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).
						

							
A stone is dropped from a height of 19.6 m what is the time taken by the stone to travel the metre of the path
 
						

							Since the stone is dropped from a height of 19.6 mIt`s initial velocity(u)=0m/s ,. S=19.6mAnd acceleration(a)=9.8m/s^2According to 2nd equation of motionS=ut +1/2 a t^219.6= 0×t + 1/2 × 9.8 × tOr, 19.6= 4.9tTherefore t = 19.6/4.9 = 4Hence the time taken by stone =4 seconds
						

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