Physics

Session

Kinematics - 2

Session Opener

REST !or

MOTION

Session Objectives

Session Objective

1. Rest and motion

2. Distance and displacement

3. Uniform and non-uniform motion

4. Velocity

5. Acceleration

6. Equations of motion

Rest and Motion

Body at rest : Position constant w.r.t. fixed point as time increases

Fixed Point : Origin of a coordinate system

Y

O900

x X

r

yP (position)

(origin)

12 2 2r ox oy

oytan

ox

ox r cos

oy r sin

What is motion ?Change in position of an objectwith time, with respect to a given co ordinate system.

Motion

Actual distance traveled :Curve P0P1P2P3P4.

Displacement : Straight line P0P4 directed from P0 to P4

P0(t=0)

P1(t=t1)P2(t=t2)

P3(t=t3)P4(t=t4)

x

y

Distance and Displacement

1. Distance Displacement.

1

2

x

to

2. Distance = Displacement (If direction remains same.)

3.Distance is always 0 or+ve. Displacement can be +ve,0 or –ve.

4. Distance always increases with motion.

Average Speed

0 1 2 3 4

4 0

actual dis tance cov eredAverage speed

time taken

Curve lengthP PP P Ptime int erval (t t )

P0(t=0)

P1(t=t1)P2(t=t2)

P3(t=t3)P4(t=t4)

x

y

Courtesy:www.physicsclassroom.com

Instantaneous Speed

When time is infinitesimal (=t 0), distance is infinitesimal(=s 0)

Instantaneous speed = Limt 0

st

t

s

tt

s

Po

P1

s

Speed : Scalar

Unit m/s

Dimension LT-1

Average Velocity

Average velocity is defined as displacement divided by time taken.

tr

ttrr

v12

12av

Nature : vector

Dimension : [LT-1]

Unit : m/savv

Displacement and average velocity are in same direction

r

t2t1t

2r

1r

Instantaneous Velocity

If t= (t2 – t1)is extremely small (t 0)

limt 0

rv

t

v is instantaneous velocity

v is a vector.

Unit of v : m/s

r

t1t

1r

2t

2r

2t

2r

Uniform Motion (One dimension)

Equal displacement (x) traveled in equal time interval (t)

1 0 02 1

1 2 1

x x x xx x... v

t 0 t t t 0

v is constant.

x = x0 + vt

If x0= 0 at t = 0 , x = vt

x0 x1 x2 x3 x4 x

0 t1t2

t3 t4t

Class Exercise

Class Exercise - 3

Graph in the figure below shows the variation of displacement with respect to time for a particle in one-dimensional motion. Which of the following represents the velocity-time graph of the particle in motion?

0 5t10 2015

x

Options is in next slide

Class Exercise - 3

0 5t10 2015

v

(a)

0 5t10 2015

v

(b)0 5

t10 2015

v

(c)0 5

t10 2015

v

(d)

Solution - 3

For t = 0 to 5

dxv ( ve and cons tant)

dt

dxt 5 to 15 v ( 0)

dt

dx

t 15 to 20 v ( ve and cons tant)dt

Hence answer is (c)

Class Exercise - 8

An object travels half the distance with v1, with v2 for half of the remaining time and with v3 for the remaining half of the time. If the object never reverses the direction of motion, find the average velocity during the motion.

1 2 32v v v

4

Solution :

Average Acceleration

tv

ttvv

a12

12av

Change in velocity divided by the time interval during which the change occurs.

Nature : vector

unit : m/s²

Dimension: [LT-2]

ava

v

t2t1t

2v

1v

Non uniform motion (constant acceleration)

Instantaneous acceleration

v(final velocity) u(initial velocity)a

t(time taken)

FFFFFFFFFFFFFFFFFFFFFFFFFFFF

t

v

v

u

t1 t2

t=t2-t1For constant acceleration

limt 0

va

t

v = u + at Equation of motion (1)

Class Exercise

Class Exercise - 1

Which graph represents increasing acceleration?

v

AB

C

t

(a) A (b) B (c) C (d) None of these

Solution - 1

Hence answer is (b).

