031 kinematics final

8/3/2019 031 Kinematics Final

1/29

KinematicsIITJEE Syllabus:

Kinematics of one and two dimensions (Cartesian Coordinates only), Projectiles, Circularmotion (Uniform and non-uniform), Relative velocity.

MOTION IN ONE DIMENSION

DisplacementThe displacement of a particle is defined as the difference between its final position and its

initial position. We represent the displacement as x. x = xfxi

The subscripts i andfrefer to be initialandfinalpositions. These are not necessarily the positions

from which the particle starts its motion nor where its motion ceases. The i and fdesignate the

particular initial and final positions we are considering out of the entire motion of the object. Note

the order : final position minus initial position . Whenever we calculate delta anywhere, we

always take the final value minus the initial value.

Average Velocity and Average Speed

The average velocity of an object travelling along the xaxis is defined as the ratio of itsdisplacement to the time taken for that displacement.

vav=if

if

tt

xx

t

x

= (1)

The average speed of a particle is defined as the ratio of the total distance travelled to the time

taken.

Average speed =ttravelleddistanceTotal

Note that velocity andspeedhave different meanings.

Example : 1

A bird flies toward east at 10 m/s for 100 m. It then turns around and flies at 20 m/s for 15

s. Find

(a) its average speed

(b) its average velocity

PIE EDUCATION, Corporate Office: 44A/2,Jassum House, Sarvapriya Vihar, New Delhi -16. Ph: 26527444, 26527455. Tele Fax:26962662

Also visit us at: www.pieeducation.com


2/29


3/29

vav= 67.2s75

m200==

t

xm/s

Since 2.67 2

1(4+2), we see that the average velocity is not, in general, equal to the

average of the velocities.

Average Acceleration is defined as the ratio of the change in velocity to the time taken.

aav =t

v=

if

if

tt

vv

(2)

Instantaneous Velocity is defined as the value approached by the average velocity when the time

interval for measurement becomes closer and closer to zero, i.e. t 0. Mathematically

v(t) =t

xlimvlim

0tav

0t =

The instantaneous velocity function is the derivative with respect to the time of the displacement

function.

v(t) =dt

dx(t)(3)

Instantaneous Acceleration is defined analogous to the method for defining instantaneous

velocity. That is, instantaneous acceleration is the value approached by the average acceleration as

the time interval for the measurement becomes closer and closer to zero.

The Instantaneous acceleration function is the derivative with respect to time of the velocityfunction

a(t) =dt

dv(t)(4)

Example: 3

The position of a particle is given by

x = 40 5t5t2, wherex is in metre and tis in second(a) Find the average velocity between 1 s and 2 s

(b) Find its instantaneous velocity at 2 s

(c) Find its average acceleration between 1 s and 2 s

(d) Find its instantaneous acceleration at 2 s

Solution(a) At t= 1 s;xi= 30 m

t= 2 s;xf= 10 m

vav= 20123010 =

=

if

if

ttxx m/s

(b) v = tdt

dx105=

At t= 2 s; v = 510(2) = 25 m/s(c) At t= 1 s; vi = 510 (1) = 15 m/s

t= 2 s; vf = 510 (1) = 25 m/s

vav= 1012

)15(25=

=

if

if

tt

vvm/s2



FS2K5-PH-3 PIE EDUCATION FUNDAMENTALS KINEMATICS


4/29

(d) a =dt

dv= 10 m/s2

GRAPHICAL INTERPRETATION OF DISPLACEMENT,

VELOCITY AND ACCELERATION

Average Velocity

The average velocity between two points in a given time interval can be obtained from a

displacementversus time graph by computing theslope of the straight linejoining the coordinates

of the two points.

Instantaneous Velocity The instantaneous velocity at time t is the slope of the tangent line drawn to the

positionversustime graph at that time.

O t

x

Fig.(3) The x versus t graph of a particle whose velocity is not

constant. The slope of the chord joining two points on the

curve is the average velocity for that interval. The

instantaneous velocity is the slope of the tangent at that point.

