# Thread: Direct and Inverse Variation

1. ## Direct and Inverse Variation

A car's stopping distance varies directly with the speed it travels, and inversely with the friction value of the road surface. If a car takes 60 feet to stop at 32 mph, on a road whose friction value is 4, what would be the stopping distance of a car traveling at 60 mph on a road with a friction value of 2?

MY WORK:

Let d = distance
Let k = constant of variation
Let s = speed
Let f = friction

I came up with the equation d = ks/f

I then proceeded to do the following:

60 = 32k/4

240 = 32k

240/32 = k

8 = k

Now that I found k, I proceeded to find the distance:

d = 60(8)/2

d = 240 feet

The correct answer is 225 feet.

Where did I go wrong?

2. Originally Posted by sologuitar
A car's stopping distance varies directly with the speed it travels, and inversely with the friction value of the road surface. If a car takes 60 feet to stop at 32 mph, on a road whose friction value is 4, what would be the stopping distance of a car traveling at 60 mph on a road with a friction value of 2?

MY WORK:

Let d = distance
Let k = constant of variation
Let s = speed
Let f = friction

I came up with the equation d = ks/f

I then proceeded to do the following:

60 = 32k/4

240 = 32k

240/32 = k

8 = k

Now that I found k, I proceeded to find the distance:

d = 60(8)/2

d = 240 feet

The correct answer is 225 feet.

Where did I go wrong?

Hi sologuitar,

$\frac{240}{32}\ne 8$

3. ## I see...

Originally Posted by masters
Hi sologuitar,

$\frac{240}{32}\ne 8$
I see that it was a simple error on my part.
We know that 240/32 = 7.5 not 8. I now see where the error occured.