May 2011 - Career Point

For (x – a) 2 + (y – b) 2 = r 2 , the motion is in a circularpath with centre at (a, b) and radius r22x y2 +2 = 1 is equation of an ellipsea bx × y = constant give a rectangular hyperbola.Note : To decide the path of motion of a body, arelationship between x and y is required.Area under-t graph represents change in velocity.Calculus method is used for all types of motion (a = 0or a = constt or a = variable)s = f(t)Differentiate w.r.ttimev = f(t)integrate w.r.t.timeDifferentiate w.r.ttimea = f(t)integrate w.r.t.timeS stand for displacement2dv dv d sa = v = = ds dt 2dtdx dyAlso v x = ⇒ vy = dt dt2dva x = x d x dv 2 y d y= and ay = =dt 2dtdt 2dtThe same concept can be applied for z-co-ordintae.Projectile motion :Pucosθu ucosθusinθ ucosθθucosθucosθQ θuusinθgg FFgFgFProjectile motion is a uniformly accelerated motion.For a projectile motion, the horizontal component ofvelocity does not change during the path becausethere is no force in the horizontal direction. Thevertical component of velocity goes on decreasingwith time from O to P. At he highest point it becomeszero. From P to Q again. the vertical component ofvelocity increases but in downwards direction.Therefore the minimum velocity is at the topmostpoint and it is u cos θ directed in the horizontaldirection.The mechanical energy of a projectile remainconstant throughout the path.the following approach should be adopted for solvingproblems in two-dimensional motion :Resolve the 2-D motion in two 1-D motions in twomutually perpendicular directions (x and y direction)Resolve the vector quantitative along thesedirections. Now use equations of motion separatelyfor x-direction and y-directions.If you do not resolve a 2-D motions in two 1-Dmotions in two 1-D motion then use equations ofmotion in vector formv r = u r + at ; s r = ut + 21 ar t2; v r . v r – u r . u = 2 a r s rs = 21 ( ur + vr )tWhen y = f(x) and we are interested to find(a) The values of x for which y is maximum forminimum(b) The maximum/minimum values of y then we mayuse the concept of maxima and minima.Problem solving strategy :Motion with constant Acceleration :Step 1: Identify the relevant concepts : In moststraight-line motion problems, you can use theconstant-acceleration equations. Occasionally,however, you will encounter a situation in which theacceleration isn't constant. In such a case, you'll needa different approacha x =dυxdtd = dt2⎛ dx ⎞ d x⎜ ⎟ =2⎝ dt ⎠ dtStep 2: Set up the problem using the following steps:You must decide at the beginning of a problemwhere the origin of coordinates are usually amatter of convenience. If is often easiest to placethe particle at the origin at time t = 0; then x 0 = 0.It is always helpful to make a motion diagramshowing these choices and some later positions ofthe particle.Remember that your choice of the positive axisdirection automatically determines the positivedirections for velocity and acceleration. If x ispositive to the right of the origin, the v x and a x arealso positive toward the right.Restate the problem in words first, and thentranslate this description into symbols andequations. When does the particle arrive at acertain point (that is, what is the value of t)?where is the particle when its velocity has aspecified value (that is, what is the value of xwhen v x has the specified value)? "where is themotorcyclist when his velocity is 25m/s?"XtraEdge for IIT-JEE 26 MAY 2011