# Calculations in regards to a car coming to a stop, help is needed.

1. May 23, 2012

### connormaphone

I have been given a physics assignment which requires that I go over the physics of why 40km/h is a better speed to be followed in a school zone rather than 50km/h. (must include calculations for stopping distances etc.)

I used the formula Vo ^ 2 / 2 x μ x g.
It should give me the stopping distance if I am correct, Vo being the inital velocity, μ the friction coefficient and g for gravity. I first tried using 40km/h for Vo, .7 for μ and 9.8 for g.

40^2/2x.7x9.8

I use this and end up with 116.61...kilometres, this is evidently wrong and I am not sure what I am doing wrong as it should come out to 20-30 metres.
Thanks.
(P.S I am in grade 11, so go easy on me )

2. May 23, 2012

### PhanthomJay

You forgot to convert km/hr into m/s. It is essential that units are consistent.

3. May 23, 2012

### connormaphone

Oh that gives me an answer of 13.72, I feel stupid now. Thanks for the help!

4. May 23, 2012

### PhanthomJay

13.72 m? Show how you arrived at that answer.

5. May 23, 2012

### connormaphone

40000^2/2x.7x9.8
Just realized this is wrong as well. I don't even know.

6. May 23, 2012

### connormaphone

Actualy, just made it 11.1 m/s which does equal 40km/h, I put it into the equation and ended up with approximately 9 metres.

7. May 24, 2012

### PhanthomJay

Now, you're talking! Looks Good!
And welcome to PF!

8. May 24, 2012

### Staff: Mentor

Your starting formula is incorrect. The μg should be in the denominator. That is the deceleration. The larger the deceleration, the shorter the stopping distance.

Chet

9. May 24, 2012

### PhanthomJay

It is in the denominator. Parentheses would help, but I think it is in proper PEMDAS, per my recollect.

10. May 25, 2012

### connormaphone

Got it, thanks for the help guys!