Question

Velocity - displacement graph of a particle moving in a straight line is as shown in figure

• A. magnitude of acceleration of particle is decreasing
• B. magnitude of acceleration of particle is increasing
• C. acceleration versus displacement graph is straight line
• D. acceleration versus displacement graph is parabola

Answer 1

Answer: (A), (C)

The correct options are
A magnitude of acceleration of particle is decreasing
C acceleration versus displacement graph is straight line

We know that acceleration $a=v\frac{dv}{ds}$
From the figure, we can see that
$v$ is decreasing but magnitude of $\frac{dv}{ds}$ or slope is constant
Hence magnitude of $a$ is decreasing.
Let the slope of the given $v-s$ graph be $-A$.
Slope is negative as $\frac{dv}{ds}$ is negative.
$\Rightarrow v=-As+B$ and $\frac{dv}{ds}=-A$
Here $A$ and $B$ are positive constants.
$\therefore a=(-As+B)(-A)$
$=A^{2}s-AB$
This looks similar to $y=mx+c$
i.e., $a–s$ graph is a straight line with positive slope and negative intercept.