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SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.2a) You can see that 0 ≤ t ≤ 40, so dividing this interval into four subintervals isn't too difficult. You get the following four intervals: \[[0,10], [10,20], [20,30], [30,40].\] So, the length of each subinterval is 10. The midpoints of these intervals are 5, 15, 25, and 35, respectively. \[\int_0^{40}v(t)\;dt\approx\frac{1}{10}\left(v(5)+v(15)+v(25)+v(35)\right)\]

Bladerunner1122
 one year ago
Best ResponseYou've already chosen the best response.0How do I do b c d?

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.2b) 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 increasedecrease 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.

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.2c) 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).

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.2d) If f(t) is a velocity function, then average velocity over [a,b] would be given by \[v_{avg} = \frac{f(b)f(a)}{ba}\]
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