<p>Like what? I think the hard part is just figuring out what to use, what equals what, etc., but after that it’s not so bad.</p>
<p>Need some help on this problem:</p>
<p>**Two people are in a boat that is capable of a maximum speed of 5 KM per hour in still water, and wish to cross a river 1 KM wide to a point directly across from their starting point. If the speed of the water in the river is 5 KM/h , how mcuh time is required for crossing? **</p>
<p>The answer is 10 hours…can someplease please explain how?</p>
<p>who else is ready to fail tomorrow?</p>
<p>^
Why? What to you is hard about C Mech? It seems so constrained to such few topics that it is easy.</p>
<p>^^Me!</p>
<p>E&M at least. Determined to pass mechanics.</p>
<p>At t=0, person A explains to person B that they are better off just ditching the boat and getting across the old fashioned way - not swimming, but walking. Average speed of a walking person is about 5 km/h I think, so since we know that the bridge is 22 km downstream, and 1 km across, and to get back up to the point where they started is another 22 km, they have walked a total of 45 km. 45 km / 5 km/h = 9 h. Of course, we know that person A and person B are not physically fit to do this all in one go, so we conclude that they take a one hour break after they cross the bridge. Therefore, it takes them a total of 10 hours to get to the other side.</p>
<p>@ ChemE - I actually thought you had the answer, but your just trying to be a smartass. haha almost got me. Anyone have the real explanation?</p>
<p>Actually, I just don’t see how it’s possible for them to get across in a completely straight line. For them to have a positive x-velocity, they will also have a negative y-velocity, causing them to go down the river and missing their target. So, instead of posting that, I made up “what really happened.”</p>
<p>I remember that problem from one of our practice tests. It’s impossible. To cross a river you need some competent of the velocity directed perpendicular to the current, but you also need part of the velocity directed against the current to make sure you don’t go down stream. Since the entire velocity vector must be directed up stream, you cannot cross directly to the other side.</p>
<p>@ medicaldoc23 </p>
<p>check this document:
[AP</a> Physics Vectors and Two-Dimensional Motion](<a href=“http://www.docstoc.com/docs/20817309/AP-Physics-Vectors-and-Two-Dimensional-Motion/]AP”>http://www.docstoc.com/docs/20817309/AP-Physics-Vectors-and-Two-Dimensional-Motion/)</p>
<p>I agree. So where did the 10 hours come from? See above :)</p>
<p>anyone have the released practice exam? I don’t understand number 16 
I keep on getting answer choice D instead of E</p>
<p>@ mulberrypie</p>
<p>F = ma</p>
<p>so a= F/m= -1/2</p>
<p>integrate the acceleration from t=0 to t=t to get v(t)- v(0) = (-t^2/4)</p>
<p>which means that v(t) = v(0) + (-t^2/4)= 4-(t^2/4)
solve for v(t)=0 to get 4</p>
<p>34 (1998). An object is released from rest at time t = 0 and falls through the air, which exerts a resistive force such that the acceleration a of the object is given by a = g - bv, where v is the object’s speed and b is a constant. If limiting cases for large and small values of t are considered, which of the following is a possible expression for the speed of the object as an explicit function of time?
(A) v = g(1 e-bt)/b<br>
(B) V = (ge^(ht))/b<br>
(C) v = gt - bt2<br>
(D) v = (g + a)t/b<br>
(E) v = v0+ gt, v0 != O</p>
<p>I really don’t know how to approach this one :x… help please? 
thanks in advance!</p>
<p>An0maly, thanks for the help!!</p>
<p>What is the curve for a 5 on the mechanics and the electromagetism tests?</p>
<p>@ agpyn3w</p>
<p>dv/dt = g-bv</p>
<p>dv/(g-bv) = dt</p>
<p>integrate the left side from 0 to v and the right from 0 to t to get</p>
<p>1/-b * ln((g-bv)/g) = t</p>
<p>ln((g-bv)/g) = -bt take the e for both</p>
<p>(g-bv)/g = e^(-bt)</p>
<p>g-bv = g*e^(-bt)</p>
<p>g(1-e^(-bt))= bv</p>
<p>(g/b)* (1-e^(-bt))=v</p>
<p>which is choice a</p>
<p>^You can take a risky but educated guess at A because the behavior of things falling with air resistance, charging a capacitor, the rise of induced current all follow the format: max (1-e^-t/something)</p>
<p>Can anyone help me on the **1988 MC Exam: Problems 8,9,27,29,30,35 **</p>
<p>d’oh, thanks TariqM</p>
<p>can’t believe it has only been 4 days and I’ve forgotten my calculus lol</p>