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The velocity (in $\mathrm{km} / \mathrm{h}$ ) of a plane flying into an increasing headwind is $v=50(12-t),$ where $t$ is the time (in h). How far does the plane travel in a 2.0 -h trip?
$1100 \mathrm{km}$
Calculus 2 / BC
Chapter 26
Applications of Integration
Section 1
Applications of the Indefinite Integral
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Okay, so we're told the velocity equation V. Is equal to 50 times 12 minus T. And the way that we're gonna figure out the um distance this plane has traveled after two hours is we're going to use the fact that the integral of our velocity equation is equal to our displacement equation S. F. T. And then once we have our displacement equation we can just plug in two for tea sits, teas and hours to figure out how far our plane has traveled. So the integral of 50 Times 12- T. It's equal to 50 times the integral of 12- T. And this should be should be a DT here. There should have been a DT here as well but it's okay so now we can find the integral, the integral of 12 is just 12 T. And the integral of negative T. Is negative T squared divided by two. So we have 50 times 12 t minus t squared divided by two. And so now if we um Distribute this 50 across we're gonna this is equal to 600 T -25 t square. And so now all we have to do is we plug in and there would have been a plus see here too. But we can ignore that since we weren't given any initial um displacement or anything like that. So see here we can assume is zero. So now that we have our displacement equation, we can say this is equal to S. Of T. We can plug in two for tea to figure out how far a plane has traveled after two hours. So Running at 600 times 2 -25 Time just two squared. So 600 times two is 1200. And then we have 25 times four Just 100. So then we get displacement of 1100 and this is in km. So after a two hour trip are playing his um Our plane has traveled 1100 km
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