You can
put this solution on YOUR website!For both of them,
, where
= distance
= rate
= time
So, for Carlos,
And for Juanita,
----------------------
The chart would look like this:
-------------------------------
Distances -- rates -- times
-------------------------------
-- d[c] -- -- r[c] -- -- t[c]
-- d[j] -- -- r[j] -- -- t[j]
--------------------------------
Now I can repace some of these variables
with things that I know about them
For instance,
because
the distance is from LA to the place
where they meet. I'll call them both
-------------------------------
Distances -- rates -- times
-------------------------------
-- d -- -- r[c] -- -- t[c]
-- d -- -- r[j] -- -- t[j]
--------------------------------
Carlos travels 50 mi/hr and Juanita travels 60 mi/hr, so
-------------------------------
Distances -- rates -- times
-------------------------------
-- d -- -- 50 -- -- t[c]
-- d -- -- 60 -- -- t[j]
--------------------------------
Juanita leaves LA an hour later, so she has to cover the same
distance in 1 hour less time, so
-------------------------------
Distances -- rates -- times
-------------------------------
-- d -- -- 50 -- --t[c]
-- d -- -- 60 -- --t[c] - 1
--------------------------------
Now I can use this to find out when Juanita will pass Carlos.
The problem doesn't ask for it, but I'll do the algebra
for both of them, so
(1)
(2)
The
s are the same in (1) and (2),
so I'll set them equal to eachother
(3)
And, since
,
Carlos left LA at 8AM, so 6 hours later,
it would have been 2PM
Juanita left LA at 9AM, so 5 hours later,
It would have been 2PM, which is when she passed him
check answer:
The distances must be the same, so
(1)
mi
(2)
mi
OK
(