- Introduction
- |
- Position and Displacement
- |
- Average velocity and speed
- |
- Instantaneous velocity and speed
- |
- Acceleration
- |
- Kinematic equations for uniformly accelerated Motion
- |
- Free fall acceleration
- |
- Relative velocity
- |
- Solved Examples Part 1
- |
- Solved Examples Part 2
- |
- Solved Examples Part 3
- |
- Solved Examples Part 4
- |
- Solved Examples Part 5

- Position Distance and Displacement
- |
- Average velocity and speed
- |
- Velocity and acceleration
- |
- Uniformly accelerated motion
- |
- Relative Velocity
- |
- Kinematics Question 1
- |
- Kinematics Question 2

- 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 = v_{0}+ gt (16a)

x = v_{0}t + ½ ( gt^{2}) (16b)

v^{2}= (v_{0})^{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.

- Consider two objects A and B moving with uniform velocities v
_{A}and v_{B}along two straight and parallel tracks.

- Let x
_{OA}and x_{OB}be their distances from origin at time t=0 and x_{A}and x_{B}be their distances from origin at time t.

- For object A

x_{A}= x_{OA}+ v_{A}t (18)

and for object B

x_{B}= x_{OB}+ v_{B}t (19)

subtracting equation 18 from 19

x_{B}- x_{A}= ( x_{OB}- x_{OA}) + ( v_{B}- v_{A}) t (20)

- Above equation 20 tells that as seen from object A , object B seems to have velocity ( v
_{B}- v_{A}) .

- Thus ( v
_{B}- v_{A}) is the velocity of object B relative to object A. Thus,

v_{BA}= ( v_{B}- v_{A}) (21)

- Similarly velocity of object A relative to object B is

v_{AB}= ( v_{A}- v_{B}) (22)

- If v
_{B}= v_{A}then from equation 20

x_{B}- x_{A}= ( x_{OB}- x_{OA})

i.e., two objects A and B stays apart at constant distance.

- v
_{A}> v_{B}then (v_{B}- v_{A}) would be negative and the distance between two objects will go on decreasing by an amount ( v_{A}- v_{B}) after each unit of time. After some time they will meet and then object A will overtake object B.

- If v
_{A}and v_{B}have opposite signs then magnitude of v_{BA}or v_{AB}would be greater then the magnitude of velocity of A or that of B and objects seems to move very fast.

- In short

a u+v

b u-v

c u

d v

a u+v-w

b u-v-w

c u

d v

a u+v

b v-u

c u

d v

a u+v-w

b v-u-w

c u

d v

Check your Answers

a. 0

b. 15 m/s

c. -15 m/s

d. 20 m/s

a. 5m/s

b -5 m/s

c. 15 m/s

d -15 m/s

a. 5 m/s

b. -5 m/s

c. 10 m/s

d. -10 m/s

Check your Answers

Class 11 Mathse Class 11 Physics

- Principles of Physics Extended (International Student Version) (WSE)
- university physics with modern physics by Hugh D. Young 13th edition
- NCERT Exemplar Problems: Solutions Physics Class 11
- H.C. Verma Concepts of Physics - Vol. 1
- CBSE All in One Physics Class 11 by Arihant
- NCERT Solutions: Physics Class 11th
- New Simplified Physics: A Reference Book for Class 11 (Set of 2 Parts)
- Pradeep's A Text Book of Physics with Value Based Questions - Class XI (Set of 2 Volumes)
- Oswaal CBSE Sample Question Papers For Class 11 Physics (For 2016 Exams)

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