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Two racers in adjacent lanes move with velocity functions $v_{1}(t) \mathrm{m} / \mathrm{s}$ and $v_{2}(t) \mathrm{m} / \mathrm{s}$, respectively. Supposethat the racers are even at time $t=60 \mathrm{s}$. Interpret the value of the integral$$\int_{0}^{60}\left[v_{2}(t)-v_{1}(t)\right] d t$$in this context.

$$t=0$$

Calculus 1 / AB

Calculus 2 / BC

Chapter 6

APPLICATIONS OF THE DEFINITE INTEGRAL IN GEOMETRY, SCIENCE, AND ENGINEERING

Section 1

Area Between Two Curves

Integrals

Integration

Applications of Integration

Area Between Curves

Volume

Arc Length and Surface Area

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uh, obvious. Basically, how would show my work to explain this problem? Um, so that's what they give you. And you have to explain. The work is if you were to actually work out this math, Um, well, the anti derivative of velocity is position. I like doing X, but some people do P, I'll do p for because it's just variable position. Um, my position of the first person, and that's from 0 to 60. So as you plug in P two of 60 minus p one of 60 um, minus p two at time, zero whips, supposed to write down zero. And I just wrote down that, um minus the position of the first racer at zero. Now, here's the thing Is they told you that at times 60 there, even, Well, if they're even, like 100 m or David, look at the units. Um, if there, even in the subtraction is equal to zero. So basically, what it boils down to, um is we're just left with this thing. Um, no, I guess I'll distribute that minus in P to zero plus pew one of zero. Um, what it boils down to is, you know, we could switch this around, cause addition is community is, um What that equation shows us is the lead that the razor would have at that time. So, you know, based on Alex written this number minus this number. So you could say the first racer, um, has a lead. Well, I guess we could still be zero. So I need to be careful how I commit to that answer. But it shows you the first racers lead at T equals zero at time zero. And that's really what? That equation represents our expression.

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