286 algebra De mystif ieD

We begin with two cars/cyclists/runners, etc. moving in the same direction.

The rate at which the distance between them is changing is the difference in

their rates. For example, if a car is traveling northward on a highway at 60 mph

and another car behind the first traveling at 70 mph, then the rate at which the

distance between them is decreasing is 10 mph.

EXAMPLE

A bicyclist starts at a certain point and rides at a rate of 10 mph. Twelve

minutes later, another bicyclist starts from the same point in the same

direction and rides 16 mph. How long will it take for the second cyclist to

catch up with the first?

When the second cyclist begins, the first has traveled:

rt= d

(^10) ()^1260 = 2
miles
(Converting 12 minutes to hours gives us^1260 hours.) Because the cyclists are
moving in the same direction, the rate at which the distance between them
is decreasing is 16 – 10 = 6 mph. Then, the question boils down to “How
long will it take for someone traveling 6 mph to cover 2 miles?”
Let t represent the number of hours the second cyclist is traveling.
drt
t
t
t

=

=
26
2
6
1
3
It will take the second cyclist 31 of an hour, or 20 minutes, to catch up the
first cyclist.
A car passes an intersection heading north at 40 mph. Another car passes
the same intersection 15 minutes later heading north traveling at 45 mph.
How long will it take for the second car to overtake the first?
In 15 minutes, the first car has traveled^40 ()^1560  = ^10 miles. The second car
is gaining on the first at a rate of 45 – 40 = 5 mph. So the question becomes
“How long will it someone traveling 5 mph to cover 10 miles?”
EXAMPLE
A bicyclist starts at a certain point and rides at a rate of 10 mph. Twelve