2 particles one with constant velocity of 50m/s and the other with uniform acceleration of 10m/s2 start moving simultaneously from the same place in the same direction.They will be at a distance of 125 m from each other after?ans.5(1+root 2)s

Dear Umme,

We can do it by relative motion concepts. From frame of 1st particle, u21 = -50 and a21 = 10.
s21 = 125 = u21 t + 1/2 a21 t2 = -50t + 1/2 10 t2
=> 25 = -10t + t2
Solving: t = 5 + 5 sqrt(2) = 5 (1 + sqrt(2)) seconds

Please feel free to post as many doubts on our discussion forum as you can. If you find any question difficult to understand - post it here and we will get you the answer and detailed  solution very  quickly.

We are all IITians and here to help you in your IIT JEE preparation.All the best.

Regards,
Askiitians Experts,
Gokul  Joshi