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
$\frac{dv}{dx}=-\frac{v_0}{x_0}$
Multiplying both sides of the above equation by $\text{v}$ , we get
$v\frac{dv}{dx}=-\frac{v_0}{x_0}\times v=-\frac{v_0}{x_0}[-\frac{v_0}{x_0}x+v_0]$ From (i),
$\therefore a=\frac{v_0^2}{x_0^2}x-\frac{v_0^2}{x_0}$ ...(ii) $[\because a=v\frac{dv}{dx}]$
On comparing equation (ii) with equation of a straight line $y=mx+c$ 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 ${}^{{}^{\prime }}{\theta }^{{}^{\prime }}$ 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.