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(male narrator) In this video,
we will look at solving distance problems
where one individual has to catch up with another.
Because he has to catch up, he has a head start...
one has a head start.
The person with the head start,
we will add the head start to his or her time.
When playing catch-up,
while one travels faster and one travels longer,
they both will travel the exact same distance
by the time the catch-up has occurred.
In math, we use equals for the same.
They have equal distances.
We will make the distances equal to each other.
So for example, in this problem, Raquel leaves a party,
and 4 hours later, Nick leaves to catch up with her.
Let's organize our information in a table...
rate times time, equals the distance.
For Raquel, her rate is 5 miles per hour.
For Nick, his rate is 7 miles per hour.
Raquel left first, and so, she has the head start of 4 hours.
We will add 4 to Raquel's time; while Nick's time is simply t.
Multiplying together gives us:
5 times t, plus 4, for Raquel's distance;
and 7 times t for Nick's distance.
If Raquel's distance is 5 times t, plus 4;
and Nick's distance is 7t, but they're both the same,
we can set those distances equal to each other:
5 times t, plus 4, equals 7t.
That gives us an equation we can quickly solve
by first distributing the 5: 5t plus 20, equals 7t.
Start solving by getting the variables
all on one side of the equation.
Subtracting 5t from both...
will give us 20 equals 2t.
Finally, dividing by 2 will tell us
the amount of time Nick needs to catch up: 10 hours.
Let's try another example where we see one person
trying to catch up with the other.
In this problem, Trip leaves... Trey leaves on a trip.
Julian leaves until she catches up to pass him.
Let's organize our information again in a table.
We know that rate times time, equals a distance.
For Trey, his rate is 20 miles per hour.
For Julian, her rate--or speed-- is 30 miles per hour.
Trey left first and has a head start of 2 hours,
so we will add 2 to Trey's time,
while Julian's time is simply t.
To find their distances, we multiply those together
to get 20, times t, plus 2, for Trey's distance;
and 30, times t,
for Julian's distance.
If Trey is 20, times t, plus 2; and Julian is 30 times t,
but they're both the same distance,
we can set those equations equal to each other:
20 times t, plus 2, is equal to 30 times t.
We can now start solving this equation
by distributing the variable... or the 20 through:
20 times t, plus 40, equals 30t.
Start by getting the variable on the same side
by subtracting 20t from both.
This gives us 40 equals 10t.
We finally solve for t by dividing both sides by 10.
t is equal to 4:
4 hours...it will take Julian to catch up with Trey.
When the distances are the same, we set them equal,
and that can help us solve these problems.