Physics Help Forum Quick Relative motion with Vectors Question!
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 Sep 21st 2009, 03:39 PM #1 Junior Member   Join Date: Aug 2009 Posts: 18 Quick Relative motion with Vectors Question! Three forces acting on an object are given by F1 = ( - 2.05 i - 2.40 j ) N, F2 = ( 5.50 i ) N, and F3 = ( - 46.0 i + 46.0 j ) N. The object experiences an acceleration of magnitude 3.75 m/s2.(a) What is the direction of the acceleration? 1-45.7 (from the positive x axis) I got this answer but is wrong? can someone tell me why? (b) If the object is initially at rest, what is its speed after 12.0 s? 3 ??m/s (c) What are the velocity components of the object after 12.0 s? ( ?? i + ?? j ) m/s Someone please help me with b and c. thanks
 Sep 21st 2009, 11:15 PM #2 Physics Team   Join Date: Feb 2009 Posts: 1,425 Add the components to get the resultant R = -42.55 i + 43 j. Note F2 = 5.5 i + 0 j If you draw this you will find it lies in the second quadrant as the x coord is - ve and y +ve . Note that 43 / 42.55 gives tan of the angle between R and the -ve y axis. Thus tan inv (43 / 42.55) will give you this angle in radians. (Pi -this angle ) will give you the angle (+ve) with respect to the +ve X axis say theta. The net force on the object, R will be given by $\displaystyle R\ =\sqrt {(-42.55)^2\ +\ (43)^2}$ . (b) You have the accln a (3.75) with initial vel u = 0 as it starts from rest and the time t Use $\displaystyle v\ =\ u\ +\ a\ t$ to get the final speed. (c) The x component will be v cos theta and the y component will be v sin theta. Last edited by physicsquest; Sep 21st 2009 at 11:18 PM.
 Sep 22nd 2009, 09:31 AM #3 Junior Member   Join Date: Aug 2009 Posts: 18 Thanks for helping me and explaining the problem.

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