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a body covers 200cm in first 2sec and 220 cm in next 4sec with deceleration velocity of the body at the end of seventh second is  
Asked by chinnarisharanya1979 | 16 May, 2021, 13:45: PM
answered-by-expert Expert Answer
From the given information , we consider that the object starts from rest and
accelerated for 2 seconds and has travelled 200 cm in accelerated motion
 
Acceleration a is detrmined from the equation , S = (1/2) a t2 ,
 
where S is the distance covered in t seconds
 
200 = (1/2) a × 2 × 2   , hence we get a = 100 cm/s2 .
 
velocity after 2 seconds = a × t = 100 × 2 = 200 cm/s
 
It is given that for next 4 seconds the object has travelled 220 cm  with deceleration ar .
 
We get deceleration ar from the relation , S = ( u t  ) - [ (1/2) ar t2 ]
 
220 = ( 200 × 4 ) - [ (1/2) × ar × 4 × 4 ] 
 
8 ar = ( 220 - 800 ) = 580   or   ar = 72.5 cm/s2
 
After 6 seconds if deceleration continues , then velocity v  after 7 seconds is determined as
 
( deceleration has started after 2 seconds from starting time and after initial velocity 200 cm/s )
 
v  =  u - ( ar t )  = 200 - ( 72.5 × 5  ) = -162.5 cm/s
Answered by Thiyagarajan K | 16 May, 2021, 14:55: PM
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