The illumination provided by a car's headlight varies inversely as the square of the distance from the headlight

Question

The illumination provided by a car's headlight varies inversely as the square of the distance from the headlight. A car's headlight produces an illumination of 3.75 (15/4) Foot-candles at a distance of 40 feet. What is the illumination when the distance is 50 feet?

Answer 1

Write out the statement of the illumination law:

 

I = k/d²

 

I is the illumination

k is the constant of proportionality

d is the distance.

 

Plug in what you know to find k:

 

3.75 = k / 40²

3.75 = k / 1600

6000 = k

 

Now use this to find the illumination at 50 feet:

I = 6000 / 50²

I = 6000 / 2500

I = 2.4

 

 

OR, you can avoid bothering to find k.

 

The second distance is 50/40 times the first distance, or 1.25.  Since the illumination varies INVERSELY as the SQUARE of the distance, the illumination at the second distance will be 1.25² times fainter than at the first distance.  1.25² is 1.5625, so divide 3.75 by 2.25 = 2.4.

 

Answer 2

Let I = illumination and D = distance

 

Since I varies inversely as D², I = k/D² for some constant, k.

 

I = 3.75 when D = 40, so 3.75 = k/(40)².

 

Therefore, k = 3.75(40)² = 6000

 

So, I = 6000/D²

 

When D = 50, I = 6000/(50)² = 2.4 foot-candles