CBSE Class 11-science Answered
signs for free falling bodies
Asked by yashpinto | 09 May, 2011, 05:24: PM
Expert Answer
When we throw a body upward, the acceleration is against the gravity so it is taken as minus therefore the equation which we know comes out to be
, and when ut is less than , then distance comes out to be negative sign.
For the problem:
Soln: The displacement of the body in first 2 sec =200cm,
Let the initial velocity = u(say) ,
Acceleration = a(say), time =2sec
Substitute the value of in above equation, we get
= 2u +2a ; 2(u+a) =200
Therefore u+a = 100 - - - - - - - - - - (1)
Given that the body travels 220cm in next 4sec.That is from the start it displaces 200+220 = 420 cm in 6sec.
Displacement =420 cm,
time =6 sec,
substitute theses values in the equation ,
6(u+3a) = 420 ; u+3a = 70 - - - - - - - - - - (2)
Solving equations (1) and (2) or Eq (2) Eq(1)
we get 2a= -30 ; a=-15 cm/,
Substitute value of a in Eq(1) we get u-15 = 100,
u = 115 cm/sec.
The velocity at the end of second v=u+a
we get v= 115+(-15)(6) =115-90 =25cm/sec.
Therefore final velocity v=25cm/sec.
Let the initial velocity = u(say) ,
Acceleration = a(say), time =2sec
Substitute the value of in above equation, we get
= 2u +2a ; 2(u+a) =200
Therefore u+a = 100 - - - - - - - - - - (1)
Given that the body travels 220cm in next 4sec.That is from the start it displaces 200+220 = 420 cm in 6sec.
Displacement =420 cm,
time =6 sec,
substitute theses values in the equation ,
6(u+3a) = 420 ; u+3a = 70 - - - - - - - - - - (2)
Solving equations (1) and (2) or Eq (2) Eq(1)
we get 2a= -30 ; a=-15 cm/,
Substitute value of a in Eq(1) we get u-15 = 100,
u = 115 cm/sec.
The velocity at the end of second v=u+a
we get v= 115+(-15)(6) =115-90 =25cm/sec.
Therefore final velocity v=25cm/sec.
Answered by | 10 May, 2011, 07:24: PM
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