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## Algebra (all content)

# Worked example: Rate problem

CCSS.Math:

Sal solves two related proportion word problems about a squirrel crossing the road and a car that approaches it. He does that using dimensional analysis. Created by Sal Khan.

## Video transcript

A squirrel is running across
the road at 12 feet per second. It needs to run 9 feet
to get across the road. How long will it take the
squirrel to run 9 feet? Round to the nearest
hundredth of a second. Fair enough. A car is 50 feet away
from the squirrel-- OK, this is a high-stakes word
problem-- driving toward it at a speed of 100
feet per second. How long will it take
the car to drive 50 feet? Round to the nearest
hundredth of a second. Will the squirrel make
it 9 feet across the road before the car gets there? So this definitely is
high stakes, at least for the squirrel. So let's answer
the first question. Let's figure out
how long will it take the squirrel to run 9 feet. So let's think about it. So the squirrel's
got to go 9 feet, and we want to figure
out how many seconds it's going to take. So would we divide or
multiply this by 12? Well, to think about that, you
could think about the units where we want to get an
answer in terms of seconds. We want to figure
out time, so it'd be great if we could multiply
this times seconds per foot. Then the feet will cancel out,
and I'll be left with seconds. Now, right over here, we're
told that the squirrel can run at 12 feet per second,
but we want seconds per foot. So the squirrel, every second,
so they go 12 feet per second, then we could also say 1
second per every 12 feet. So let's write it that way. So it's essentially
the reciprocal of this because the units are
the reciprocal of this. So, it's 1 second
for every 12 feet. Notice, all I did is I took this
information right over here, 12 feet per second, and I wrote
it as second per foot-- 12 feet for every 1 second,
1 second for every 12 feet. What's useful about this is
this will now give me the time it takes for the
squirrel in seconds. So the feet cancel
out with the feet, and I am left with 9 times
1/12, which is 9/12 seconds. And 9/12 seconds
is the same thing as 3/4 seconds, which is
the same thing as 0.75 seconds for the squirrel
to cross the street. Now let's think about the car. So now let's think
about the car. And it's the exact same logic. They tell us that the
car is 50 feet away. So the squirrel is trying
to cross the road like that, and the car is 50 feet
away coming in like that, and we want to figure out if
the squirrel will survive. So the car is 50 feet away. So it's 50 feet away. We want to figure out the
time it'll take to travel that 50 feet. Once again, we would
want it in seconds. So we would want
seconds per feet. So we would want to multiply
by seconds per foot. They give us the speed
in feet per second, 100 feet per second. And so we just have
to realize that this is 100 feet for every 1 second,
or 1/100 seconds per feet. This is once again
just this information, but we took the
reciprocal of it, because we don't
want feet per second, we want seconds per feet. And if we do that,
that cancels with that, and we're left with
50/100 seconds. So this is 50/100
is 0.50 seconds. And so now let's answer the
question, this life and death situation for the squirrel. Will the squirrel make
it 9 feet across the road before the car gets there? Well, it's going to take the
squirrel 0.75 seconds to cross, and it's going to take the
car only half a second. So the car is going to get to
where the squirrel is crossing before the squirrel
has a chance to get all the way across the road. So unfortunately for the
squirrel, the answer is no.