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let initial velocity = u & accleration is a then
distance travelled in 4 sec = S6
using , s=ut + at2/2
S6 = 4u + a36/2
= 6u+18a
distance covered in 2 sec = S2 = 2u + 4a/2
=2u+2a = 200 (given)
u+a = 100 ........................1
now distance covered in 2 to 6 sec is given 220cm
S6-200 = 250
6u + 18a = 450
u+3a = 75 .................2
from 1& 2
a = -25/2 & u=112.5
velocity after 5 sec = V5
V5 = u + at
=112.5 - (25/2)*5=50cm/sec
Regards
Arun (askIITians forum expert)
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