1. Direct Proportions

I'm stuck on this math problem:

"The stopping distance d of an automobile is directly proportional to the square of its speed v. A car required 75 feet to stop when its speed was 60 miles per hour. Find a mathematical model that gives the stopping distance d in terms of its speed v"

I'm not sure how to approach this problem. I tried plugging converting the 60 mph to ft (318,800ft) and then plugging it into $D=KV^2$, and then solving for K to get the constant term needed to write the equation.

Whenever I solve K, I get "7.37946E-10", and plug it into $D=(7.37946E-10)(V^2)$, which is apparently wrong according to the program I am using. Any help?

2. Originally Posted by JoeF107
I'm stuck on this math problem:

"The stopping distance d of an automobile is directly proportional to the square of its speed v. A car required 75 feet to stop when its speed was 60 miles per hour. Find a mathematical model that gives the stopping distance d in terms of its speed v"

I'm not sure how to approach this problem. I tried plugging converting the 60 mph to ft (318,800ft) and then plugging it into $D=KV^2$, and then solving for K to get the constant term needed to write the equation.

Whenever I solve K, I get "7.37946E-10", and plug it into $D=(7.37946E-10)(V^2)$, which is apparently wrong according to the program I am using. Any help?
Well, mph is "miles per hour." The speed you want is in ft/s, not feet. I'm getting your v to be 88 ft/s. That should fix things up for you.

-Dan

3. Originally Posted by topsquark
Well, mph is "miles per hour." The speed you want is in ft/s, not feet. I'm getting your v to be 88 ft/s. That should fix things up for you.

-Dan
Hmmm...

I got $d=(v^2)(.009685)$, which the program is still telling me is wrong O_o.

However, I was able to use that exact same formula to correctly solve for a distance given a velocity 104 ft/s. do you think I am still doing something wrong, or do to think there may be a glitch in the program I'm using.