Chord x

t

Tangent line v = dt

dx

vav=t

x

O t

x

v > 0

v = 0

v < 0v = 0

Fig.(4) The sign of the instantaneous velocity depends on the slope of

the tangent.

Average AccelerationThe average acceleration between two points in a time interval is equal to the slope of the chord

connecting the points on a velocity versus time graph.

Instantaneous Acceleration

The instantaneous acceleration at time t is the slope of the tangentdrawn to the velocity versus

time graph.

t

aav =t

v

O t

v

Fig.(5) On a v-versus-tgraph, the slope of the line joining two pointsis the average acceleration for that time interval. The

instantaneous acceleration at a given time is the slope of the

tangent at that time.

a =

dt

dv

v

O t

v

Fig.(6) The sign of the instantaneous acceleration depends on the

slope of the tangent.

a = 0

a < 0

a = 0

a > 0





5/29

Displacement from Velocity Time Graphs

Given a velocity versus time graph, the

displacement during an interval between time tiand tf is the area bounded by the velocity curve

and the two vertical lines t = ti and t = tf, asshown in the Fig.(7 a).

t

v

t

vo

O ti tf

Fig.(7 a) The area under the v-versus-tcurve is

the displacement x = vot

t1

v

tO

Fig.(7 b) For each segment of motion, the velocity is a

different constant. The displacement x1

during the tth

interval is the area v1t1. So thetotal displacement is

x= v1 t1 + v2 t2 + v3 t3 + v4 t4

v1

v2

v4

v3

t2 t3 t4

v

tO ti tf

Fig.(7 c) When v versus t graph is a smoothcomplex curve. the area under the curve

may be obtained using integration.

Velocity from Acceleration Graphs

Given an accelerationversustime graph, the change in velocity between t = tiand t = tf is thearea bounded by the acceleration curveand the vertical lines t= tiand t= tf.

t

a

t

ao

O ti tf

Fig.(8 a) The area under the a versus tcurve is

the change in velocity V = aot

a

tO ti tf

Fig.(8 b) When a versus t graph is a smoothcomplex curve. the area under the curvemay be obtained using integration.





6/29

Table Variation of Displacement (x), velocity (v) and acceleration (a) with respect to time for

different types of motion.

Displacement Velocity Acceleration

1. At rest

O t

c

x

x= c

O t

v

O t

a

2. Motion with constant

velocity

O t

x

x= vot+xo

xo

O t

v

vo

O t

a


acceleration

O t

x

x= vot+(1/2)aot2

O t

v

v = aot+ vo

vo

O t

a

ao


deceleration.

O t

x

x= vot- (1/2)aot2

O t

v

vov = vo aot

Ot

a

ao

Example: 4

At t= 0 a particle is at rest at the origin. Itsacceleration is 2 m/s2 for the first 2 second

and 2 m/s2 for the next 2 s. Plot the

x versus tand v versus tgraphs.SolutionIt is given thatx = 0 and v = 0 at t= 0. Theacceleration versus time graph is plotted in

Fig. (9 a). The velocity at t= 2 s is equal tothe sum of velocity at t = 0 and the area

under the accelerationtime graph between

t= 0 and t= 2 s.

a

+2

t(s)

2

m/s2

O 2 4

(a)

Fig.(9 a) Example (4)





7/29

3

v

(m/s)

t(s)

4

O 1 2 4

Fig. 9(b)

3 t(s)O 1 2 4

4

8

x(m)

Fig. 9(c)

v2 = v0 + [ ]2t

0tgraphofArea

==ta

Since at t= 0: velocity is zero, vo= 0

v2 = 0 + (2) (2) = 4 m/sSimilarly velocity at t= 4s is

v4 = v0 + [ ]4t

0tgraphofArea

==ta = 0 +2 (+2) + 2 (2) = 0

Note that area below thexaxis is taken as negativeThe displacement at t= 2 s is

x2 =x0 + [ ]st

ttv

2

0graphofarea

==

x2 = 0 + 2

1

(2) (4) = 4 m

and x4 = 0 +2

1(4) (4) = 8 m

MOTION WITH CONSTANT ACCELERATION

In this section, we are going to derive equations relating to displacement, velocity, acceleration

and time.

