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Deceleration A car slows down with an acceleration of $a(t)=-15 \mathrm{ft} / \mathrm{s}^{2} .$ Assume that $v(0)=60 \mathrm{ft} / \mathrm{s}$ and $s(0)=0.$a. Determine and graph the position function, for $t \geq 0.$b. How far does the car travel in the time it takes to come to rest?

a. $s(t)=-\frac{15}{2} t^{2}+60 t$b. $120 \mathrm{ft}$

Calculus 2 / BC

Chapter 6

Applications of Integration

Section 1

Velocity and Net Change

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So here we have a function of acceleration for a car that is accelerating at negative 15 feet per second per second. We know that at time zero the velocity 60 and the position is zero. So what we're gonna do first is we're gonna figure out our formulas are functions for velocity and position. So we're gonna start with the of tea. So v f t to find the FT Rupeni anti derivative of a hefty, which is negative 15 t plus c And then to figure out what see is we're gonna make this whole thing cool to 60 and we're gonna make t equal to zero because of the of zero. If we do that, this cancels and C is equal to 60. Thus, our function is negative. 15 t plus 60 doing the same thing for the position as of tea. We know that at time zero it's equal to zero. So r c will most likely V equals zero. Let's figure that out. We have negative 15 t squared over two plus 60 t plus. See if we make this whole thing into equal to zero, this is zero. This is zero. His cancels this cancels and see is again equal to zero. So we get negative 15 t squared over two plus 60 t From there, it asks us to graph the put, uh, positive function. If you were to do that, you would get a large function on large quadratic function, where one is equal to 52.5 and it's kind of keeps on going up. So you should try like that. And yet so it asks us also, how far does the car travel in the time it takes to come to arrest? So I coming to arrest implies that it is having a velocity of zero. So to figure that up, let's make our velocity equals zero and figure out what our time is when the equals zero. If we do that, find a 60 here minus 60. Here we get 60 equals 15 t so t has to be equal to four. So if he is equal to four, we plug that back into our position function. So s of four goals negative. 15 times four square over two, plus 60 times four Foursquare is 16 but divided by two is eight. So we get negative 15 times eight, which is 120 plus 240 which is equal to 120 feet, and that's it.

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