Kinematics1

definition

Motion in One Dimension

If a particle is moving in a single direction throughout its journey then it is said to be moving in one dimension. For example, an ant moving along X-axis. Motion is described in terms of displacement (x), time (t), velocity (v), and acceleration (a). Velocity is the rate of change of displacement and the acceleration is the rate of change of velocity.

law

Motion in two and three dimension

Motion in two dimension: Motion in a plane is described as two dimensional motion.
Example: An ant moving on the top surface of a desk is example of two dimensional motion. Projectile and circular motion are examples of two dimensional motion.

Motion in three dimension: Motion in space which incorporates all the X, Y and Z axis is called three dimensional motion.
Example: Movement of gyroscope is an example of three dimensional motion.

definition

Point Objects

Point object is an expression used in kinematics, it is an object whose dimensions are ignored or neglected while considering its motion. A point object refers to a tiny object which is calculated or counted as dot object to simplify the calculations. A real object can rotate as it moves.

result

Locating Position Vector of a point Object

If we join the origin to the position of the particle by a straight line and put an arrow towards the position of the particle, we get the position vector of the particle.
For example, the points A, B and C are the vertices of a triangle, with position vectors a, b and c respectively.

definition

Polar Coordinate System

The polar coordinate system is a 2D coordinate system in which each point on a plane is determined by a distance from a reference point (origin) and an angle (

θ

) from a reference direction. The polar coordinates

r

(the radial coordinate) and

θ

(the angular coordinate, often called the polar angle) are defined in terms of Cartesian coordinates by

x = r \cos θ

y = r \sin θ

where

r

is the radial distance from the origin, and is the counterclockwise angle from the x-axis. In terms of

x

and

y

r = \sqrt{x^{2} + y^{2}}

θ = \arctan \frac{y}{x}

definition

Displacement

Displacement of object is equal to the length of the shortest path between the final and the initial points. Its direction is from the initial point to the final point. It is a vector quantity.
For example, if a body moves along a circle of radius

r

and covers half the circumference, then displacement is given by

s = 2 r

definition

Distance v/s displacement

S.No	Distance	Displacement
1.	It is the length of actual path traveled	It is the length of shortest distance between final and initial points
2.	It is a scalar	It is a vector

Distance is more than or equal to the magnitude of displacement.

definition

Calculating distance in one-dimensional motion

Total distance traveled in one-dimension can be found by adding (or integrating) the path lengths for all parts of motion. Note that every path length is greater than 0.
Athletes race in a straight track of length

200

m and return back.The total distance traveled by each athlete is

200 \times 2

400 m

definition

Displacement

In one dimensional motion displacement of the of the object will be shortest distance between final and initial point.
Example:
Displacement of a particle in a circular motion would be zero when it reaches to the starting point.

definition

Distance in two-dimensions

Total distance traveled in two-dimensions can be found by adding (or integrating) the path lengths for all parts of motion. Note that every path length is greater than zero.
A man walk

4

km north, then

3

km east and then

2

km south. Then the total distance covered by him is given by

4 + 3 + 2 = 9

definition

Displacement of an Object Moving in two dimensions

Displacement of the object can be calculated in a two dimensional motion as shown in the figure.

example

Displacement of an Object Moving in two dimensions where direction changes after finite distance

definition

Calculating average speed

Average speed is found by first finding the total distance covered by the object and dividing it by the total time taken in travelling the distance.
Example:
A boy covers a circle of radius

100

m in

100

s.
The total distance covered by him will be

d = 2 π r = 200 π m

Average speed will be

v = \frac{d}{t} = \frac{200 π}{100} = 2 π m / s

example

Calculating Average Velocity

Velocity of a particle is given by:

v = \frac{d x}{d t}

average velocity

= \frac{t o t a l d i s p l a c e m e n t}{t o t a l t i m e t a k e n}

Example: A body covers first half of distance with a speed

v_{1}

and second half of distance with a speed

v_{2}

, to calculate its average velocity
Let total distance be

s

.So, body covers

\frac{s}{2}

each with velocities

v_{1}

and

v_{2}

.
Now, average speed

= \frac{t o t a l d i s t a n c e}{t o t a l t i m e t a k e n}

\Rightarrow v_{a v g} = \frac{s}{\frac{\frac{s}{2}}{v_{1}} + \frac{\frac{s}{2}}{v_{2}}} = \frac{2 v_{1} v_{2}}{v_{1} + v_{2}}

definition

Instantaneous speed

Instantaneous speed can be defined as the rate of change of distance with respect to time.

v = \frac{d s}{d t}

Note:
1. Instantaneous speed is always greater than or equal to zero and is a scalar quantity.
2. For uniform motion, instantaneous speed is constant.

definition

Instantaneous velocity

Instantaneous velocity is defined as the rate of change of displacement with time,where the period of time is narrowed such that it reaches zero.

