A train moving with constant velocity 'V', a carriage on a road parallel to the track. at distance 'd', from it is observed to move so as to appear always in line with a more distant fixed object whose distance from the track is 'D'. What is velocity of the carriage?

V (1 - \frac{d}{D})

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d \frac{V}{2} (1 - \frac{d}{D})

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V (1 - \frac{2 d}{D})

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V (1 - \frac{d}{2 D})

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Solution

The correct option is A $V (1 - \frac{d}{D})$
$y = D t a n θ$
$x = (D - d) t a n θ$
$\frac{d x}{d t} = v_{1}$
$\frac{d y}{d t} = V$
From above equtions
$\frac{v_{1}}{V} = \frac{D - d}{d}$
$v_{1} = V (1 - \frac{d}{D})$

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A train moving with constant velocity 'V', a carriage on a road parallel to the track. at distance 'd', from it is observed to move so as to appear always in line with a more distant fixed object whose distance from the track is 'D'. What is velocity of the carriage?

The correct option is A V(1−dD)y=Dtanθx=(D−d)tanθdxdt=v1dydt=V From above equtionsv1V=D−ddv1=V(1−dD)

The correct option is A $V (1 - \frac{d}{D})$
$y = D t a n θ$
$x = (D - d) t a n θ$
$\frac{d x}{d t} = v_{1}$
$\frac{d y}{d t} = V$
From above equtions
$\frac{v_{1}}{V} = \frac{D - d}{d}$
$v_{1} = V (1 - \frac{d}{D})$