Increasing a Increasing slope of

v-t curve. By observation, we find that the velocity is increasing at an increasing rate. So acceleration is increasing.

dva

dt

Class Exercise - 9

An object starts from rest. It accelerates at 2 m/s2 till it reaches its maximum velocity. Then it retards at 4 m/s2 and finally comes to rest. If the total time taken is 6s, find vmax and the displacement of the object.

Solution - 9

O6

Vm ax

t1 t2

Vmax = 2t1

Vmax = 4t2

s = Area of triangle

t1 = 4 s, t2 = 2 s

Hence, Vmax = 8 m/s

1

s 6 8 24 m2

2 1

2

1

1 2

4t 2t

t 1t 2

t t 6

Class Exercise - 2

A particle is thrown vertically upwards with velocity v. It returns to the ground in time T. Which of the following graphs correctly represents the motion?

v

T2

T

t

(a)v T

2

t

– v

(b)

v

T2

tT

(d)v

T2

t

–v

(c)

Solution - 2

The acceleration is constant (= –g). So slope has to be negative throughout the motion and velocity varies between v and –v.

Hence answer is (c).

Class Exercise - 5

If a particle has an initial velocity of

and an acceleration of ,

its speed after 10 s is

7 2

(a) 10 units (b) 7 units

3 i 4 j

0.4 i 0.3 j

(c) units (d) 8.5 units

Solution - 5

v u at

Hence, v 3 i 4 j 0.4 i 3 j 10

7 i 7 j

Speed v 7 2 m/s

Hence answer is (c).

One dimensional equations of motion

Distance s = area under v-t graph = ½ (u+v)t

But, s = vavgt Hence, vavg = (u+v)/2

t

v

v

u

t1 t2

t=t2-t1

Using equation of motion (1)

s = ut + ½ at2 Equation of motion (2)

v uAs t , equation of motion (2)

agives

2 2v u 2a.s Equation of motion (3)

One dimensional equations of motion

v = u + at

v² = u² + 2as

21s ut at

2

n

2n 1S u a

2

Distance traveled in nth second

Class Exercise

Class Exercise - 6

A particle moves along the X-axis as x = u(t – 3) + a(t – 3)2, then

Which of the following are true?

(a) initial velocity of particle is u at t = 0

(b) acceleration of particle is a

(c) at t = 3 the particle was at origin

(d) the particle may have negative velocity

Solution - 6

The observation of displacement has started at time t = 3 s, after the object has actually started. So if it represents the time for which the object has traveled and s be the displacement after the observation has started, then general form is

21s ut at

2

2compare with s u(t – 3) a(t – 3)

Acceleration 2a, initial velocity at t 3s u

Class Exercise - 10The magnitude of maximum acceleration, retardation of an object is ‘a’ m/s2. What is the minimum time taken by the object to cover a displacement ‘s’ if it starts from rest and finally comes to rest?Solution :

The minimum time would be when the acceleration is at maximum and deceleration is also maximum. Half the time accelerating at a and the rest of the time deceleration at ‘a’.

1 t sHence a t s t 2 sec

2 2 a

Uniform and Non-Uniform Motion

Let us see a comparison of uniform and non-uniform motion

Courtesy:www.physicsclassroom.com

Class Exercise

Class Exercise - 7

An object has one-dimensional motion. If V = 6t + 4t3, then

(a) what is the distance covered from t = 3 s to t = 5 s?

(b) when is the acceleration < 0 for the first time?

5

3

3

(a) (6t 4t) dt 592

dv(b) Since is never zero

dt

So acceleration is never negative.

Solution :

Class Exercise - 4A particle moves in a straight line, starting from rest. The acceleration of

the particle is given by

What is the distance traveled by the particle in the time interval 0 to seconds.

21

a sintt 1

Solution - 4

2

dv 1sin t

dt t 1

t t

20 0

1dv sin t dt

t 1

t

0

1v t cos t

t 1

1

v t 2 cos tt 1

[Check that v(0) = 0]

Solution - 4

dx tNow v(t)

dt

dx 12 cos t

dt t 1

x

0 0

1dx 2 cos t dt

t 1

0x [2t sin t log t 1 ]

x = 2 – log( + 1)

Thank you