Question

Question 2 2 Question 2 LO1, PO1 . A position of a particle from an origin is given by dy --3y=1 dt Where v and 1 are in meter and second respectively. Solve the position of the particle using initial conditions 1 - 1.1 - 1. [12 marks] 11) Calculate total distance travelled from 1 = 0 to 1 = 1. [8 marks] Ene 99+

Answer 1

Given, tay - 3y = +² here, t second y displacemed 2 so, dy rate of Change of distance = velocity det assume de av t v 34=+2 2 3y = vt t 2 Or, * = (y) fue 2 then, a - 3 If we compare it with s=ut that U = Me, a - 2 Us velocity (initial) tal m above equations So Put y=1 m/82 an acceleration and L = 18 x 1 - 12 x (132 > اس 4 3 1 - 4 mis

v2 =u²+ 2a8 here, sy displacement &5=4 = 1 n (given) = 4 m/s final velocity, V- Womis acceleration a x² = (2-12 * (2 x 4 Initial velocity. V it 2 x s = ut t 1 at? Using motion formular - 2/3 m / 82 • So final velocity will v' = 2/3 m 18 which is of the initial velocity of pelticle. ii) Distance fleavelled from t-o to t=1 again, Using motion formula x² = 16 9 x 1 u = 9 m/s V* =* = m/s at t= 1 sec, a = alo U = m/s 4x1 + x s - 1 m