properties of graph

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1Motion In a Straight Line

Properties of Graph1. Main quantity: on x-axis

2. Dependent quantity: on y-axis

3. Slope of graph:y axisx axis

4. Area covered with x- axis by graph: y axis x axis

e.g. For uniform motion equal distance travel by object in equal time, so graph between dis-tance and time

Slope of graph = tan st

Speed

svt

tan

svt

Slope of distance- time graph represent speed of the object.

Calculation of distance by speed-time graph

If speed of an object is v, distance travel in dt time

ds v.dtSo total distance travelled from time t

1 to t

2

2

1

t

tS ds

2

1

ttS vdt

If an object move with uniform speed from time t1 to t

2, then speed-time graph

A BV

t1 t2t



distance travel from time t1 to t

2 = area of AB t

2t1

area = length × width = V × (t2 – t

1)

Note : In speed-time graph, area covered with x-axis shows distance.Average Velocity

Vav = tx

= if

if

ttxx

the average velocity is equal to the slope of the line (chord)

x

Q

P

tO

xf

xi

ti tf

joining the points corresponding to P and Q on the x-t(position-time) graph.

Instantaneous Velocity

The instantaneous velocity at P is the slope of the tangent at P in the x t (position-time) graph .

t 0

x dxv limt dt

When the slope of the x t graph is positive, v is positive (as at

the point A in figure). At C, v is negative because the tangent has

negative slope. The instantaneous velocity at point B (turning point)is zero as the slope is zero.



Motion with uniform velocityxt graph is a straight line of slope v

x

O

xi

t v is positive

slope = v

x

O

xi

tv is negative

slope = v

as velocity is constant, v t graph is a horizontal line.

v

O

u

t

positive velocity

v

O

u

t

negative velocity

at graph coincides with time axis because a = 0 at all time instants.

Some extra points about graphsStraight line-equation, graph slope (+ve, –ve, zero slope)

y = mx + c (equation of a straight line)slope = mintercept = c on the yaxis.

m = slope = tan = dy

dx

Cx

+ve slope

y

Cx

slope = 0

y

C

x

–ve slope

y

Parabolic curve-equation, graph

x

y

y = kx2 x

y

y = –kx2



x

y

x = ky2 x

y

x = –ky2

Where k is a positive constant.

Position – Time Graphs for a Moving Object

(i) If an object is at rest or stationary, its (ii) If an object is in uniform motion along a straight

position will not change with time. Then line starting from, the origin O, the position

the position (x)-time (t) graph for the (x) time (t) graph is a straight line inclined to

stationary object is a straight line AB time axis. Greater the slope of x – y graph,

parallel to time axis. the greater is the velocity

O

x

larger positive v

smaller positive v

zero v

small negative vt

(iii) If an object is moving with a constant

negative velocity starting from the positive

position, then the position time graph of

this motion is a straight line AB inclined

to time axis.

(iv) Displacement of the body becomes negativeafter time t and then increases in magnitudewith time. It indicates that the body returningfrom position A, moves past the original posi-tion B and then moves towards C with auniform velocity.



(v) The position-time curve OA represents anincreasing velocity or an accelerated motion.For constant acceleration, the “position-timegraph is a parabola bending upwards.

(vi) The position-time curve OA represents adecreasing velocity or deceleration. Foruniform deceleration, the position-time graphis a parabola bending downwards.

(vii) If an object is in non-uniform motion along a straight line e.g. a car starting from rest at timet = 0 second from the origin, moves along a straight path. It picks up some speed, then moveswith constant speed for some time After that the brakes are applied and the car comes to rest.

x

tD

O

F

x0

E F

Graphs which are not possible

(i) Graph is a straight line parallel to distanceaxis. It represents infinite speed which isnot possible.

(ii) The distance covered by a body cannotdecrease with the increase of time. So thedistance-time graph of this type is notpossible.

(iii) The Graph is not possible because it represents two different positions of the body at the sameinstant which is not possible.



AccelerationThe rate of change of velocity is called acceleration

Acceleration = Change in velocity

Time interval

Type of change in velocity

Only direction is change Only magntiude is change Direction and magnitude both are change

acceleration is perpendicular to acceleration is parallel or acceleration has two components, one is

velocity anti parallel to velocity parallel or antiparallel and other isperpendicular

e.g. Uniform circular motion e.g. Motion under gravity e.g. Parabolic Motion

SI unit : ms–2

Acceleration is a Vector Quantity.

