<p>Ok, this a multiple choice question form 1998; and I can't figure it out for the life in me :(</p>
<p>HELP!!!</p>
<p>A particle moves in the xy-plane with coordinates given by
x = Acos(wt) and y=Asin(wt)</p>
<p>where A = 1.5 meters, and w=2.0 radians per second. What is the magnitude of the particles acceleration? </p>
<p>THE CORRECT ANSWER IS 6.0m/s^2...BUT HOW?</p>
<p>Thanks a lot for any help!!!</p>
<p>disclosure: i only took physics B (4) but i'm in calculus...and i own in that :)</p>
<p>looks like vectors. Find (dx/dt) and (dy/dt). That's your velocity vector. But for speed, find the magnitude of that. </p>
<p>that would be:</p>
<p>Square-root ((dx/dt)^2 + (dy/dt)^2)</p>
<p>.............................................</p>
<p>dx/dt = -3sin(2t)
dy/dt = 3cos(2t)
square-root(9(cos(2t))^2 + 9(sin(2t))^2)</p>
<p>^lol, I can't believe I didn't do that...I kept finding dy/dx, and completely overlooked the "magnitude part"...lol..</p>
<p>by-the-way, you found velocity; not acceleration, but I got the concept</p>
<p>|a| = sqrt((x'')^2 + (y'')^2) = 6</p>
<p>THANKS MUCH!!!</p>
<p>Another (non-calc) way of doing this one: if you recognize that the particle is going in a circle, with</p>
<p>r = A = 1.5 m</p>
<p>and</p>
<p>v = r * w = (1.5 m) * (2 rad/sec) = 3 m/sec</p>
<p>then</p>
<p>a = v^2 / r = (9 / 1.5) m/sec^2 = 6 m/sec^2</p>
<p>a = r*w^2 (derived from calculus). to get r, we have by definition r = sqrt(x^2 + y^2) = sqrt(A^2 * cos^2(wt) + A^2 * sin^2(wt)) = sqrt(A^2(1)) = A.</p>
<p>So a = r*w^2 = (3/2)(4) = 6 m/s^2</p>