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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.