Dimensional formula [M0L T–2].

Direction of acceleration is same as change in velocity.

Acceleration is positive if velocity is increases and is negative if the velocity decreases. The negativeacceleration is also called retardation or deceleration.

Average acceleration It is the ratio of the total change in velocity to the total time taken.

Average acceleration

av

total change in velocity va

total time taken t

The slope of straight line joining two points on velcoity-time graph is the average acceleration of theobject between these two points.

The average acceleraton can be positive or negative depending

upon the sign of slope of velocity-time graph It is zero if the change invelocity of the object in the given interval of time is zero.

Check Point : A car is moving in the positive x direction in 20 ms–1 andcomes up behind a truck and is unable to pass. The car slows to 15 ms–1

in a time 2 s. Calculate car's average acceleration.

Sol.: Change in car's velocity, 22 1v v v 15 20 5 ms

Note that the change in velocity is negative because the car's velocity in the positive x direction isdecreased by 5 ms–1.

Time interval, 2 1t t t 2 s



Average acceleration, 2v 5a 2.5 mst 2

The average acceleration is negative because the car's velocity along the positive x direction isdecreased.

Instantaneous acceleration The acceleration of the object at a given instant of time is called itsinstantaneous acceleration.

t

v dva limt dt

Instantaneous acceleration is also the slope of the tangent to the velocity-time graph at given time.

Note: If an object is moving with uniform acceleration, then, instantaneous acceleration = uniformacceleration = average acceleration.

Check Point: The position of a particle moving along a straight line is given by:

x = 2 – 5t + 6t2

Find the acceleration of the particle at t = 2s.

Sol. x = 2 – 5t + 6t2 or 2dx d 2 5t 6t 5 12tdt dt

2

2d x da 5 12t 12

dtdt

Note: If velocity change with displacement then acceleration.

dva

dt

dv dxa .

dt dx

dva v

dx

Graphs in uniformly accelerated motion (a 0)* x is a quadratic polynomial in terms of t. Hence x t graph is a parabola.

xi

x

a > 0

t0

xi

x

a < 0

t0

x-t graph



* vt graph is a straight line of slope a.

v

ua is positive

slope

= a

t0

v

u

a is negative

slope = a

t0

v-t graph* at graph is a horizontal line because a is constant.

a

apositiveacceleration

t0a

a

negativeacceleration

0

a-t graph

Velocity Time-Graph of an Accelerated Motion(i) When an object is moving with zero (ii) When an object is moving with constant positive

acceleration acceleration, having zero initial velocity.

(iii) When an object is moving with positive (iv) When an object is moving with constant

constant acceleration having some initial velocity. negative acceleration, having positive initial velocity.

(v) When an object is moving with uniform negative (vi) The v-t graph represents a body projected upwards

acceleration having negative initial velocity. with an initial velocity u. The velocity decreases with

time (negative uniform acceleration), becoming zero after

certain time t. Then the velocity becomes negative and

increases in magnitude, showing body is returning to

original position with positive uniform acceleration.



(vii) When an object is moving with increasing (viii) Smaller changes in velocity are taking place in

acceleration, having zero initial velocity. equal intervals of time. So the v-t graph

bending downwards represents a decreasing

acceleration.

.

(ix) The area between the velocity-time graph and the time-axis gives the displacement. The v-tgraph represents variable acceleration.

(i) For a body projected upwards, When the bodymoves, its speed decreases uniformly, becoming zeroat the highest point. As the body moves down, itsspeed increases uniformly. It returns with the samespeed with which it was thrown up.

Different Types of Speed-Time Graphs(ii) For a ball dropped on the ground from a certain

height, As the ball falls, its speed increases. As theball bounces back, its speed decreases uniformly andbecomes zero at the highest point.



Position-Time Graph for Accelerated Motion(i) When an object is moving with (ii) When an object is moving with negative

uniform positive acceleration. acceleration.

(iii) When object is moving with zero acceleration.

F

TIME

POSITION

xO

O

E

Acceleration - time graph(i) Constant acceleration (ii) Uniformly increasing acceleration

tan = 0 is constant.

dtda

= 0 0º < < 90º tan > 0

Hence, acceleration is constant. da

dt = tan = positive constant

Hence, acceleration is uniformly increasing

with time.(iii) Uniformly decreasing acceleration (iv) Variable acceleration



> 90º For a body moving with variable tan is constant and negative. acceleration, the a -t graph is a curve.

dtda

= negative constant The area between the a -t graph and the

Hence, acceleration is uniformly time-axis gives the change in velocity,decreasing with time Change in velocity

= Area 1 - Area 2 + Area 3

Note : The area under a-t graph gives the change in velocity.* The area between the v-t graph gives the distance travelled by the particle, if

we take all areas as positive.* Area under v-t graph gives displacement, if areas below the t-axis are taken

negative.