\vec{v} = \frac{d \vec{s}}{d t}

Note:
1. Direction of instantaneous velocity at any time gives the direction of motion of particle at that point of time.
2. Magnitude of instantaneous velocity equals the instantaneous speed. This happens because for an infinitesimally small time interval, motion of a particle can be approximated to be uniform.

example

Average Velocity

For a body moving with uniform acceleration

a

, initial and final velocities in a time interval

t

are

u

and

v

respectively.

Then, its average velocity in the time interval

t

will be given by:

S = u t + \frac{1}{2} a t^{2}

v = u + a t

Average velocity

= \frac{T o t a l d i s p l a c e m e n t (s)}{T o t a l t i m e (t)}

= \frac{u t + \frac{1}{2} a t^{2}}{t} = u + \frac{1}{2} a t

Substitute,

u = v - a t

in the above expression,
We get average velocity

= v - a t + \frac{1}{2} a t = v - \frac{1}{2} a t

definition

Acceleration

Acceleration is defined as the rate of change of velocity with time. It is a vector quantity. Its unit is

m / s^{2}

.
Constant speed does not guarantee that acceleration is zero. For example a body moving with constant speed in a circle changes its velocity every instant and hence its acceleration is not equal to zero.

definition

Average Acceleration

Average acceleration is the change in velocity divided by an elapsed time. For instance, if the velocity of a marble increases from 0 to 60 cm/s in 3 seconds, its average acceleration would be 20 cm/s

^{2}

. This means that the marble's velocity will increase by 20 cm/s every second.

definition

Acceleration as a rate of change of velocity with respect to displacement

Acceleration is defined as

a = \frac{d v}{d t}

∴ a = \frac{d x d v}{d x d t}

a = \frac{v d v}{d x}

Example:
If the velocity of a particle is given by

v = (180 - 16 x)^{1 / 2} m / s

, then find its acceleration.
Solution:

a = \frac{d v}{d t} = \frac{d x}{d t} \frac{d v}{d x} = v \frac{d v}{d x}

v = {(180 - 16 x)}^{\frac{1}{2}} v^{2} = 180 - 16 x

Differentiate both sides w.r.t

x

, we get:

2 v \frac{d v}{d x} = - 16 v \frac{d v}{d x} = a = - 8 m / s^{2}

formula

Understand Acceleration as Second Derivative of Displacement

\vec{a} = \frac{d^{2} \vec{x}}{d t^{2}}

Where

x

is a displacement vector.

formula

Definition of Jerk

Rate of change of acceleration is defined as jerk.
jerk =

\frac{d \vec{a}}{d t}

example

Problems Involving First Equation of Motion

A body is moving with an initial velocity of

2 \hat{i} m / s

and acceleration of

- 0.5 \hat{i} m / s^{2}

. Find the time when the velocity of the particle becomes zero.
Final velocity,

v = 0

v = u + a t

0 = 2 \hat{i} - 0.5 \hat{i} t

t = 4 s

definition

Problems on second equation of motion

If initial velocity of a particle is

2 m / s

and it travels

10 m

1 s

then what will be its acceleration?
Since

s = u t + 1 / 2 a t^{2}

10 = 2 \times 1 + 1 / 2 a \times 1^{2}

a = 16 m / s^{2}

example

Problems on Third Equation of Motion

Let the initial velocity of the particle is

5 m / s

. It is accelerating at an acceleration of

1 m / s^{2}

. Its final velocity is

10 m / s

. Then the distance traveled by this particle will be given by:

v^{2} = u^{2} + 2 a s

10^{2} = 5^{2} + 2 \times 1 \times s

s = 37.5 m

example

Solving problems involving more than one equation of motion

Let the initial velocity of a particle is

5 m / s

and it is accelerated for

2 s

with an acceleration of

1 m / s^{2}

. The distance covered by the particle during this motion can be estimated by:

v = u + a t

v = 5 + 1 \times 2^{2}

v = 9 m / s

Now

9^{2} = 5^{2} + 2 \times 1 \times s

s = 28 m

example

Finding Stopping Distance

Let the initial velocity of the particle is

10 m / s

. It is retarding at a deceleration of

1 m / s^{2}

. Its final velocity is

0 m / s

. Then the stopping distance by this particle will be given by:

v^{2} = u^{2} + 2 a s

0^{2} = 10^{2} - 2 \times 1 \times s

s = 50 m

Stopping time will be:
from

v = u + a t

0 = 10 - 1 \times t

t = 10 s

example

Equation of trajectory

Equation of trajectory can be found for a body whose displacement components are given as a function of time by eliminating the variable of time.
Example:
A radius vector of a point

