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A girl of height $0.9 m$ is walking away from the base of a lamp-post at a speed of $1.2 m/s$. If the lamp is $3.6 m$ above the ground, find the length of her shadow after $4$ seconds. I calculated the ration the length of lamp-post and height of the girl which is $4:1$. But I do not know how to link the speed to shadow

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2 Answers 2

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After 4 seconds the girl will be at $4.8m$ away (she moves $1.2 m/s$)

Now let's draw a straight line from the lamp to the ground that passses by the girls head

$y = Ax + B$

when $x=0$ $y=3.6$ therefore $B=3.6$

when $x=4.8$ $y=0.9$ therefore $A=-0.5625$

So the equation becomes $y=-0.5625x+3.6$

Now to find the shadow, let's first find the intersection of that above equation to the ground, in other words when it intersects with the equation $y=0$

you can simply solve $y=-0.5625x+3.6=0$ to find $x = 6.4$

last step

$shadow = 6.4 - 4.8$

$shadow = 1.6m$

enter image description here

Generalization

the girl's x position's equation with respect to time is $x_1 = 1.2t$

the straight line's equation according to the girl's location can be generalized to :

$$y=Ax+B$$ at $x=0$ $y=3.6$ therefore $B = 3.6$

at $x = 1.2t$ $y = 0.9$ therefore $A = -\frac{2.7}{1.2t}$

so : $$y=-\frac{2.7}{1.2t}x+3.6$$

the intersection of this equation with $y=0$ can be found at $x_2 = \frac{3.6*1.2t}{2.7}$

Last:

shadow = $x_2 - x_1$

$$(\frac{3.6*1.2}{2.7}-1.2)t$$ $$\frac{1.2}{3}t$$

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Hint: $\dfrac{x}{x+4.8} = \dfrac{0.9}{3.6}$, where $x$ is her shadow.

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  • $\begingroup$ All thanx to Mr.Thales for his sweet theorem ;) $\endgroup$
    – chouaib
    Oct 14, 2014 at 7:18

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