A steel ball is released from rest a distance | vedclass

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

Question

(D) The area under a velocity-time graph represents the displacement of the object.In the given graph,area $X$ represents the displacement of the ball

Vedclass · Accepted Answer

The ball rises and falls through the same distance between the impact $1$ and $2$.

Vedclass · Answer

(D) The area under a velocity-time graph represents the displacement of the object.In the given graph,area $X$ represents the displacement of the ball as it rises from the surface after the $1^{st}$ impact until it reaches its maximum height.Area $Y$ represents the displacement of the ball as it falls from that maximum height back to the surface before the $2^{nd}$ impact.Since the ball rises and falls through the same vertical distance between two consecutive impacts,the magnitude of the displacement for the upward motion must equal the magnitude of the displacement for the downward motion.Therefore,the areas $X$ and $Y$ are equal because the ball rises and falls through the same distance between the $1^{st}$ and $2^{nd}$ impact.

$A$ steel ball is released from rest a distance above a rigid horizontal surface and bounces several times. The diagram shows how its velocity varies with time. Which statement correctly explains why the areas $X$ and $Y$ are equal?

Similar Questions

The velocity-displacement graph of a particle is shown in the figure.
$(a)$ Write the relation between $v$ and $x$.
$(b)$ Obtain the relation between acceleration and displacement and plot it.

The velocity-time graph of an object moving along a straight line is shown in the figure. What is the distance (in $m$) covered by the object between $t = 0$ and $t = 4 \; s$?

Draw the $x-t$ graph for an object moving with zero acceleration.

What is the maximum acceleration of the train during the journey in $km \ h^{-2}$?

The position $x$ of a particle with respect to time $t$ along the $x$-axis is given by $x = 9t^2 - t^3$,where $x$ is in metres and $t$ is in seconds. What will be the position of this particle when it achieves maximum speed along the $x$-direction?

Vedclass Test Series

Exam Paper Generator

Online Exam Module