A

relative to the origin varies with time

t

\vec{r} = a t \hat{i} - b t^{2} \hat{j}

, where

a

and

b

are positive constants and

\hat{i}

and

\hat{j}

are the unit vectors of the

x

and

y

axes.Solution:
Given,

y = - b t^{2}

and

x = a t

Now,

y = - b t^{2}

Multiply and divide RHS by

a^{2} t^{2}

i.e.

x^{2}

which gives us

y = \frac{- b x^{2}}{a^{2}}

example

Total distance traveled when velocity and accelerations are in opposite directions

If a particle has initial velocity as 10m/s and it's retarding at a rate of

- 1 m / s^{2}

then distance travel by this in one second is:

s = u t + 1 / 2 a t^{2}

s = 10 \times 1 - 1 / 2 \times 1 \times 1^{2}

s = 9.5 m

example

Motion in different acceleration for different time intervals

A particle started its motion from rest with an acceleration of

1 m / s^{2}

for 2s and then continued it for next 1s with an acceleration of

2 m / s^{2}

. The distance traveled during this will be:
After 2s final velocity is:

v = u + a t

v = 0 + 1 \times 2

v = 2 m / s

Now this is the initial velocity for the second half of the motion.

s = u t + 1 / 2 a t^{2}

s = 2 \times 1 + 1 / 2 \times 2 \times 1^{2}

s = 3 m

Distance traveled in first half is:

s = 1 / 2 a t^{2}

s = 1 / 2 \times 1 \times 2^{2}

s = 2 m

Hence total distance traveled

= 3 + 2 = 5 m

example

Motion involving more than one uniform acceleration

Example: A particle is moving along x-axis with zero initial velocity for

2 s

with an acceleration of

1 m / s^{2}

. After this it takes a

90

degree turn and starts moving along Y-axis with same acceleration. What will be distance covered by this particle along Y-direction after

4 s

of motion.

Solution:
Final Velocity along x-axis is:
from

v = u + a t

v = 2 \times 1 = 2 m / s

s = u t + 1 / 2 a t^{2}

s = 2 \times 2 + 1 / 2 \times 1 \times 2^{2}

s = 6 m

formula

Distance Traveled in nth second

Distance traveled in

n^{t h}

second is given by:

S_{n} = u + 1 / 2 a (2 n - 1)

example

Equations of Motion under free fall

In free fall initial velocity of the particle is zero.
Therefore, equations of motion are:

v = g t .

h = \frac{g t^{2}}{2}

and

v^{2} = 2 g s

formula

Maximum Height of a Projectile

In vertically upward projection case

θ = 90^{0}

Therefore maximum height for an object projected vertically upwards is

\frac{v^{2}}{2 g}

Example: Find the maximum height reached and the time taken in reaching there by a particle thrown at a speed of

10 m / s

?

Solution:

H = \frac{v^{2}}{2 g}

H = 10^{2} / 20 = 5 m

Time taken in reaching height:

t^{2} = 2 g / H

t^{2} = 2 \times 10 / 5

t = 2 m / s

example

Problems on freely falling objects

Example: A stone projected vertically up from the top of a cliff reaches the foot of the cliff in

8 s

. If it is projected vertically downwards with the same speed, it reaches the foot of the cliff in

2 s

. Then what is its time of free fall from the cliff?

Solution:A to B

\overset{V I α C}{\to}

(with

u

)

\Rightarrow H = u (

) - \frac{1}{2} g (8)^{2} = 8 u - 32 g - (1)

A to B (with

u

downwards)

\Rightarrow H = u (2) - \frac{1}{2} g (2)^{2} = 2 u - 2 g - (2)

free fall,

e q^{n} (1) - e q^{n} (2)

t = \sqrt{\frac{2 H}{g}} \Rightarrow 6 u = 30 g

u = 5 g

So,

H = 10 g - 2 g = 8 g

(from 2)
hence,

t = \sqrt{\frac{2 \times 8 g}{g}} = 4 s e c .

