A horse runs along a circle with a speed of $20 km/h$ . A lantern is at the centre of the circle. A fence is there along the tangent to the circle at the point at which the horse starts. The speed with which the shadow of the horse moves along the fence at the moment when it covers $\frac{1}{8}$ of the circle (in $km/h$ ) is

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40.0

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40.000

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40.00

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Solution

$tan θ = \frac{x}{r}$
$\Rightarrow x = r tan θ$

$\Rightarrow \frac{d x}{d t} = r {sec}^{2} θ \frac{d θ}{d t}$
$\Rightarrow \frac{d x}{d t} = r ω {sec}^{2} θ = v {sec}^{2} θ$
When, $θ = \frac{2 π}{8}$
$[∵ full circle = 2 π]$
$\frac{d x}{d t} = v {sec}^{2} (\frac{π}{4}) = 2 v = 40 km/h$

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tanθ=xr ⇒x=rtanθ ⇒dxdt=rsec2θdθdt ⇒dxdt=rωsec2θ=vsec2θ When, θ=2π8 [∵full circle=2π] dxdt=vsec2(π4)=2v=40 km/h

$tan θ = \frac{x}{r}$
$\Rightarrow x = r tan θ$

$\Rightarrow \frac{d x}{d t} = r {sec}^{2} θ \frac{d θ}{d t}$
$\Rightarrow \frac{d x}{d t} = r ω {sec}^{2} θ = v {sec}^{2} θ$
When, $θ = \frac{2 π}{8}$
$[∵ full circle = 2 π]$
$\frac{d x}{d t} = v {sec}^{2} (\frac{π}{4}) = 2 v = 40 km/h$