The displacement time graphs of two particles A and B are straight lines making angles of 30- and 60- respectively with the time axis- If the velocity of A is VA and that of B is VB- then the value of VA-VB is

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Finding Average Velocity

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Question

The displacement time graphs of two particles A and B are straight lines making angles of $30^{\circ}$ and $60^{\circ}$ respectively with the time axis. If the velocity of A is $V_{A}$ and that of B is $V_{B}$ , then the value of $\frac{V_{A}}{V_{B}}$ is

\frac{1}{2}

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\frac{1}{\sqrt{3}}

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\sqrt{3}

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\frac{1}{3}

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Solution

The correct option is D $\frac{1}{3}$
Slope of displacement-time graph is equal to velocity
$v_{A} = t a n (30^{o}) = 1 / \sqrt{3}$
$v_{B} = t a n (60^{o}) = \sqrt{3}$
so $\frac{v_{A}}{v_{B}} = 1 / 3$

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The displacement time graphs of two particles A and B are straight lines making angles of 30∘ and 60∘ respectively with the time axis. If the velocity of A is VA and that of B is VB, then the value of VAVB is

The correct option is D 13Slope of displacement-time graph is equal to velocityvA=tan(30o)=1/√3vB=tan(60o)=√3so vAvB=1/3

The displacement time graphs of two particles A and B are straight lines making angles of $30^{\circ}$ and $60^{\circ}$ respectively with the time axis. If the velocity of A is $V_{A}$ and that of B is $V_{B}$ , then the value of $\frac{V_{A}}{V_{B}}$ is

The correct option is D $\frac{1}{3}$
Slope of displacement-time graph is equal to velocity
$v_{A} = t a n (30^{o}) = 1 / \sqrt{3}$
$v_{B} = t a n (60^{o}) = \sqrt{3}$
so $\frac{v_{A}}{v_{B}} = 1 / 3$