We know that, adt

dv=

or dv = adtOn integrating both sides,we get

dv = adtv = at+ c1 where c1 = constant of integration

If v0 be the initial velocity of the particle then at t= 0, v = v0. This implies that c1 = v0v = v0 + at (5)

Also we know that

atvvdt

dx+== 0

or dx = (v0 + at)dt





8/29

On integrating, we get

dx = v0dt+ atdt

x = v0t+2

1at2 + c2 where c2 = constant

Ifx0 be the initial position of the particle, then at t= 0;x =x0. This implies that c2 =x0

x =x0 + v0t+2

1at2 (6)

Eliminating tfrom equations (5) and (6), we get

v2 = )(2 020 xxav + (7)

The equations (5) to (8) are called the equations of kinematics in the direction along thexaxis.

The equations of kinematics are summarised as

v = v0 + at (5)

x =x0+ v0t+

2

2

1

at (6)

v2 = )(2 020 xxav + (7)

x =x0 +2

1(v0 + v)t (8)

where x0 = Initial position coordinate

x = Final position coordinate

vo = Initial velocity

v = Final velocity

a = Acceleration (constant)

t = Elapsed time

Problem Solving Strategy1. Make a simple sketch of the situation described

2. Set up a coordinate system and clearly indicate the origin.3. (a) List the given quantities with appropriate signs.

(b) List the unknown quantities.

4. Find the equation that has the quantity you need as the only unknown. (This is not always

possible).

5. It is often helpful to obtain a rough graphical solution

6. Solve the equation (s) to find the desired unknown(s).

Example 5A car accelerates with a constant acceleration from rest to 30 m/s in 10 s. It then continues

at a constant velocity. Find

(a) its acceleration

(b) how far it travels while speeding up

(c) the distance it covers while its velocity changes from 10 m/s to 20 m/s

SolutionA sketch and coordinate system are shown in figure(a). Note thatx0= 0

(a) Givenv0= 0; v = 30 m/s; t= 10s





9/29

Unknown:a = ?x = ?From equation (5) we have

a = 3

0

+=

t

vv

m/s

2

(b) Given :

v0= 0; v = 30 m/s; t= 10 s; a = 3 m/s2

Unknown: x = ?

The position coordinate x appears as the onlyunknown in equations (6) and (8).

x =2

1at2 =

2

1(3) (10)2 = 150 m

If we had not found the acceleration in (a), we

would have to use equation (8).

x =x0+ 2

1(v0 + v) t

= 0 +2

1(0 + 30) (10) = 150 m

(c) Given: v0= 10 m/s; v = 20 m/s; a = 3 m/s2

Unknown : x0= ?;x = ?; t= ?

xo = 0, vo = 0

a

x

v

Fig. 10 (a)

v = 20m/s

xo xx

vo = 10m/s

Fig. 10 (b)

If we maintain the origin as shown in figure, we have to find x0 to this part of the trip.

However, we need only the difference x =x x0, which can be found from equation (7).

v2 = xav +220202 = 102 + 2(3) x x = 50 m

Example 6

A particle is atx = 5 m at t=2 s and has a velocity v = 10 m/s. Its acceleration is constant

at 4 m/s2. Find the initial position at t= 0Solution

Given x = 5 m; v = 10 m/s ; a = 4 m/s2; t=2 sUnknown:x0 = ?; v0= ?

