2. SithsAndGiggles

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. $\int_0^{40}v(t)\;dt\approx\frac{1}{10}\left(v(5)+v(15)+v(25)+v(35)\right)$

How do I do b c d?

4. SithsAndGiggles

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.

5. SithsAndGiggles

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).

6. SithsAndGiggles

d) If f(t) is a velocity function, then average velocity over [a,b] would be given by $v_{avg} = \frac{f(b)-f(a)}{b-a}$