definition

Problems on motion under gravity including collisions with the ground

In free fall initial velocity of the particle is zero.
Therefore,

h = 1 / 2 g t^{2}

If a particle is falling from 19.6m then after how long the particle will collide the ground:

19.6 = 1 / 2 \times 9.8 \times t^{2}

t = 2 s

example

Inderstanding displacement,velocity and acceleration vectors in projectile motion

For a projectile motion displacement can be calculated from the point of projection along the direction of projection. Velocity has two components

v c o s θ

in horizontal and

v s i n θ

in vertical direction ,in which horizontal component of velocity remains constant because of absence of force. Acceleration due to gravity is present throughout the motion in vertical direction.

diagram

Time of Flight of a Projectile

Let, time taken to reach maximum height

= t_{m}

Now,

v_{x} = v_{o} \cos θ_{o}

and

v_{y} = v_{o} \sin θ_{o} - g t

Since, at this point,

v_{y} = 0

, we have:

v_{o} \sin θ_{o} - g t_{m} = 0

Or,

t_{m} = (v_{o} \sin θ_{o}) / g

Therefore, time of flight

= T_{f} = 2 t_{m} - 2 (v_{o} \sin θ_{o}) / g

because of symmetry of the parabolic path.

diagram

Maximum Height of a Projectile

x = v_{x} t = (v_{o} \cos θ_{o}) t

and

y = (v_{o} \sin θ_{o}) t - (1 / 2) g t^{2}

The maximum height

h_{m}

is given by:

y = h_{m} = (v_{o} \sin θ_{o}) (\frac{v_{o} \sin θ_{o}}{g}) - \frac{g}{2} (\frac{v_{o} \sin θ_{o}}{g})^{2}

(for

t = t_{m}

)
Or,

h_{m} = \frac{(v_{o} \sin θ_{o})^{2}}{2 g}

formula

Range of a Projectile Motion

The horizontal distance travelled by a projectile from its initial position

(x = y = 0)

to the position where it passes

y = 0

during its fall is called the horizontal range,

R

. It is the distance travelled during the time of flight

T_{f}

. therefore, the range

R

R = (v_{o} \cos θ_{o}) (T_{f}) = (v_{o} \cos θ_{o}) (2 v_{o} \sin θ_{o}) / g

......(1)
Or,

R = \frac{{v_{o}}^{2} \sin 2 θ_{o}}{g}

......(2)
Equation 2 shows that for a given projectile velocity

v_{o}

R

is maximum when

\sin 2 θ_{o}

is maximum, i.e. when

θ_{o} = 45^{o}

.
The maximum horizontal range is, therefore

R_{m} = \frac{{v_{0}}^{2}}{g}

definition

Range of a Projectile

Range is maximum for

θ = 45^{0}

Moreover, from the expression of range:

R = \frac{u^{2} s i n 2 θ}{g}

Mathematically, it can be said that range is maximum for

θ = 45^{0}

And its maximum value will be

\frac{u^{2}}{g}

result

Angle of Projection of a Projectile

For a projectile the range and maximum height are equal. The angle of projection is :Solution:
In projectile thrown at angle

θ

Range

R

and maximum height

H

are given as :
Range

R = \frac{u^{2} (\sin 2 θ)}{g} = \frac{u^{2} 2 \sin θ \cos θ}{g}

Max Height

H = \frac{u^{2} \sin^{2} θ}{2 g}

Given

H = R

\frac{u^{2} (2 \sin θ \cos θ)}{g} = \frac{u^{2} \sin^{2} θ}{2 g}

4 \cos θ = s i n θ

\tan θ = 4

θ = 76^{0}

example

Radius of curvature of projectile at a point

Example:
A body is thrown from the surface of the Earth at an angle

α

to the horizontal with the initial velocity

v_{0}

. Assume that the air drag was neglected.Find radius of curvature of projectile. Assume acceleration due to gravity

= g

.Solution:
Since, radius of curvature

= \frac{{(v e l o c i t y)}^{2}}{n o r m a l a c c e l e r a t i o n}

At peak point A,velocity

= v_{o} \cos α

and normal acceleration

= g

R_{A} = \frac{v_{o}^{2} \cos^{2} α}{g}

definition

Magnitude of velocity at any point in projectile motion

Magnitude of velocity at any general point is:

v^{2} = v_{x}^{2} + v_{y}^{2}

diagram

Angle of Projection of a Projectile with horizontal

v_{x} = v c o s θ \Rightarrow c o n s t a n t

v_{y} = v s i n θ - g t

Clearly

v_{y}

is maximum at t=0

Hence velocity,

v_{n e t} = \sqrt{v {_{x}}^{2} + v {_{y}}^{2}}

The direction of velocity will be along the direction of motion and angle of velocity with the horizontal will be given by:

t a n θ = \frac{v_{y}}{v_{x}}

formula

Equation of Parabolic Trajectory

y = x t a n θ - \frac{g x^{2}}{2 u c o s^{2} θ}

Where u is initial velocity and

θ

is angle of projection.

