Question

The velocity displacement graph of a particle moving along a straight line is -



The most suitable acceleration-displacement graph will be

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

The equation for the given $\text{v - x}$ graph is
$v = - \frac{v_{0}}{x_{0}} x + v_{0}$ ...(i)
Differentiating the above equation w.r.t. $\text{x}$ we get
$\frac{d v}{d x} = - \frac{v_{0}}{x_{0}}$
Multiplying both sides of the above equation by $\text{v}$ , we get
$v \frac{d v}{d x} = - \frac{v_{0}}{x_{0}} \times v = - \frac{v_{0}}{x_{0}} \left[\right. - \frac{v_{0}}{x_{0}} x + v_{0} \left]\right.$ From (i),
$\therefore a = \frac{v_{0}^{2}}{x_{0}^{2}} x - \frac{v_{0}^{2}}{x_{0}}$ ...(ii) $\left[\right. \because a = v \frac{d v}{d x} \left]\right.$
On comparing equation (ii) with equation of a straight line $y = m x + c$ where $\text{m}$ is the slope of the line and $\text{c}$ is its intercept on $\text{y}$ -axis we get, $m = \frac{v_{0}^{2}}{x_{0}^{2}}$ [therefore, $\text{m}$ will always be positive]
$\Rightarrow $ The slope $\text{m}$ or $tan \theta = 0$ or $^{'} \theta ^{'}$ is an acute angle.
Also, the comparison gives -
$c = \frac{- v_{0}^{2}}{x_{0}}$ or the intercept will be negative.
These conditions are met only in this graph.