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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?
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ASSIGNMENT

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=50a*t or t=50/a to stop.
it now has went a second distance of
vf^2=Vi^2+2ad
0=50^22a 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/a1200/a(
vf^2=2*2450
vf=sqrt(4900( =70m/s in the opposite direction of the original 10m/s
check my math.posted by bobpursley
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