Question

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?

Answer 1

avg velocity =30m/s
vf^2=vi^2+2ad
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
vf^2=Vi^2+2ad
0=50^2-2a d or
d=1250/a+1200/a

now on the return trip from the max total distance to the original position
vf^2=vi^2+2ad
vf^2=2(-a)(-1250/a-1200/a(
vf^2=2*2450
vf=sqrt(4900( =70m/s in the opposite direction of the original 10m/s
check my math.