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Bob is writing his bicycle along the same path for ( 0 \leq t \leq 10 ). Bob's velocity is modeled by ( b(t) = t^3 - 6t^2 + 300 ) where ( t ) is measured in minutes and ( b(t) ) is measured in meters per minute. Find Bob's acceleration at time ( t = 5 ).

Well, acceleration—this is a velocity function right over here—so the acceleration is going to be the derivative of the velocity function with respect to time. How? What is the rate of change of velocity with respect to time? That's acceleration.

So we really just want to evaluate Bob's acceleration at ( t = 5 ). That's going to be ( b'(5) ). So let's first figure out what ( b'(t) ) is. ( b'(t) ) is equal to—we'll take the derivative here. It's pretty straightforward; just use the power rule. So it's going to be ( 3t^2 - 12t ) (2 times negative 6 is negative 12), and then the derivative of 300 is 0 since it doesn't change with respect to time.

And so ( b'(5) ) is going to be equal to ( 3 \times 5^2 - 12 \times 5 ), which is equal to ( 75 - 60 ). This is equal to ( 15 ).

And the units here—this is acceleration, so this is going to be his velocity was in meters per minute, and so this is going to be meters per minute per minute. Because remember, time is in terms of minutes. So we could write it out as meters per minute per minute, which is the same thing as meters per minute squared.

All right, let's do the next part. Based on the model ( b ) from part C, find Bob's average velocity during the interval from ( 0 \leq t \leq 10 ).

And if the notion of average velocity or average value of a function is completely foreign to you, I encourage you to watch the videos on Khan Academy on finding the average of a function. But straight to the chase, the average velocity—the average velocity is going to be the area under the velocity curve divided by our change in time.

So the area under the velocity curve from ( t = 0 ) to ( t = 10 ) of ( b(t) ) ( b(t) , dt ) divided by our change in time. So it's going to be divided by—well, you're going from 0 to 10; so ( 10 - 0 = 10 ).

And if you wanted the intuition here, it's like—well, if you know the area of something, and if you wanted to find its average height, you could just divide by its width. And that's what we're doing here. If we know the area of something, we want to figure out its average height, and so you divide by its width. That's, I guess, a very high-level intuition for where this expression came from.

And so this is going to be equal to ( \frac{1}{10} ) times the integral from ( 0 ) to ( 10 ) of ( b(t) = t^3 - 6t^2 + 300 , dt ).

And so this is going to be equal to ( \frac{1}{10} ). Take the antiderivative here; so this is going to be ( \frac{t^4}{4} - 2t^3 + 300t ).

I'm going to evaluate it at 10 and subtract from that the evaluation at 0. And so this is going to be equal to—this is going to be equal to ( \frac{1}{10} )—that same ( \frac{1}{10} ) there.

When you evaluate all of this at 10, what are we going to get? Let’s see. ( 10^4 = 10000 ), divided by 4 is ( 2500 ). Then minus ( 2 \times 10^3 ) (which is ( 2000 )), and then ( 300 \times 10 )—well, that’s plus ( 3000 )—and then you subtract all of this evaluated at zero, which is just going to be zero.

So this is going to be equal to ( 2500 - 2000 = 500 + 3000 ). This all simplifies to ( 3500 ) or ( 3,500 ).

Then you divide by ( 10 ); this is going to be ( 350 ).

This is an average velocity ( 350 ) meters per minute, and we are done.