<p>Speed in increasing when acceleration > 0.
Speed is decreasing when acceleration < 0.
Speed is constant when acceleration = 0.</p>
<p>hey, when July comes around, we should all come back to this thread and tell each other what scores we got, and see whose predictions turned out to be optimistic or pessimistic
i’m curious…</p>
<p>bioboy, both acceleration and velocity WERE negative though</p>
<p>well i hope lol</p>
<p>Isn’t what badgirls club saying for the acceleration and velocity problem incorrect? You wouldn’t add the two for distance travelled because it was already at that position. I initially did that and then changed my answer.</p>
<p>1.a is so disappointing. I wish I had read the problem right!</p>
<p>bioboy… Perhaps if you would have spent all year practicing free response (as they are almost identical every year), you would not have had to force yourself to wake up at 5.
And go and google “ask mr calculus ap ab” to go see the answers for yourself, he’s posting them now with 5 as the first one up.</p>
<p>There is more than one answer for most differentiable equations…</p>
<p>For those of you who are unsure about number 1, here it is:</p>
<p>a) v(5.5)=2sin(e^(5.5/4))= -1.453
a(5.5)=1/2<em>e^(5.5/4)</em>cos(e^(5.5)/4)= -1.358
Since both a(5.5) and v(5.5) are less than zero, the speed is increasing.</p>
<p>b) 1/(6-0)*integral(2sin(e^(x/4))+1))= 1.949
c) integral[abs(v(t)] = 12.572
d) v(t) crosses the x-axis at time t=5.1955
x(5.1955)= 2+integral[v(t),x,0,5.1955] = 14.135</p>
<p>Hey, I just posted these on the BC forum but I figured you guys might like to see this too, so here you go.</p>
<p>[2011</a> AP CALCULUS AB and BC Free Response Solutions [DRAFT]](<a href=“http://www.ilearnmath.net/help/index.php?topic=1611.msg5441#msg5441]2011”>http://www.ilearnmath.net/help/index.php?topic=1611.msg5441#msg5441)</p>
<p>Edit: The pdf attachment with the solutions is at the bottom of the post.</p>
<p>I have seen some errors in previous solutions, and want to give you the correct one.</p>
<p>g(x) = 2x+integral [f(t) from 0 to x]</p>
<p>a) g(-3)=2(-3) - integral(f(x),x,0) <== i switched the upper b. with lower one by putting a minus in front of the integral;
g(-3)=-2-9pi/4</p>
<pre><code>g’(x)= 2+f(x)
g’(-3)=2
</code></pre>
<p>b) Absolute max at x=5/2.
c) Point of inflection: x=0
d) Average rate of change: [f(3)-f(-4)] / (3+4) = (-3–1)/7 = -2/7
MVT applies when function is differentiable on the open interval and continuous on the open interval. At x=0, the function was not differentiable. Thus, the MVT did not apply.</p>
<p><a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
<p>come on anyone who did form b?</p>
<p>
</p>
<p>Yeah I did form B, it was a ***** compared to form A. You guys had it so so so easy :|</p>
<p>-Doom</p>
<p>yh, I did form B. Wt questions do u want to look at?</p>
<p>The speed was increasing because the velocity and acceleration had like signs. Just to clarify since a lot of you were saying it was decreasing.</p>
<p>I was hoping to see some familiar answers.
I discussed my answers with peers who took the same exam, but was unable to get closure.</p>
<p>Tomorrow maybe I’ll sit down and work on a solutions guidlines sort of thing for form B, I’ll post it up here if I ever get around to it.</p>
<p>-Doom</p>
<p>I know I got 5C completely right and I might’ve gotten partial for 5B. How many points is this likely to be out of 9 for that particular FRQ?</p>
<ol>
<li>a. integral from 0 to 60 of 2sin(0.03)+1.5 which = 171.81
b. (1/60)* (integral from 0 to 60 of 2sin(0.03)+1.5) = 2.86
c. ((((prime is notated with this apostrophe (‘)))))
(H<em>pie</em>r^2)’ to see the change in volume = pie(2r’h+h’r^2)
plugin values; r’ is zero since radius never changes = pie(100h’)
we need to find H’ so we find S’(60) = 3.45
plug in the values = 345pie milliliters^3/day
d. D(t) must equal zero for them to have the same change in height
M’(t)= (1/400) ( 9t^2 -60t +330)
Plug in the calculator and find where D(t) which equals M’(t)-S’(t)=0
that would be at t= 11.82</li>
</ol>
<ol>
<li>a. No since limit as x approaches 5 of r(t) does not exist. The left hand limit which equals 375 doesnt equal the right hand limit which equals 904.8.
B. (Integral from 0 to 5 of 600t/(t+3) + integral from 5 to 8 of 1000e^(-0.02x))/8 = 258.05 liters per hour.
C. r’(3)= 50 this means there is an increase of 50 in the rate of change of the rate of the draining water.
d. 12000 - integral from 0 to t of r(t)= 9000</li>
</ol>
<p>1a) i also agree that the speed is increasing because velocity and acceleration were negative</p>