Kinematics

JEE Main level, single correct, medium

115 of 308

Question 115 of 308

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

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Answer: (A) vA=vB=0.5 m/sv_A=v_B=0.5 \mathrm{~m} / \mathrm{s}

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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