Question

The given graph shows the variation of velocity (v) with position (x) for a particle moving along a straight line. Which of the following graph shows the variation of acceleration (a) with position (x)?

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

Answer 2

Given line have positive intercept but negative slope. So its equation can be written as, $v=-m x+v_{0} \quad\ldots\left(1\right)$ where $m=tan\, \theta=\frac{v_{0}}{x_{0}}$ By differentiating with respect to time we get, $\frac{d v}{d t}=-m \frac{d x}{d t}=-mv$ Now substituting the value of ??? $\frac{d v}{d t}=-m\left[-mx+v_{0}\right]=m^{2}x-mv_{0}$ $a=m^{2}x-mv_{0}$ i.e., the graph between a and x should have positive slope but negative intercept on acceleration axis.