posted by ASSIGNMENT
A particle starts from the origin with a velocity of 10 m/s and moves with a constant acceleration till the velocity increases to 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?
avg velocity =30m/s
2500=100+2a d or d=1200/a
So new issue: we apply -a to stop it first.
vf=vi+at=50-a*t or t=50/a to stop.
it now has went a second distance of
0=50^2-2a d or
now on the return trip from the max total distance to the original position
vf=sqrt(4900( =70m/s in the opposite direction of the original 10m/s
check my math.