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pishuonlain [190]
3 years ago
8

a street light is mounted at the top of a 15 foot pole. A man 6 ft tall walks away from the pole wit a speed of 7 ft/s along a s

traight path. How fast is the tip of his shadow moving when he is 40 ft from the pole
Physics
1 answer:
Mariulka [41]3 years ago
4 0

Answer:

16.3 ft/s

Explanation:

Let d=distance

and

x = length of shadow.

Therfore,

x=(d + x)

 = 6/15

So,

    15x = 6x + 6d

     9x = 6d.

x = (2/3)d.

As we know that:

dx=dt

   = (2/3) (d/dt) 

Also,

Given:

d(d)=dt

     = 7 ft/s

Thus,

d(d + x)=dt

           = (7/3)d (d/dt)

Substitute, d= 7  

d(d + x) = 49/3 ft/s.

Hence,

d(d + x) = 16.3 ft/s.

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Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

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Which the formula for constant acceleration would be > v_2^2=v_1^2 + 2as

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So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > 0.0134 mi *  \frac{5280ft}{1mi}  = 70.8 ft
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