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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