During an experiment performed in a particle accelerator, a particle A is forced into motion, along a straight path, with constant acceleration of 10 m/s2, until it enters a medium of higher viscosity so that its velocity steadies at 70 m/s.At t=0, another particle B is forced to move in opposite direction along the same path as that of first, at a distance of 8000 m, with a constant speed of 40 m/s. Determine the time and the distance travelled by the two particles before they collide. 1) 74.955 s, sA = 245 m, sB = 280 m. 2) 23.456 s, sA = 146 m, sB = 965 m. 3) 12.743 s, sA = 574 m, sB = 256 m. 4) 54.126 s, sA = 432 m, sB = 168 m. Coordinate System- The coordinate system is taken with positive axes pointing towards right. Time taken to reach 70m/s by the first particle is calculated by- v = u+at 70 = 0+10t t = 7s. DISPLACEMENT- Distance travelled in 7s by particle Ais The displacement of the particle B in the 7 s is s = 7*(-40) = -280 m TIME- Let t be the time after which the two particles collide. The relation between the distance travelled by the two particles is given by- 245+70*(t-7) = 8000-40*t t = 74.955s