Question

a particle starts from the origin with velocity 10 m/s and moves with a constant acceleration till the velocity increases 50 m/s at that instant ,the acceleration is suddenly reversed ... what will be the velocity of the particle when it returns to the starting point

Answer 1

position when it hits 50m/s: 

vf^2=vi^2+2ad 
50^2=10^2+2*a*d solve for a*d. 

Now, the reverse trip: 
vf^2=50^2 -2ad put the above in, and viola, 

vf^2=50^2-50^2+10^2=10^2 
vf=10