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A body covers 200cm in first 2 sec and 220 cm in next 4sec .what is the velocity after 7(th) second.
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 = 220
6u + 18a = 420
u+3a = 70 .................2
from 1& 2
a = -15 & u=115
velocity after 7 sec = V7
V7 = u + at
=115 - 15*7=10cm/sec
Let the initial velocity be u and the accelaration be f.For the motion of 200cm we have 200=u+1/2f(2)^2 and for 220cm we have 220=u+1/2f(6)^2..Solve those two equations and get the values of u and f..Then for distance covered in 7th sec apply the formula s=u+1/2f(2t-1).
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