<p>I had my first AP PHysics quiz today, and I think I didnt the whole thing wrong!! But here are the questions that I remember. How do you do them?</p>
<p>Define: vector a = 2.50 m and vector b = 5.00 m and 30 degrees between them.</p>
<p>(1) Calculate the the magnitude of (vector a) + vector (b).</p>
<p>(2) On a xy plane, identify with magnitude and direction the x and y compoents of vector a and vector b.</p>
<p>(3) Calculate the magnitude and direction of the resulting vector.</p>
<p>wow, me too, except mine was way harder. I'm guessing you're probably in Physics B?</p>
<p>Define: vector a = 2.50 m and vector b = 5.00 m and 30 degrees between them.</p>
<p>(1) Calculate the the magnitude of (vector a) + vector (b).</p>
<p>First of all, you need to be given direction of the vectors to calculate magnitude. Assuming that everything is positive this is how you would calculate magnitude</p>
<p>2.5+5cos30=6.8
5sin30=2.5
6.8squared+2.5squared=46.2
square root of that=7.24 which is magnitude</p>
<p>(2) On a xy plane, identify with magnitude and direction the x and y compoents of vector a and vector b.</p>
<p>Your question is confusing; they can be anywhere on the xy plane unless specified, and direction should be given.</p>
<p>(3) Calculate the magnitude and direction of the resulting vector.</p>
<p>you need to be given the position and direction of A and B vector.</p>
<p>given a=-k^2*x, and at t=0, x=x0, v=0 m/s, find x(t).</p>
<p>Masamune:
Is that a=-(k^2)(x) or a=-k^(2x) ?</p>
<p>Hmmmm....
You need law of cosines:
c^2 = a^2+b^2-2abcosC
c^2 = (2.5)^2 +(5)^2-2(2.5)(5)cos(30))
c = 3.098</p>
<p>The rest is true - you need the directions of the vectors to place the resulting vector.</p>
<p>Is the law of cosines the only way to do it? Once I got home I found out we needed the law of cosines. But my AP Physics teacher never taught us the law of cosines, and my math class never got up to trig. OMG, that is so stupid. 1 question wrong = all wrong.</p>
<p>No - it wasn't the only way - just the most direct. You could have resolved each vector into it's x and y components and then added the x components, added the y and the find the resultant, R, using Pythagorean's Theorem. R = sqrt (x^2 + y^2)</p>
<p>Unfortunately in AP, I doubt they will go over all the math you need to have in place to do the physics. So sorry to be the bearer of bad news. I would suggest getting that program into your calculator before you have the chapter test.</p>
<p>oops lemme clarify:
given a=(-k^2)*x, and at t=0, x=x0, v=0 m/s, find x(t).</p>
<p>look familiar? it should if you've analyzed springs!</p>
<p>
[quote]
Hmmmm....
You need law of cosines:
c^2 = a^2+b^2-2abcosC
c^2 = (2.5)^2 +(5)^2-2(2.5)(5)cos(30))
c = 3.098
[/quote]
</p>
<p>is this an answer to question 1 or to "given a=-k^2*x, and at t=0, x=x0, v=0 m/s, find x(t)."?</p>
<p>thanks</p>
<p>Masamune:
Well, it's harmonic motion, so we can have x = r cos(kt) or x = r sin(kt).
Since v(t)= dx/dt, and v(0)=0, we need</p>
<p>x = r cos(kt)
v = -rk sin(kt)
a = -(k^2) r cos(kt) = -(k^2) x</p>
<p>x(0) = r cos(0)= r = x0; so we can rewrite x(t) = x0 cos(kt)</p>