In this case none of the equations of kinematics yields x0 immediately. The quantity x0appears in three equations, but always with the other unknown, v0. We have to find v0first,From equation (5).

v = v0+at10 = v0+ (4) (2)

Thus v0= 18 m/s. Any of the other equations will givex0. From equation (8).

x =x0+2

1(v0+ v)t

5 =x0+2

1(18 +10) (2)

Thus,x0= 23 m





10/29

0 5 10 15 20

a = -4 m/s2 t= 2 s v = 10 m/s

Fig. 11

Example 7

CarA moves at a constant velocity 15 m/s. Another carB starts from rest just as the carApasses it . The carB accelerates at 2 m/s2 until it reaches its maximum velocity of 20 m/s.Where and when does the carA is caught by other carB ?

SolutionWhen two particles are involved in the same problem, we use simple subscripts to

distinguish the variables, as shown in figure. The motion of the carB has two phases: oneat constant acceleration and other at constant velocity. In such problems it is convenient to

use tinstead oftin the equations. The carB may or may not catch the carA during theacceleration phase. This has to be checked. We set the origin at the carB, which meansxoA=xoB= 0

vA = 15m/s vA

0

A A

BaB

BvB

x t= t1 t= t1 + t2Fig. 12

Acceleration Phase:

Let it takes a time interval t1Given: vA= 15 m/s; aB=2 m/s

2; v0B = 0; vB= 20 m/s

Unknown:xA= ?;xB= ?

vB= vOB+ aBt t= 102

20==

B

B

a

vs

At this time, the positions are given byx =x0 + v0t+2

1at2

xA = (15) (10) = 150 m

xB=2

1(2) (10)2 100 m

The car A is still ahead

Constant velocity phase:

Let it take a time interval t2Given:xoA = 150 m;xoB =

100 m; vA = 15 m/s; vB= 20 m/s. aA= aB=0

Unknown:xA= ?; t2 = ?The cars meet when they have same position

xA=xB





11/29

Here xA= 150 + 15 ( t2)xB= 100

+ 20 ( t2) 150 + 15 ( t2) = 100 + 20 ( t2)or t2 = 10 sAlso, xA=xB= 300 m

FREE FALL

Motion that occurs solely under the influence of gravity is calledfree fall.

In the absence of air resistance all falling bodies have the same acceleration due to gravity,

regardless of their sizes or shapes.

The value of the acceleration due to gravity depends on both latitude and altitude. It is

approximately 9.8 m/s2

near the surface of the earth. For simplicity a value of 10 m/s2

isbeing used in this package.

The equations of kinematics may be modified as

v = v0gt (9)

y =y0+2

1(v0+ v) t (10)

y =y0+ v0t 2

1gt2 (11)

v2 = )(2 020 yygv (12)

The signs of v and v0 are determined bytheir directions relative to the chosen +yaxis.

Note that the sign of the accelerationdoes not depend on whether the body is

going up or coming down.

ay = -g

O xFig.(13) If they axis is chosen to point upward; the accelerationof a particle in free-fall is ay = -g, where g = 10m/s

2 is the

magnitude of the acceleration due to gravity.

y

Example 8

A ball thrown up from the ground reaches a maximum height of 20 m. Find

(a) its initial velocity

(b) the time taken to reach the highest point

(c) its velocity just before hitting the ground

(d) its displacement between 0.5 and 2.5 s

(e) the time at which it is 15 m above the ground.

SolutionA coordinate system is shown in figure. Note that at

the highest point the ball is instantaneously at rest,

that is v = 0

(a) Given: y0= 0;y = 20 m; v = 0; a = 10 m/s2

Unknown:v0= ? t= ?

Both equations (9), (10) and (11) involve the other

unknown, namely t. So from equation (12),

O

y

x

a = -g

Fig.14

0 = )020()10(220 +v





12/29

Thus, 40020 =v m2/s2, and v0= 20 m/s. Since up is positive, v0= + 20 m/s

(b) Given: All information in part (a) and v0= 20 m/s

Unknown :t= ?

