In a displacement-time graph, two straight lines make angles of $60^{o}$ and $30^{o}$ with the time axis. The ratio of the velocities represented by them is

2 : 1

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1 : 3

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1 : 2

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3 : 1

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Solution

The correct option is D $3 : 1$
velocity OA
$V_{O A} = slope of O A$
$= tan 60$
$= \sqrt{3}$
velocity of AB
$V_{A B} = slope of A B$
$= tan 30$
$= \frac{1}{\sqrt{3}}$
So,
$\frac{V_{O A}}{V_{A B}} = \frac{\sqrt{3}}{\frac{1}{\sqrt{3}}} = \frac{3}{1} = 3 : 1$

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In a displacement-time graph, two straight lines make angles of 60o and 30o with the time axis. The ratio of the velocities represented by them is

The correct option is D 3:1velocity OAVOA=slopeofOA=tan60=√3velocity of ABVAB=slopeofAB=tan30=1√3So, VOAVAB=√31√3=31=3:1

In a displacement-time graph, two straight lines make angles of $60^{o}$ and $30^{o}$ with the time axis. The ratio of the velocities represented by them is

The correct option is D $3 : 1$
velocity OA
$V_{O A} = slope of O A$
$= tan 60$
$= \sqrt{3}$
velocity of AB
$V_{A B} = slope of A B$
$= tan 30$
$= \frac{1}{\sqrt{3}}$
So,
$\frac{V_{O A}}{V_{A B}} = \frac{\sqrt{3}}{\frac{1}{\sqrt{3}}} = \frac{3}{1} = 3 : 1$