A particle is moving with constant accelerati | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

Vedclass · Answer

(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:

Similar Questions

The acceleration at the end of $2 \ s$ of a particle whose motion is represented by the equation $S = 4t^3 - 8t^2 + 5t + 4$ is $......$ (in $m \ s^{-2}$)

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?

What is acceleration? What is its direction,and state its $SI$ unit.

What will be the change in the speed of a moving object if both speed and acceleration are positive or both are negative?

Vedclass Test Series

Exam Paper Generator

Online Exam Module