Either Equation (9) or equation (11) would lead to t. Since equation (11) involves a

quadratic it is quicker to use equation (9)

0 = 20 10 tThus t= 2 s

(c) Given:y0= 0;y = 0; vo=20 m/s

Unknown : v = ?; t= ?From equation (12) we have

v2 = )00()10(220 v

Thus, v = v0. At the single pointy = 0, the velocity has two values +v0 initially and v0when it lands

(d) To find the displacement, y =y2y1, we need equation (11);

y1 = 20(0.5) 5 (0.5)2 = 8.75 m

y2 = 20(2.5) 5 (2.5) 2 = 18.75 mHence y = + 10 m

(e) Given: y = 15 m;y0 = 0; v0 = 20 m/s

Unknown:t= ?

Since equations (9), (10) and (12) also contain the other unknown, we use equation (11)

15 = 0 + 20t5t2

The solutions of the quadratic equations are

t=10

1554)20(20 2 = 1 s, 3 s

Example 9A ball is thrown upward with an initial velocity of 10 m/s from a rooftop 40 m high. Find

(a) its velocity on hitting the ground.

(b) the time of flight

(c) the maximum height

(d) the time to return to roof level

(e) the time it is 15 m below the rooftop.

SolutionThe origin is assumed at the ground level

so that all positions are positive

Given:y0 = 40 m; v0 = +10 m/s

a = 10 m/s2

(a) When the ball lands, its final position

coordinate isy = 0The final velocity v appears as the only

unknown in equation (12)

v2 = 102 + 2(10) (040) = 900 m2/s2

or v = 30 m/s(b) Since v is now known, we can use

equation (9)

O

40 m

x

ay = -g

10 m/s

y

Fig. 15

30 = 10 10t





13/29

which gives t= 4 s. If v were unknown we could use equation (11)

0 = 40 +10t5t2

(c) At the maximum height v = 0, so from equation (12)

0 = (10)2 + 2 (10) (y40)

Thusy = 45 m.(d) At the roof level, the final position isy = 40 m, From Equation (11)

40 = 40 + 10t5t2

Therefore, t= 0 s, 2 s. Of course we pick t=2 s. This is just double the time needed toreach the maximum height.

(e) Again from equation (11), withy = 25 m

25 = 40 + 10t5t2

After solving, we get t= 3 s.

Example 10

Two balls are thrown towards each other, ball A at 20 m/s upward from the ground, and

two second later ball B at 10 m/s downward from a roof 30 m high.

(a) Where and when do they meet

(b) What are their velocities on impact?

SolutionRemember in this kind of problem we have to find when

before where. We need to write the general expression

for the position coordinates. The coordinate system is

shown in figure.

(a) Given :y0A= 0; v0A= +20 m/s

y0B= 30 m; v0B= 10 m/s; a = 10 m/s2.

If A has been in motion for time t, then B has been inmotion for time (t2). From equation (11),

yA= 20t5t2

yB= 30 10(t2) 5(t2)2

10 m/s

O

30 m

xy

A

B

20 m/s

ay= -g

Fig .16

They meet when yA = yB. This condition immediately leads to t= 3 s. Substituting intoeitheryAoryB givesy = 15 m

(b) Since t= 3 s, we have vA= 20 + (10) (3) = 10 m/sand vB= 10 + (10) (32) =20 m/s. Notice that A is already moving downward when itcollides with B.

MOTION IN TWO DIMENSIONS





14/29

Whatever we have studied in the kinematics

of one dimensional motion the same is

applicable for motion in two and threedimensions.

In two dimensions the position vectorrof aparticle whose coordinate are (x,y) is

r=x i +y j

If the particle moves from P1 at position r1to

P2 at position r2, as shown in figure its

displacement is given by

r= r2r1= x i + y j

P2

path followed

P1

rr2

r1

O x

y

Fig.(17) Displacement of a particle

The student must carefully note the

correct direction of r.

r is the vector that must be added tothe initial position r1 to give the final

position r2,

i.e. r2 = r1 + rThe average velocity is defined as theratio of the displacement over the time

interval

vav= t

r

O x

y

Fig.(18) The instantaneous velocity v is directedalong the tangent to the path, but its

magnitude is not the slope of that line.

dt

drv =

t+ tr

t

t

rvav

=

or v=dt

dr = vx i + vy j (13)

where vx=dt

dyv

dt

dxy =and

The direction ofv is along the tangentto the path.

