
body dropped from the top of a tower and after 'n' seconds another stone is thrown downwards with velocity ' v '. They can meet after a time of :
{ (gn/2 - v) / (gn - v) }n

bus accelerates at a constant rate 'a' from rest. man standing at a distance of 'd' away from the bus should run at a minimum velocity of v =

(2ad) to get to the bus.

body uniformily accelerares from rest at a rate a1 for time t1 and then at a2 for t2, at a3 for t3 .........
avg. acceleration is : (a1t1 + a2t2 + a3t3 + ....) / (t1 + t2 + t3 +....)
I will be back.....