Check Point : Describe the motion shown by the following velocity-time graphs.

(a) (b)

Solution : (a) During interval AB: velocity is +ve so the particle is moving in +vedirection, but it is slowing down as acceleration (slope of v-t curve)is negative. During interval BC: particle remains at rest as velocity iszero. Acceleration is also zero. During interval CD: velocity is -ve sothe particle is moving in -ve direction and is speeding up asacceleration is also negative.

(b) During interval AB: particle is moving in +ve direction with constantvelocity and acceleration is zero. During interval BC: particle is mov-ing in +ve direction as velocity is +ve, but it slows down until itcomes to rest as acceleration is negative. During interval CD: velocityis -ve so the particle is moving in -ve direction and is speeding upas acceleration is also negative.

Check Point : The accompanying figure shows the velocity v of a particle moving on a coordinateline.

-4

2

(m/s)

(a) When does the particle move forward? move backward? Speed up? slow down?(b) When is the particle's acceleration positive? Negative ? zero?(c) When does the particle move at its greatest speed ?(d) When does the particle stand still for more than an instant?

Ans : (a) (0, 1)s & (5,7)s(1, 5)s(1, 2) s & (5, 6) s



(0, 1) s & (3, 5) s & (6, 7) s(b) (3, 6) s

(0, 2) s & (6, 7) s(2, 3) s & (7, 9) s

(c) 0 s & (2, 3) s(d) (7, 9)s

Derivation of Equations of Uniform Accelerated Motion from Velocity - Time GraphConsider an object moving with u initial veloctiy with uniform acceleration at time t = 0 s.

After time t final velocity of object becomes v.

Velocity time graph of this motion is a straight line AB.

Slope of graphdvtandt

BCtanAC

V utan

t 0

V utant

...(i)

Slope of velocity - time graph represents acceleration,

tan = a ...(ii)

By Eq. (i) and Eq. (ii)V ua

t

V u at

V u at (First Equation of Motion)

In time-velocity graph covered area with x-axis represents displacement

x Area of ACB + Area of rectangle ACDO

1x AC BC OD DC2

1x t V u ut2

1x ut V u t2

V u at

1x ut u at u t2

21x ut at2

(Second Equation of Motion)



Square of First Equation of Motion

v u at (Square)

2 2v (u at)

2 2 2 2v u a t 2uat

2 2 21v u 2a ut at2

21s ut at2

2 2v u 2as (Third Equation of Motion)

Displacement travelled in nth second

Consider an object moving with a acceleration for n seconds. If Sn and S

n-1 are the

displacement of object in n and n - 1 seconds.

Distance travel in n seconds

2n

1S un an

2

Distance travel in (n - 1) seconds

2n 11S u n 1 a n 12

Distance travel in nth second 22n n 1

1 1S S un an u n 1 a n 12 2

2 21 aun an un u n 2n 12 2

2 21 1 1 1S un an un u an a 2n a2 2 2 2

2 21 1 1S un an un u an an a2 2 2

1S u an a2

1S u a 2n 12

Derivation of Equations of Motion by Calculus Method(i) First Equation of Motion

Acceleration dv

a=dt

dv = adt

By integration



v t

u 0dv adt

v t

u 0v a t

v – u = a (t – 0)

v – u = at

v = u + at ...(i) (First Equation of Motion)

(ii) Second Equation of Motion

Velocitydx

vdt

By First Equation v = u + at

dxu at

dt

By Integration

0

x t t

x 0 0dx u dt a t dt

0

t2x t

x 00

tx u t a

2

20

1x x ut at

2

0x - x = s (Displacement) 21s ut at2

...(ii) (Second Equation of Motion)

(iii) Third Equation of Motion

dv dv dx dva v

dt dx dt dx

adx vdv

By intergration, 0

v x

u xvdv adx

0

v2x

xu

v a x2

2 2

0v u a(x x )

2



2 20v u 2a(x x )

2 20v u 2a(x x )

2 2v u 2as ...(iii) (Third Equation of Motion)

properties of graph

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