A particle is moving with constant accelerati | vedclass

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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 $v$.

Column-$I$	Column-$II$
$(1)$ Acceleration is positive	$(a)$ Speed of the particle decreases
$(2)$ Acceleration is negative	$(b)$ Speed of the particle increases
	$(c)$ Speed of the particle keeps changing

$A$ 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:

Similar Questions

Match the items in Column-$I$ with the items in Column-$II$ correctly.

Column-$I$ Column-$II$

$(1)$ Acceleration is positive $(a)$ Speed of the particle decreases

$(2)$ Acceleration is negative $(b)$ Speed of the particle increases

$(c)$ Speed of the particle keeps changing

If a particle is moving along a straight line with increasing speed,then:

Let $v$ and $a$ denote the velocity and acceleration respectively of a particle in one-dimensional motion. Which of the following statements is correct?

When are the average and instantaneous accelerations equal?

$A$ particle starts from rest. Its acceleration $(a)$ versus time $(t)$ graph is as shown in the figure. The maximum speed of the particle will be.....$m/s$.

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