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Standard XII

Mathematics

Angle between Pair of Straight Lines

A point moves...

Question

A point moves in xy-plane according to equation $x = a t, y = a t (1 - b t)$ where $a$ and $b$ are positive constants and $t$ is time. The instant at which velocity vector is at $π / 4$ with acceleration vector is given by:

\frac{1}{a}

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\frac{1}{b}

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\frac{1}{a} + \frac{1}{b}

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(a + b) (a^{2} + b^{2})

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Solution

The correct option is B $\frac{1}{b}$
$a_{x} = 0 a_{y} = - 296$
when $a_{y}$ make an angle $\frac{π}{4}$
then $v_{x} = v_{y} a t = a t (1 - b t) t = \frac{1}{b}$

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A point moves in xy-plane according to equation x=at,y=at(1−bt) where a and b are positive constants and t is time. The instant at which velocity vector is at π/4 with acceleration vector is given by:

The correct option is B 1bax=0ay=−296when ay make an angle π4then vx=vyat=at(1−bt)t=1b

A point moves in xy-plane according to equation $x = a t, y = a t (1 - b t)$ where $a$ and $b$ are positive constants and $t$ is time. The instant at which velocity vector is at $π / 4$ with acceleration vector is given by:

The correct option is B $\frac{1}{b}$
$a_{x} = 0 a_{y} = - 296$
when $a_{y}$ make an angle $\frac{π}{4}$
then $v_{x} = v_{y} a t = a t (1 - b t) t = \frac{1}{b}$