Λparticle starts from rest at x=-2.25 m and moves along the x -axis with the v-t graph as shown

step 1 In this problem at t is equal to 0 x is equal to minus 2 .25 meter Now area at t is equal to 0 t

Question

$\Lambda$ particle starts from rest at $x=-2.25 \mathrm{~m}$ and moves along the $x$ -axis with the $v-t$ graph as shown. The time when the particle crosses the origin is (a) $0.5 \mathrm{~s}$ (b) $1 \mathrm{~s}$ (c) $1.5 \mathrm{~s}$ (d) $2 \mathrm{~s}$

   $\Lambda$ particle starts from rest at $x=-2.25 \mathrm{~m}$ and moves along the $x$ -axis with the $v-t$ graph as shown. The time when the particle crosses the origin is
(a) $0.5 \mathrm{~s}$
(b) $1 \mathrm{~s}$
(c) $1.5 \mathrm{~s}$
(d) $2 \mathrm{~s}$

Practice Problem in Physics for the JEE Main and Advanced

Abhay Kumar 2015 Edition

Chapter 2, Problem 56

Instant Answer

Solved by Expert Narayan Hari

01/15/2022

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$\Lambda$ particle starts from rest at $x=-2.25 \mathrm{~m}$ and moves along the $x$ -axis with the $v-t$ graph as shown. The time when the particle crosses the origin is (a) $0.5 \mathrm{~s}$ (b) $1 \mathrm{~s}$ (c) $1.5 \mathrm{~s}$ (d) $2 \mathrm{~s}$

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Key Concepts

Velocity-Time Graph

This is a graphical representation where velocity is plotted against time. It provides a visual tool to analyze how an object's velocity changes over time, and the area under the curve represents the displacement during the time interval.

Displacement from Integration

Displacement can be determined by integrating the velocity function over time. This calculation, equivalent to finding the area under the velocity-time graph, is fundamental in kinematics for relating velocity to position.

Kinematics with Initial Conditions

In problems of motion, knowing the initial conditions, such as the starting position and velocity, is crucial. They set the baseline for calculating future positions when integrating the velocity to find the net displacement.

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