- Introduction
- |
- Position and Displacement
- |
- Average velocity and speed
- |
- Instantaneous velocity and speed
- |
- Acceleration
- |
- Motion with constant acceleration
- |
- 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.

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

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

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

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

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

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

Velocity of train w.r.t to driver=Velocity of train w.r.t earth -Velocity of driver w.r.t earth

so =0

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

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

Class 11 Maths page Class 11 Physics page

- 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)

- ncert solutions for class 6 Science
- ncert solutions for class 6 Maths
- ncert solutions for class 7 Science
- ncert solutions for class 7 Maths
- ncert solutions for class 8 Science
- ncert solutions for class 8 Maths
- ncert solutions for class 9 Science
- ncert solutions for class 9 Maths
- ncert solutions for class 10 Science
- ncert solutions for class 10 Maths