# Force of an auto while driving

1. Jan 18, 2009

### afg_91320

1. The problem statement, all variables and given/known data
The force felt on an auto due to its movement through still air varies approx. as the square of the car's speed. (the force is also known as air resistance) assume the force varies exactly as the square of the speed. around town at 30mi/hr a car experiences an air resistance force of 100lbs. what size force would you expect the car to face while traveling at 65mi/hr?

2. Relevant equations
car travels at 65mi/hr
at 30 mi/hr its air resistance force is 100 lbs.

3. The attempt at a solution
ive been looking at this problem for the last hour, and i dont know what good the weight will help me with. ive tried conversion but it gets me nowhere. i dont want the answer, i just want to know how should i approach the problem, and how do i start!

2. Jan 18, 2009

### tiny-tim

Welcome to PF!

Hi afg_91320! Welcome to PF!

This is a dimensions problem.

So write an equation with a "proportional" sign in the middle, instead of an equals sign …

Force ~ (a function of v) …

multiply v by 65/30, and what happens?

3. Jan 18, 2009

### afg_91320

ok-here is what i got:

65/30 = 100/v <-- equal sign means propotional
65v=3000
v=3000/65
v=46.153
but the answer in the back is 4.7 x 10^2 lb.
what did i do wrong?? :(

4. Jan 18, 2009

### mgb_phys

Remember they said the drag was proportional to v2

5. Jan 18, 2009

### afg_91320

so that would make it:
65/30 = 100/v^2
65v^2=3000
v^2=46.153
v=6.8 <--that is incorrect.

6. Jan 18, 2009

### mgb_phys

drag is prop to v2

so drag1/drag2 =v12/v22

7. Jan 18, 2009

### afg_91320

^^how does that help me?

this is what i did:
v1^2/v2^2=65/30
(100)^2/v2^2=65/30
65v^2=300,000
v^2=4615.384
but its still wrong!

can anyone show me the correct steps on how to do it! i have 3 more prblms similar to this! help me with this one, and ill be able to the others correctly!

8. Jan 18, 2009

### mgb_phys

First make an estimate of the answer - so you know if your calculator got it right.
65mph is a little more than twice 30mph, so since force goes up by 4 if speed goes up by 2 we expect a force of around 400-500 lb

Now d1/2 = v1^2/v2^2
d1/100 = 65^2/30^2
d1 = 100 * (65^2/30^2) = 100 * (4225/900) = 100 * 4.69 = 470 lbs

9. Jan 18, 2009

### afg_91320

oohhh!! i got it! i was just doing my math all wrong! thanks!