### Antiderivatives r us

Today in class, we did an optimization problem which took half of the class. The rest of class was spent on antiderivative problems. I'll show what the antiderivative problem looked like because it is something new that we haven't done before.

Solution:

First you have to start with the acceleration function because that is what the problem gives you.

a (t) = -8

Now find the antiderivative.

v (t) = -8t +C

Solve for when t=0 and the velocity equals 66

v (t) = -8(0) +C = 66

C = 66

v (t) = -8t +66

Now find the antiderivative of the velocity function to find the position function.

s(t) = -4t^2 +66t +C

When t =0, the position is 0, so solve for t=0

s(0) = -4(0)^2 +66(0) +C = 0

C = 0

s(t) = -4t^2 + 66t

Solve for t when the function v(t) = 0

0 = -8t +66

0 = -2(4t -33)

t = 33/4 = 8.25

Now evaluate the position function for when t = 8.25

s(8.25) = -4(8.25)^2 +66(8.25)

s(8.25) = -272.25 + 544.5

= 272.25 ft

The next scribe is Chris

*A train is traveling at 66 ft/sec when the brakes are applied. If the train decelerates at 8 ft/sec^2. How far did the train travel between the time the brakes were applied and when it stopped?*Solution:

First you have to start with the acceleration function because that is what the problem gives you.

a (t) = -8

Now find the antiderivative.

v (t) = -8t +C

Solve for when t=0 and the velocity equals 66

v (t) = -8(0) +C = 66

C = 66

v (t) = -8t +66

Now find the antiderivative of the velocity function to find the position function.

s(t) = -4t^2 +66t +C

When t =0, the position is 0, so solve for t=0

s(0) = -4(0)^2 +66(0) +C = 0

C = 0

s(t) = -4t^2 + 66t

Solve for t when the function v(t) = 0

0 = -8t +66

0 = -2(4t -33)

t = 33/4 = 8.25

Now evaluate the position function for when t = 8.25

s(8.25) = -4(8.25)^2 +66(8.25)

s(8.25) = -272.25 + 544.5

= 272.25 ft

The next scribe is Chris

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