The instantaneous acceleration is the

ratio of change of the velocity with

respect to time.

a = jiv

yx aadt

d+= (14)

where ax =dt

dva

dt

dvyyx =;

Note that one cannot determine

acceleration directly from the path of theparticle. One needs to know how eachcomponent of the velocity varies as

function of space and time.

O x

y

Fig.(19) Possible directions of acceleration of a particle

traveling along a curved path.

a

a

a

path

General equations of kinematics for constant acceleration

v = v0 + at (15)

r= r0 + v0t+2

1at2 (16)





15/29

r= r0 +2

1(v0 + v)t (17)

Fortwodimensional motion in the plane, thex andy components of these equations are:vx = v0x+ axt vy = v0y + ayt

x =x0+ v0x+2

1axt

2 y =y0+y0yt+2

1ayt

2

x =x0+2

1(v0x+ vx)t y =y0+

2

1(v0y+ vy)t

)(2 022 xxavv xoxx += )(2

22oyoyy yyavv +=

Example : 11

A particle moves one quarter of a circular

path of radius 20 m in 10 s. Its initial

position is given by ri = (20 m)i and itsfinal position is rf= (20m)j .

(a) Find its displacement r and averagevelocity vav

(b) Find the magnitude of its average velocity

and its average speed.

Solution

(a) The displacement is given by

ijrrr 2020if ==

or )m(2020 jir +=

and the average velocity is

Fig.(20) Average speed depends on total distance traveled, butaverage velocity depends only on the location of the

initial and final points.

x

y

S

ri

rf r

jijir

v 2210

2020+=

+=

=

tav (m/s)

(b) The magnitude of the average velocity is

83.2)2()2(|| 22 =+== avvavv

m/s

Average speed = s/m14.310

)]20(2[4

1

TakenTime

travelleddistanceTotal=

=

PROJECTILE MOTION

A projectile motion near the surface of the earth consists of two independent motions, a

horizontal motion at constant speed and a vertical one subject to the acceleration due to

gravity.

In order to deal with problems in projectile motion, one has to choose a coordinate system

and clearly specify the origin. If the x axis is horizontal and the yaxis points verticallyupward, then

ax= 0 and ay = gOne can easily assume the origin such that the initial horizontal coordinate is zero.

i.e. xo= 0

The equations of kinematics for projectile motion are





16/29

x = voxt (18)

vy= voygt (19)

y =yo+ voyt2

1gt2 (20)

22oyy vv = 2g(yyo) (21)

Example 12

A ball is projected horizontally at 20 m/s from a cliff of height 45 m

(a) Find its time of flight

(b) Find its horizontal rangeR (the horizontal displacement from the point of firing).

SolutionThe origin is assumed to be at the base of the cliff

Given:xo=0;yo= 45 m; vox= 20 m/s and voy= 0The coordinates at a latter time are given by

x = 20 t (i)

x = 45 5t2 (ii)(a) When the ball lands, its vertical coordinate is zero

i.e. y = 0 From (ii) , we get

0 = 45 5t2 or t= 3 sNote that the time of flight does not depend on the value of the horizontal component ofthe initial velocity. If the ball were dropped from the same height it would have reached

the ground in 3 s.

(b) To find the horizontal range we use the horizontal velocity and the time of flight.

