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JEE Main 20127 MayOnline
Exam
JEE Main
Year
2012
Date
7 May
Mode
Online
Subject
Physics
Chapter
Motion in One Dimension

JEE Main 2012 · 7 May · Online · Physics · Motion in One Dimension

The graph of an object's motion (along the x- axis) is shown in the...

The graph of an object's motion (along the x−x- axis) is shown in the figure. The instantaneous velocity of the object at points AA and BB are vAv_A and vBv_B respectively. Then The graph of an object's motion (along the x- axis) is shown in the figure. The instantaneous velocity of the object at

Options

  1. AvA=vB=0.5 m/sv_A=v_B=0.5 \mathrm{~m} / \mathrm{s}
  2. BvA=0.5 m/s<vBv_A=0.5 \mathrm{~m} / \mathrm{s} < v_B
  3. CvA=0.5 m/s>vBv_A=0.5 \mathrm{~m} / \mathrm{s}>v_B
  4. DvA=vB=2 m/sv_A=v_B=2 \mathrm{~m} / \mathrm{s}

Correct answer

A. vA=vB=0.5 m/sv_A=v_B=0.5 \mathrm{~m} / \mathrm{s}

Step-by-step solution

Instantaneous velocity v=ΔxΔtv=\frac{\Delta x}{\Delta t} From graph, vA=ΔxAΔtA=4m8s=0.5 m/sv_A=\frac{\Delta x_A}{\Delta t_A}=\frac{4 m}{8 s}=0.5 \mathrm{~m} / \mathrm{s} and vB=ΔxBΔtB=8m16s=0.5 m/sv_B=\frac{\Delta x_B}{\Delta t_B}=\frac{8 m}{16 s}=0.5 \mathrm{~m} / \mathrm{s} i.e., vA=vB=0.5 m/sv_A=v_B=0.5 \mathrm{~m} / \mathrm{s}
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