The displacement-time graph for two particles A and B are straight lines inclined at angles of 30- and 60- with the time axis- The ratio of velocities of vA - vB is

The slope is given by tan θ. Therefore, v_Av_B = tan θ_Atan θ_B = tan 30^∘tan 60^∘ = 1√(3)√(3) = 13 ...

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Standard XIII

Physics

Position Time Graph

The displacem...

Question

The displacement-time graph for two particles A and B are straight lines inclined at angles of $30^{\circ}$ and $60^{\circ}$ with the time axis. The ratio of velocities of $v_{A}$ : $v_{B}$ is

1 : 2

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1 : $\sqrt{3}$

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$\sqrt{3}$ : 1

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1 : 3

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Solution

The correct option is D
1 : 3

The slope is given by $t a n θ$ . Therefore,

$\frac{v_{A}}{v_{B}} = \frac{t a n θ_{A}}{t a n θ_{B}} = \frac{t a n 30^{\circ}}{t a n 60^{\circ}} = \frac{\frac{1}{\sqrt{3}}}{\sqrt{3}} = \frac{1}{3}$

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The displacement-time graph for two particles A and B are straight lines inclined at angles of 30∘ and 60∘ with the time axis. The ratio of velocities of vA : vB is

The correct option is D 1 : 3 The slope is given by tan θ. Therefore, vAvB=tanθAtanθB=tan30∘tan60∘=1√3√3=13

The displacement-time graph for two particles A and B are straight lines inclined at angles of $30^{\circ}$ and $60^{\circ}$ with the time axis. The ratio of velocities of $v_{A}$ : $v_{B}$ is

The correct option is D
1 : 3

The slope is given by $t a n θ$ . Therefore,

$\frac{v_{A}}{v_{B}} = \frac{t a n θ_{A}}{t a n θ_{B}} = \frac{t a n 30^{\circ}}{t a n 60^{\circ}} = \frac{\frac{1}{\sqrt{3}}}{\sqrt{3}} = \frac{1}{3}$