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November 01, 2025, 03:20:10 pm
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Author Topic: shadow rate  (Read 748 times)  Share 

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Martoman

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shadow rate
« on: July 12, 2010, 12:21:09 am »
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A person 2 meters tall is walking away from a streetlight 8 meters high at a rate of 50 cm per second.

At what rate is the length of the person's shadow changing?

So I set up the triangles. i know

i'm calling the length of the shadow L and the distance from the man and the pole x.

Similar triangles being:

Doing the mathy thing:

so

I want and -ve as its getting smaller

So ?!?

Or is this just pain wrong?
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moekamo

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Re: shadow rate
« Reply #1 on: July 12, 2010, 02:16:10 am »
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the rate of change of distance is in cm, but u have the heights of the man and pole in metres, i think it has something to do with that...
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Martoman

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Re: shadow rate
« Reply #2 on: July 12, 2010, 02:19:12 am »
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ah god.damn.it is the method right though?
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moekamo

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Re: shadow rate
« Reply #3 on: July 12, 2010, 02:42:25 am »
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yea i think so, do you have the answer?
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Martoman

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Re: shadow rate
« Reply #4 on: July 12, 2010, 02:44:19 am »
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nah which is why i'm posting it to verify if what i've done is right. So long as the method stands up to scrutiny. I don't care about the numbers, its the thought processes that matter.  :smitten:
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kakar0t

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Re: shadow rate
« Reply #5 on: July 12, 2010, 03:48:26 am »
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thats what she said :D

ily similar triangles!
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Chavi

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Re: shadow rate
« Reply #6 on: July 12, 2010, 10:34:33 am »
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Firstly, you need to ensure all figures are in the same unit. you can't have dx/dt = 50 in cm, and Dl/dx in meters.

So,  
and the rest is correct:
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