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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 16th May 2021, 1:45 PM
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 Expert 16th May 2021, 2:55 PM
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Tags: velocity