Which of the following curves possibly repres | vedclass

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(A) Analysis of the given graphs for one-dimensional motion:$(A)$ Phase vs Time graph: $A$ linear relationship $\phi = kt + C$ is physically possible

Vedclass · Accepted Answer

$A, B$ and $D$ only

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(A) Analysis of the given graphs for one-dimensional motion:$(A)$ Phase vs Time graph: $A$ linear relationship $\phi = kt + C$ is physically possible for a particle undergoing simple harmonic motion $(x = A \sin(kt + C))$. Thus,this represents a possible $1D$ motion.$(B)$ Velocity vs Displacement graph: $A$ circular path $v^2 + x^2 = R^2$ represents the phase space trajectory of a simple harmonic oscillator. This is a valid representation of $1D$ motion.$(C)$ Velocity vs Time graph: The graph shows a circle,which implies that for a single value of time,there are two possible values of velocity. Furthermore,the graph extends into the negative time axis,which is physically impossible as time cannot be negative. Thus,this is not a possible representation.$(D)$ Total distance vs Time graph: The graph shows total distance increasing with time. Since total distance is a non-decreasing function of time for any moving particle,this is a physically possible representation of $1D$ motion.Therefore,graphs $(A), (B),$ and $(D)$ represent possible one-dimensional motions.

Which of the following curves possibly represent one-dimensional motion of a particle? Choose the correct answer from the options given below.

Similar Questions

At time $t=0$, a particle leaves the origin and moves in the positive direction of the $X$-axis. If the velocity of the particle varies as $v(t)=v_0(1-t/t_0)$, where $|v_0|=10 \ m/s$ and $t_0=10 \ s$, then the distance covered by the particle during the first $20 \ s$ is: (in $m$)

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Look at the graphs $(a)$ to $(d)$ carefully and state,with reasons,which of these cannot possibly represent one-dimensional motion of a particle.

In what different ways can the velocity of an object be changed?

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