Question

The velocity–displacement graph of a particle is shown in the figure.

The acceleration – displacement graph of the same particle is represented by:

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The correct option is D

Explanation for the correct answer:-

Option (D)

The slope of the given graph,

$m=\frac{-v_0}{x_0}$

So, $v=\left(\frac{-v_0}{x_0}\right)x+v_0$ and

$a=\frac{vdv}{dx}$

$a=\left[\left(\frac{-v_0}{x_0}\right)x+v_0\right]·\frac{d}{dx}\left[\left(\frac{-v_0}{x_0}x+v_0\right)\right]$

$a=\left[\left(\frac{-v_0}{x_0}\right)x+v_0\right]·\left[\left(\frac{-v_0}{x_0}+0\right)\right]$

$=\frac{v_0^2}{x_0^2}x-\frac{v_0^2}{x_0}$

Now, comparing with the standard equation of a straight line $(y=mx+c)$

Here, $m=+veandc=-ve$

In the case of option C,

1. From the above derivation we can see we got the slope from the equation of a straight line.
2. The graph of acceleration($a$) versus displacement($x$) is a straight line having a positive slope = ${\left(\frac{v_0}{x_0}\right)}^2$ and negative intercept = $-\frac{v_0^2}{x_0}$.
3. From the figure, we can see that, $v$ is decreasing but the magnitude of $\frac{dv}{ds}$ or slope is constant.
4. Hence magnitude of acceleration($a$) is decreasing.
5. In option C we can see the acceleration is constant.
6. Hence it is incorrect.

Thus, the correct answer is Option (D).