Relative velocity and free fall acceleration

7. Free fall acceleration

  • Freely falling motion of any body under the effect of gravity is an example of uniformly accelerated motion.
  • Kinematic equation of motion under gravity can be obtained by replacing acceleration 'a' in equations of motion by acceleration due to gravity 'g'.
  • Value of g is equal to 9.8 m.s-2.
  • Thus kinematic equations of motion under gravity are
    v = v0 + gt                          (16a)
    x = v0t + ½ ( gt2 )          (16b)
    v2 = (v0)2 + 2gx           (16c)
  • The value of g is taken positive when the body falls vertically downwards and negative when the body is projected up against gravity.

8. Relative velocity

  • Consider two objects A and B moving with uniform velocities vAand vB along two straight and parallel tracks.
  • Let xOAand xOB be their distances from origin at time t=0 and xAand xB be their distances from origin at time t.
  • For object A
    xA = xOA + vAt               (18)
    and for object B
    xB = xOB + vBt                (19)
    subtracting equation 18 from 19
    xB - xA = ( xOB - xOA ) + ( vB - vA) t                          (20)
  • Above equation 20 tells that as seen from object A , object B seems to have velocity ( vB - vA) .
  • Thus ( vB - vA ) is the velocity of object B relative to object A. Thus,
    vBA = ( vB - vA )                (21)
  • Similarly velocity of object A relative to object B is
    vAB = ( vA - vB )                (22)
  • If vB = vA then from equation 20
    xB - xA = ( xOB - xOA )
    i.e., two objects A and B stays apart at constant distance.
  • vA > vB then (vB - vA ) would be negative and the distance between two objects will go on decreasing by an amount ( vA - vB ) after each unit of time. After some time they will meet and then object A will overtake object B.
  • If vA and vB have opposite signs then magnitude of vBA or vAB would be greater then the magnitude of velocity of A or that of B and objects seems to move very fast.

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