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a) You can see that 0 ≤ t ≤ 40, so dividing this interval into four sub-intervals isn't too difficult. You get the following four intervals: \[[0,10], [10,20], [20,30], [30,40].\] So, the length of each sub-interval is 10. The midpoints of these intervals are 5, 15, 25, and 35, respectively. \[\int_0^{40}v(t)\;dt\approx\frac{1}{10}\left(v(5)+v(15)+v(25)+v(35)\right)\]
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b) Here you have to look for when v(t) is increasing and decreasing. I presume you know about the first derivative test, and that acceleration, a(t), is the derivative of velocity, v(t). When a(t) < 0, v(t) is decreasing; when a(t) > 0, v(t) is increasing. So, when velocity changes from increasing to decreasing (or decreasing to increasing), this would indicate a(t) = 0 at some t. For example, one instance in this situation is over [0,15]. Over [0,10], it looks like v(t) is increasing (in general; it's completely possible that velocity is oscillating), but when t = 15, v(t) drops down from 9.5 to 7.0 mpm. The increase-decrease indicates at least one instance of zero acceleration. If you take a look at the tick marks on the table, it looks like someone has already pointed out where the sign changes of v(t) occur.
c) Velocity is given by the function \[f(t) = 6+\cos{\frac{t}{10}}+\sin{\frac{7t}{40}}.\] To find the acceleration at t = 23, you must find f '(23), since f '(t) = a(t).
d) If f(t) is a velocity function, then average velocity over [a,b] would be given by \[v_{avg} = \frac{f(b)-f(a)}{b-a}\]

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