Question

For timet > 0, the acceleration of a particle moving along the x-axis is given by a (t) = cost+t. At time t = 0, the velocity of the particle is 8 and the position of the particle is 0. What is the position of the particle at time t = ?

Answer 1

is a (t) = costat 0 acceleration along a axis = da = costat. - (1) The auxiliary ear of tha's differential ear is the ul 20 is mizo a) m = 0,0 ...C.f = (a+te2 Jet where a and Co are] 1 = (1+t(2) I arbitory constant I How for particular integral (P.2), a koBl+t) where Da = bn cosd + bet - 6054 + 5 (5") (operating on)

W WW. - cost at e (again operating 8') Therefore general solution of (1) is X(t) = (of tp.? ano => x64) = (4+4(2) + 43 - copt — (2) velocity doilet) a Gt th t sint Güven at tao , diye de 78 1. =) 8=6toto 2001 (2=8 at tso act) = 0 =) 0 = a +0 -1 3 = 1 now putting the value of a and ca We get 20+) = (1++) + 42 - Copt at taula X(014) = (1+8)+6)xt-Cop () = (1428) te which is the positions of particle at taula.