What is the speed with which the shadow of the horse move along the fence at the moment when it covers $1/8$ of the circle in km/hr?

Question

A horse runs along a circle with a speed of 20 km/hr.A lantern is at the center of the circle.A fence is along the tangent to the circle at the point at which the horse starts.What is the speed with which the shadow of the horse move along the fence at the moment when it covers $1/8$ of the circle in km/hr?

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I tried to attempt the question,but it confuses me what is the importance of the lantern put at the center of the circle.Does it impact the speed of the shadow anyway?I could not solve it,please help me.

Answer 1

As the other answer already pointed out, the position of the shadow from the starting point is given by $x=R\tan\theta$. The time derivative of this position will give the speed $v$ of the shadow. $$v=\frac d{dt}\left(R\tan\theta\right)=R\sec^2\theta\frac{d\theta}{dt}$$ Let $\omega=\dfrac{d\theta}{dt}$ then we have that $v=R\omega\sec^2\theta$

Since the horse is running at a speed of 20 km/h we get $R\omega=20$. Finally $v=20\sec^2 \dfrac{\pi}4=40$ $ km/h$

NOTE: The question assumes that the shadow formed is due to the lantern, so the shadow will always be on the straight line joining the horse and the lantern. So it being on the centre just enables you to find where the shadow is.

Answer 2

A picture usually helps. You know how fast the horse moves around the circle. Just find out how fast the projection of the horse's shadow on the fence moves...

Hint:

Given that the horse has moved an angle $\theta$. The length of the shadow projection is,

$$L=\tan(\theta(t)) \cdot r$$

Where $\theta(t)$ is the angle at time $t$ and $r$ is the circle's radius.

Answer 3

Lantern at circle center

$ y = R \tan\theta$

$ \dot y = V_{shadow} = R \sec^{2}\theta \dfrac{d\theta}{dt} = 2 V_{horse} = 40 \frac{km}{hr} , \because \theta = \pi/4. $