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(C) Let the constant acceleration be $a = k$,where $k$ is a positive constant.We know that acceleration $a = v \frac{dv}{ds}$.Substituting $a = k$,we

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(C) Let the constant acceleration be $a = k$,where $k$ is a positive constant.We know that acceleration $a = v \frac{dv}{ds}$.Substituting $a = k$,we get $k = v \frac{dv}{ds}$.Rearranging the terms,we have $k \, ds = v \, dv$.Integrating both sides with initial conditions $s=0$ at $v=0$ (assuming the particle starts from rest):$\int_0^{s} k \, ds = \int_0^{v} v \, dv$$ks = \frac{v^2}{2}$$v^2 = 2ks$This equation represents a parabola opening along the $s$-axis. Since $s = \frac{v^2}{2k}$,the graph of $s$ versus $v$ is a parabola symmetric about the $s$-axis,opening towards the positive $s$-direction. Looking at the provided options,Graph $C$ represents this relationship where $s$ increases quadratically with $v$.

$A$ particle is moving with constant acceleration,whose direction is the same as that of its instantaneous motion. The distance $(s)-$velocity $(v)$ graph for the particle is:

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