The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Safe Passing Distance

Date: 07/30/99 at 10:39:38
From: k.m.
Subject: Time and distance

If a car travels at 50 mph and a passing car travels at 55 mph, what 
is the distance that the 55 mph car takes to get around the 50 mph 
car? What is the formula? (So I can also figure it out for a 50 mph 
car passed by a 70 mph car, etc.)

Date: 07/30/99 at 13:41:19
From: Doctor Rick
Subject: Re: Time and distance

Hi, K. M.

Now there's a good practical question. When you're on a 2-lane road 
and you can only see ahead 1000 feet, is it safe to pass?

You need to put a distance figure on what "get around" means. That is, 
say we require that the passing car starts 5 car lengths back and ends 
up 5 car lengths ahead; it has to move up 12 car lengths relative to 
the 50 mph car. How long is an average car - say, 15 feet? That means 
the passing car needs to go 15' * 12 = 180 feet relative to the passed 

For now let's work with variables instead of specific numbers:

     D = distance passing car moves while passing

     d = distance passing car needs to move relative to passed car

     v1 = speed of passed car in mph

     v2 = speed of passing car in mph

The speed of the passing car relative to the passed car is v2-v1. 
Using the rate equation,

     distance = speed * time

we get

     d = (v2 - v1) * t

     t = d / (v2 - v1)

How far does the passing car move in this time? Use the rate equation 

     D = v2 * t

     D = d * v2 / (v2 - v1)

Looking at the dimensions in this equation, we see that it doesn't 
matter whether the speeds are measured in mph or feet per second or 
whatever; speed divided by speed is dimensionless. The dimensions of D 
will be the same as the dimensions of d (namely, feet).

Now we can plug in some numbers.

      d = 180 feet
     v1 = 50 mph
     v2 = 55 mph

      D = (180)(55)/(55 - 50)
        = 1980 feet

Looks like it isn't safe to pass if you can only see 1000 feet ahead, 
unless you can accelerate to at least 60 mph. In an emergency you 
could pull in front of the passed car after going maybe 8 car lengths, 
or 120 feet, at which point you would have traveled 1320 feet at 55 

And that reminds me: we are assuming that the passing car is already 
going 55 mph. If it started going 50 mph behind the other car, we'd 
need to factor in the rate of acceleration, and it would take longer 
to pass.

- Doctor Rick, The Math Forum   
Associated Topics:
High School Physics/Chemistry

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.