving along a straight line is shown in figure. Identify the correct statement regarding the acceleration ($a$) of particle:

• A. is constant.
• B. increases linearly with $x$.
• C. increases parabolically with $x$.
• D. None of these

Answer 1

The correct option is B increases linearly with $x$.

Acceleration of the particle can be written as:
$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}.............i$

where;
$\frac{dv}{dx}=$ slope of velocity Vs displacement graph
$v=$ velocity of particle


From the $v-x$ graph, it is a straigh line relationship. Moreover, slope of graph i.e $\tan \theta =+ve$ because $\theta <{90}^{\circ }$.
Hence, velocity ($v$) INCREASES linearly with respect to displacement $x$ or directly proportional relation exist between $v$ and $x$
$\therefore v\propto x..........ii$
$\Rightarrow \frac{dv}{dx}=\text{slope of v-x graph}\therefore \frac{dv}{dx}=\tan \theta =+\text{constant}$..........$iii$
As $a=v\frac{dv}{dx}$
Combining Eq. $i$, $ii$, $iii$ we can infer that
$\Rightarrow$acceleration $a$ is varying in direct proportionality with velocity $v$:

$\therefore a\propto +v$
also from graph of $v-x$:
$v\propto +x$
This ultimately implies that $a\propto x$

$\therefore$ acceleration will vary in direct proportionality with displaceement ($+x$) and due to +ve sign of slope, It will increase.

$\Rightarrow$Hence, $a-x$ graph will increase linearly.