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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.

Relative velocity Conceptual questions

A. A gun is mounted on a train roof .Train is traveling with the velocity v in north direction.A car is also moving on a parallel track with train with velocity w in north direction.Two bullet are fired from the muzzle of the gun . Take north as positive and south as negative
1. Bullet one is fired in the north direction with the muzzle velocity u.Find the velocity of the bullet as seen from the observer on the earth
a u+v
b u-v
c u
d v

2. find the velocity of the bullet as seen from the observer on the moving car
a u+v-w
b u-v-w
c u
d v

3.Second Bullet is fired in the south direction with the muzzle velocity u.Find the velocity of the bullet as seen from the observer on the earth
a u+v
b v-u
c u
d v

4.find the velocity of the second bullet as seen from the observer on the moving car
a u+v-w
b v-u-w
c u
d v

Solutions
1) Velocity of bullet w.rt to train=u
Velocity of train = v
Velocity of bullet w.rt to train=Velocity of bullet w.r.t earth -Velocity of train w.r.t earth
So Velocity of bullet w.r.t earth =Velocity of bullet w.rt to train+Velocity of train w.r.t earth
=u+v
2) Velocity of bullet w.r.t earth =u +v
Velocity of Car w.r.t earth =w
Velocity of bullet w.rt to car=Velocity of bullet w.r.t earth -Velocity of car w.r.t earth
=u+v-w
3) Velocity of bullet w.rt to train= -u
Velocity of train = v
Velocity of bullet w.rt to train=Velocity of bullet w.r.t earth -Velocity of train w.r.t earth
So Velocity of bullet w.r.t earth =Velocity of bullet w.rt to train+Velocity of train w.r.t earth
=-u+v= v-v
4) Velocity of bullet w.rt to train=-u
Velocity of train w.r.t earth =v
velocity of car w.r.t earth  =w
Velocity of bullet w.rt to car=Velocity of bullet w.r.t earth -Velocity of car w.r.t earth
= v-u-w

B.A train is moving in the west direction with a velocity 15m/s.A monkey runs on the roof of the train against its motion with a velocity 5m/s with respect to train .Take the motion along west as positive
5.Velocity of train relative to its driver
a. 0
b. 15 m/s
c. -15 m/s
d. 20 m/s

6. What is the velocity of train with respect to monkey
a. 5m/s
b -5 m/s
c. 15 m/s
d -15 m/s

7. find the velocity of monkey with respect to ground
a. 5 m/s
b. -5 m/s
c. 10 m/s
d. -10 m/s

Solutions
5)
Velocity of train w.r.t to driver=Velocity of train w.r.t earth -Velocity of driver w.r.t earth
so =0
6)
Given
velocity of train = 15m/s
velocity of monkey w.r.t train=-5 m/s
Now velocity of monkey w.r.t train=-velocity of train w.r.t monkey
So velocity of train w.r.t monkey = 5 m/s
7)  velocity of train = 15m/s
velocity of monkey w.r.t train=-5 m/s
velocity of monkey w.r.t train = velocity of monkey with respect to ground   - velocity of train
-5 =velocity of monkey with respect to ground   - 15
velocity of monkey with respect to ground   =10 m/s


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