R = voxt= (20) (3) = 60 m

Fig.(21) The horizontal motion of a projectile is at constant velocity while the

vertical motion occurs at constant acceleration. The vertical componentof the motion of a ball projected horizontally is the same as that of a

ball that is simply dropped.

vox t= 1s t= 2 s t= 3 s

t = 1 s

t= 2 s

t= 3 s

10 m/s

20 m/s

30 m/s 30 m/s

20 m/s

20 m/s

20 m/s

20 m/s

10 m/s

y

Example 13

A ball is projected from the ground with an initial velocity voat an angle above thehorizontal

(a) Find the time of flight

(b) Determine the horizontal rangeR





17/29

(c) Determine the maximum height obtained by the ball

(d) Obtain an expression for the trajectory of projectile

SolutionThe origin is assumed at the point of projection. The instantaneousx andy coordinates of

the ball are given byx = (v0 cos ) t (i)

y = (v0 sin ) t2

1gt2 (ii)

(a) To find the time of flight, we puty = 0 in equation (ii)

0 = (v0 sin ) t2

1gt2

or t=g

v sin2 0..(iii)

(b) To find the horizontal range we substitute (iii) in (i)

R = (vo cos )

g

v sin2 0 or R =g

vo 2sin2

(iv)

(c) The maximum heightHfor the projectileis given by

gHvv oyy 222=

Since vy=0, therefore,

H=g

v

g

voy

2

sin

2

220

2 = (v)

(d) To find the trajectory of the projectile we

have to expressy in terms ofx. Thereforeeliminating tfrom (i) and (ii), we get

vo

y

vox

voy

voy= vosin

vy

vox

vox

vox

vy

Fig.(22) The path of a projectile is parabolic. The path issymmetrical about the highest point only if the particlelands at the same level from which it was fired.

voy = vo cos

y =x tan ( )

2

20 cos2

xv

g

(vi)

IMPORTANT1. The time of flight is given by

T=g

vo sin2(22)

2. The horizontal range is given by

R =g

vo 2sin2

(23)

Equations (22) and (23) are valid only when the projectile returns to the initial vertical

level.

For a given initial speed vo, the range is a maximum when sin 2 =1, that is when = 45.For a given velocity vosame range occurs at two angles of projection, viz.

= 45





18/29

Fig.(23) The horizontal ranges for angle of projection (45o

- ) and

(45o

+ ) are equal.

45

o





19/29

3. The maximum height of the projectile is

H=g

vo

2

sin22

(24)

4. The trajectory of a projectile is a parabola

y =xtan 2

1

2

20 )cos(

xv

g

(25)

Example: 14

A hunter with a blowgun wishes to shoot a monkey hanging from a branch. The hunter

aims right at the monkey, not realizing that the dart will follow a parabolic path and thus

fall below the monkey. The monkey, however seeing the dart leave the gun, lets go of the

branch and drops out of the tree, expecting to avoid the dart. Show that the monkey will be

hit regardless of the initial velocity of the dart so long as it is great enough to travel the

horizontal distance to the tree before hitting the ground.Solution

Let the horizontal distance to the tree bex and the original height of the monkey beH, as

shown in Fig. (24). Then the dart will be projected at an angle given by tan =H/xIf there were no gravity, the dart would reach the height H in the time t taken for it to

travel horizontal distancex:

y =yoyt=H in time t=oxv

xwith no gravity

y = voyt2

1gt2

or y =H 2

1gt2

This is lower thanHby2

1gt2, which

is just the amount the monkey falls in

this time. The initial velocity of the

dart is varied so that for large v0 thetarget is hit very near its original

height and for small v0 it is hit justbefore it reaches the floor.

Dart y

1/2gt2

H= voyt

voy

vox

Fig.(24) Example 14

CIRCULAR MOTION

In its simplest kind, a particle moves in a circular path ofconstant radius with constantspeed. Although the magnitude of the velocity vector remains constant but the directionof the velocity vector changes direction continuously as shown in Fig. (25). Thus a

particle is continuously accelerated. Since the acceleration produces a change only in the

direction of the velocity vector, therefore, it must always be at right angles to the

direction of motion. It is because any component in the direction of the motion would

produce a change in speed of the particle.





20/29

Fig.(25)(a) Velocity vector (tangent to the circle) represents the direction of a body willmove if it is not pulled toward the centre.

