gibbenergy Posted November 12, 2005 Share Posted November 12, 2005 I don't know how to solve this one. Can anyone help me? A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks from the pole with a speed of 5ft/s along the straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole? Link to comment Share on other sites More sharing options...

Garfield Posted November 14, 2005 Share Posted November 14, 2005 [math]\frac{15*5}{6} = 12.5\frac{ft}{s}[/math]? Link to comment Share on other sites More sharing options...

Sisyphus Posted November 14, 2005 Share Posted November 14, 2005 You've got some similar triangles. Therefore: length of shadow:man's height::distance from end of shadow to pole:pole's height Since both the man and the pole have constant heights with a constant ratio of 6:15, so too will the ratio of man's distance to shadow's distance be constant, with the same ratio. So if the man is walking at 5ft/s, the shadow is moving at (15/6)(5ft/s)=12.5ft/s. EDIT: Since this is in the calculus forum, I guess you want a calculus solution. You can write the total length (y) as a function of the man's distance from the pole (x) with y=f(x)=15x/6=5x/2. Also, you can write the man's distance as a function of time: x=g(t)=5t. Combine for f(g(t)) to get total distance as a function of time, y=(15/6)(5t)=75t/6=12.5t. Then you just take its derivative at t=8 to find the ft/s there, which in this case will always just be 12.5. Link to comment Share on other sites More sharing options...

Garfield Posted November 15, 2005 Share Posted November 15, 2005 I'm having doubts about 12.5 ft/s... Post height = [math]H[/math] Object height = [math]h[/math] Shadow speed = [math]V[/math] Object speed = [math]v[/math] [math]\frac{H}{h}=\frac{V}{V-v} =>V=\frac{Hv}{H-h}[/math] Then it should be [math]8.(3)\frac{ft}{s}[/math] Is this correct? Link to comment Share on other sites More sharing options...

Sisyphus Posted November 15, 2005 Share Posted November 15, 2005 Er, yes... I was a tad mixed up there. The ratio is NOT total distance:man's distance::pole height:man's height but RATHER total distance:total distance - man's distance::pole height:man's height So: The ratio of the shadow to the whole is 2 to 5 (simplified from 6 to 15). Therefore the ratio of the man's distance to the whole is 3 to 5. The speed is constant, so the ratio of the man's speed to the shadow's speed is also 3 to 5. The man's speed is 5ft/s. Therefore, the shadow's speed is 25/3. I put the blame entirely on you. Link to comment Share on other sites More sharing options...

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