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Several Ways to Solve a Rate-Time-Distance Problem

Date: 02/14/2006 at 01:17:41
From: Al
Subject: When and where will the car pass the bus?

A bus leaves point A travelling at a speed of 10 mph.  22 minutes 
later a car leaves point A travelling in the same direction at 40 mph.
At what distance will the car and the bus meet, and how much time will
have elapsed?  Is there a formula for a problem like this?



Date: 02/14/2006 at 10:02:53
From: Doctor Greenie
Subject: Re: When and where will the car pass the bus?

Hi, Al --

The basic formula is the familiar

  distance = rate * time

but few math problems are solved simply by applying formulas.  We need 
to find a way to use the formula to solve our particular problem.

I see several ways to approach your problem.  Let me show you three of 
them; you can see which of them makes most sense to you and then you 
can use that method the next time you run into a problem like this.  
Or better yet, try to understand all three methods so you can choose 
to apply any one of them to a particular problem--because any one of 
the approaches might be obviously easier to apply to any particular 
problem.

Probably the traditional approach to the problem is using algebra and 
the basic distance/rate/time formula.  To begin with, we need to 
remember to keep our units straight--because the speeds are given in 
mph and the time difference is given in minutes.  22 minutes is 22/60 
of an hour, so we can say

  let t = time car travels (hours)
  then t + 22/60 = time bus travels (hours)

We have defined expressions for the times the two vehicles travel; and 
the problem tells us the two speeds. So our basic formula

  distance = rate * time

tells us

  40t = distance car travels
  10(t + 22/60) = distance bus travels

We want to know when the distances traveled by the two vehicles are 
the same because when the car catches the bus they will each have gone
the same distance from point A, so we can write and solve an equation
which says just that:

  40t = 10(t + 22/60)

Solving this equation is then basic algebra:

  40t = 10t + 22/6
  30t = 22/6
    t = 22/180 = (7 1/3)/60 (hours) = 7 1/3 minutes

A different use of the basic formula to solve the problem involves 
finding how far the bus has gone before the car starts and then using 
the DIFFERENCE of the speeds to determine how long it will take the 
car to catch up to the bus.  The thinking behind this approach is 
that, after the car starts, the bus is still moving, so the car is 
CATCHING UP TO the bus at a rate equal to the DIFFERENCE between the 
two speeds.

The distance the bus travels before the car starts is

  10(22/60) = 22/6 miles

The car is catching up to the bus at the rate of 30mph.  To find out 
how long it takes for the car to catch up, we use the following 
variation of the basic formula:

  time = distance/rate

  t = (22/6)/30 = 22/180 (hours) = ... = 7 1/3 minutes

The third method I easily see for solving this problem uses the ratios 
of the speeds instead of the speeds themselves.  The thinking behind 
this approach is that, since the car travels 4 times as fast as the 
bus, the bus will take 4 times as long as the car to go a particular 
distance.  Again (as in the first method above) we start with

  let t = time car travels (hours)
  then t + 22/60 = time bus travels (hours)

Then, since we know the bus takes 4 times as long to go a particular 
distance (since the car at 40 mph is 4 times as fast as the bus at 10
mph), we have

  (t + 22/60)   4
  ----------- = -
       t        1

  4t = t + 22/60
  3t = 22/60
   t = 22/180 (hours) = ... = 7 1/3 minutes

I hope I haven't confused you by showing several different approaches
to your problem.  Different readers are going to find different ones
of these methods easier to understand and to use; I like to give them
that choice.

And I personally like to have different methods to choose from.  One 
method or another might be easier to apply to any particular problem.
Or I might find for some reason that I can't get a particular method
to work out, and I can then use another method to solve the problem. 
Or I can even get a confidence check in my answer if I can solve a
particular problem by more than one method and get the same answer.

I hope all of this helps.  Please write back if you have any further 
questions about any of this.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
Middle School Algebra
Middle School Word Problems

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