definition

Path length of a projectile

For a projectile projected with an initial velocity

u

and initial angle with the horizontal

θ

, length of path for the projectile is given by:

l = \frac{u^{2} \cos^{2} θ}{2 g} [2 \sec θ \tan θ + l n | \frac{1 + \sin θ}{1 - \sin θ} |]

example

Projectile Projected from certain height above the ground

Example: A stone is projected from the top of a tower with velocity

20 m / s

making an angle of elevation of

30^{o}

with the horizontal. If the total time of flight is

5 s

and

g = 10 m / s^{2}

, then find the maximum height attained by this projectile.

Solution:

S_{y} = u t - \frac{1}{2} g t^{2}

h = S_{y} = 20 \sin 30^{o} \times 5 - \frac{1}{2} \times 10 \times (5)^{2}

h = 50 - 125 = - 75 m

(minus sign just indicates that the displacement is in downward direction)

h_{m a x} = \frac{u^{2} \sin^{2} θ}{2 g} = \frac{(20)^{2} \sin^{2} 30^{0}}{2 \times 10} = 5 m

Hence, maximum height attained by projectile

= h + h_{m a x} = 75 + 5 = 80 m

definition

Average Velocity of Projectile

The average velocity of a projectile between the instant it crosses one third the maximum height. It is projected with

u

making an angle

θ

with the vertical.
There will be a pair of points for which vertical velocities at the same height are in opposite direction and therefore their average sum

= 0

It is the horizontal velocity which is uniform and hence

v_{a v} = u_{x} = u \cos θ

For a general point:
Displacement in Y-direction:

y = u s i n θ \times t - \frac{g t^{2}}{2}

Displacement in X-direction:

x = u c o s θ \times t

Now in order to calculate average velocity:
Average Velocity =

\frac{N e t D i s p l a c e m e n t}{T o t a l t i m e}

definition

Relative Position of a Projectile with respect to a vertical obstacle

Let the co-ordinate of the pick point of the vertical obstacle is (x, y).
And at any instant of time, position of the projectile is (

v c o s θ

), (

v s i n θ

),
Position relative to the point at the obstacle is: (

v c o s θ - x

), (

v s i n θ - y

)
This is relative position of a projectile with respect to a vertical obstacle.

example

Monday, July 31, 2017

Kinematics1

Motion in One Dimension

Motion in two and three dimension

Point Objects

Locating Position Vector of a point Object

Polar Coordinate System

Displacement

Distance v/s displacement

Calculating distance in one-dimensional motion

Displacement

Distance in two-dimensions

Displacement of an Object Moving in two dimensions

Displacement of an Object Moving in two dimensions where direction changes after finite distance

Calculating average speed

Calculating Average Velocity

Instantaneous speed

Instantaneous velocity

Average Velocity

Acceleration

Average Acceleration

Acceleration as a rate of change of velocity with respect to displacement

Understand Acceleration as Second Derivative of Displacement

Definition of Jerk

Problems Involving First Equation of Motion

Problems on second equation of motion

Problems on Third Equation of Motion

Solving problems involving more than one equation of motion

Finding Stopping Distance

Equation of trajectory

Total distance traveled when velocity and accelerations are in opposite directions

Motion in different acceleration for different time intervals

Motion involving more than one uniform acceleration

Distance Traveled in nth second

Equations of Motion under free fall

Maximum Height of a Projectile

Problems on freely falling objects

Problems on motion under gravity including collisions with the ground

Inderstanding displacement,velocity and acceleration vectors in projectile motion

Time of Flight of a Projectile

Maximum Height of a Projectile

Range of a Projectile Motion

Range of a Projectile

Angle of Projection of a Projectile

Radius of curvature of projectile at a point

Magnitude of velocity at any point in projectile motion

Angle of Projection of a Projectile with horizontal

Equation of Parabolic Trajectory

Path length of a projectile

Projectile Projected from certain height above the ground

Average Velocity of Projectile

Relative Position of a Projectile with respect to a vertical obstacle

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