Show that the radial acceleration is zero when θ = 1/sqrt(2) radians A particle moves outward along a spiral. Its trajectory is given by r = A0, where A is a constant.A = (1/n) m/rad. 0 increases in time according to 0 = ať²/2, where a is a constant.(a) Sketch the motion, and indicate the approximate velocity and acceleration at three different points. Hints: a) The trajectory is given, so to sketch velocity and acceleration vectors at three different pointsfind the velocity and acceleration (for acceleration a see hints given in part b).v = ř f + rê ð and a = (* – rở ) î + (rë + 2řė) ôðThe components of velocity and acceleration in polar coordinates should give a feeling about magnitude and direction of vectorsv and a as function of position. You do not need to be perfect in the sketch , just give e feeling (i.e., do not calculate) how themagnitude and direction of these vectors change as function of position . Remember velocity is always in the tangential directionwith the trajectory, which in our case is a spiral one and not a circular onefrom the formula of the acceleration in polar coordinates prove that5a²f²b)a = (* – ri) î + (rë + 2řė) ôð = (¤2л2л5a?f) ê +2л() (1 – 20) F + 50 ê]a=(2л

Given, trajectory is r=Aθ where, A=1πm/rad and θ=αt22 . Here α is constant. So r=1π×αt22=αt22π Then…

Answered: Show that the radial acceleration is…

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