Question

If the displacement of a particle and corresponding time be plotted along $y$ and $x$ axes respectively, then which of the following statements is (are) correct?
A) The curve may lie in the fourth quadrant.
B) The curve may drop as time passes.
C) The curve may exhibit peaks corresponding to maxima.
D) All of the above.

Answer 1

Displacement is the shortest vectorial distance between the final and the initial point of the motion of the particle. With small instances of time, this can be measured as the shortest vectorial distance from the starting point to the point it has just reached that moment.

Complete step by step answer:
Step 1:
The displacement is said to be plotted along the $y$ axis and the time is said to be plotted along the $x$ axis.
In the fourth quadrant, the signature of the $x$ axis is negative but the signature of $x$ axis is positive.
From the definition, you can see that displacement is a vector, hence the direction has a crucial role in setting the signature. If the final point of the journey is in the opposite direction to the motion of the journey, then the direction of the displacement vector is negative.
Thus with time, you can have negative displacements as mentioned in option (A).

Step 2:
The displacement is only dependent on the initial and the final position, not the distance covered.
So, you may think of such a condition that the particle at first starts moving to a distance and after a certain time it starts moving backward.
Hence, with the time you can see, at first, the particle has an increased displacement and after a certain time, the displacement starts reducing.
For such cases, you can easily find the drop in the curve plotted by the displacement and time as mentioned in (B).

Step 3:
Let the displacement plotted along the $y$axis is represented by $S$ and the time plotted along the $x$ axis is represented by $t$.
Hence, you can express the tangent by the derivative of $S$ with respect to $t$.
So for being a maxima you need to satisfy the following conditions
$\dfrac{{dS}}{{dt}} = 0$ and $\dfrac{{{d^2}S}}{{d{t^2}}} < 0$
Physically this can be interpreted as a point in the motion where the particle attins the velocity $\dfrac{{dS}}{{dt}} = v = 0$ and then starts to decelerate eventually, $\dfrac{{{d^2}S}}{{d{t^2}}} = a < 0$.
You can realize that this is a possible scenario as mentioned in option (C).

If the displacement of a particle and corresponding time be plotted along $y$ and $x$ axes respectively, then the correct statement is (D) All of the above, which are (A) The curve may lie in the fourth quadrant, (B) The curve may drop as time passes and (C) The curve may exhibit peaks corresponding to maxima.

Note:
The displacement of the particle is negative indicates the particle has eventually reached the opposite side from where it was moving with respect to the starting point. The negative displacement does not mean a negative distance in any way and doesn’t be afraid – negative distance does not exist. The velocity of a particle reaches zero when the displacement becomes zero or the change of the displacement with time becomes zero for a certain time. The change in velocity is negative implies a negative acceleration only.