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.

Distance travelled in 7s by particle Ais

The displacement of the particle B in the 7 s is
s = 7*(-40)
= -280 m

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 =