(b) Two velocity vectors are subtracted to find the change in velocityv = vf- vi(c) The change v in velocity vectors, when placed at the average position,

between initial and final points toward the centre of the circular motion.

(a)

(c)(b)

vi

vf

vivf

v

vi

vf

v

= 0

g

Fig.(26) Comparison of uniform circular motion with straight line motion and projectile motion(a) In straight line motion the acceleration vector is either parallel or antiparallel to the velocity vector.

(b) In projectile motion: 90o

< < 180o

during upward motion.

0 < < 90o during downward motion and = 90o at the highest position.(c) In uniform circular motion acceleration is always perpendicular to the velocity vector.

(a) (c)(b)

v

a

g

180o

vg

v v

g

vg

90o < 90o

v

= 90o

Let us consider a particle which moves in a circular path of radius rand constant speed vas shown in Fig. (27). In a time tthe line joining the particle and the centre subtends an

angle which is given by

(radian) =r

s=

lengthRadius

lengthArc

On differentiating both sides, we get

dt

ds

rdt

d 1=

or = v/r (26)





21/29

where is the angular speedFrom Fig.(25 b), = ( v/v) |vf| = |vi| = vSince s = v t = r t =

v

r

Thus, acceleration, a =r

v

t

v 2=

Note that this acceleration point inward

toward the centre of the circle. It is called

centripetal acceleration (means centre

seeking)

Fig.(27) A particle moves with constant speed in

a circular path of constant radius r.

x

y

O

r

r

t= t s

t= 0

Alternatively the magnitude of the centripetal acceleration can be found using calculus

and vector notation.

r=xi +yj

r= rcos i + rsin jr(t) = rcos ti + rsin tj

v(t) =dt

dr= r sin ti + r cos tj

v= r

a(t) =dt

tdv )(= r 2 cos t i r 2

sin tja(t) = 2r(t)a = ac= 2r

ac=r

v 2

Fig.(28) A particle moves with constant angular

speed in a circular path of constant radius

y

r rsin t

rcos t

= t

To summarize, an object moving in uniform circular motion at radius rand constant angular

speed has1. a constant linear speed v = r tangent to the circle and2. a centripetal acceleration ac= 2r= v2/rdirected inward toward the centre of the circle

Example: 15

A stone tied to the end of a string 75 cm long is whirled in a horizontal circle with aconstant speed. If the stone make 15 revolutions in 25s. What is the magnitude and

direction of acceleration of the stone ?

SolutionRadius of the circular path, r= 75 cm = 0.75 m

Number of revolutions per second n =1s6.0

25

15 =

Now, v = 2 nror v = 2(3.14) (0.6) (0.75) = 2.83 m/s

Magnitude of acceleration, a =r

v 2=

75.0

)83.2( 2= 10.68 m/s2





22/29

The direction of acceleration is radially inwards.

Tangential and Normal Components of Acceleration in Two Dimensions

1. The velocity vector is always tangential to the path

2. The acceleration vector may have two components: one tangential to the path and one

perpendicular to the path.

(a) The component of the acceleration parallel to the path is due to a change in speed.

When the speed is increasing, the tangential component atpoints in the samedirection as the velocity; when the speed is decreasing, the tangential component

points opposite to the velocity.

(b) When the path of an object curves, there is a component of the acceleration

perpendicular to the velocity. This component of the acceleration acpoints toward

the inside of the curve.

(c) The total acceleration is the vector sum of the tangential and centripetalcomponents.

RELATIVE VELOCITY

The motion of any body has to be described relative to some frame of reference, such as

the ground. Sometimes it becomes necessary to study the motion of one body relative to

another body which is also moving relative to the ground. Let us study the relative motion

in general.

Consider the particle P which is being observedfrom two frames A and B. Its positions with

respect to frame A and B are rPA and rPB,

respectively.

The position vector triangle shown in the

Fig. (29) explains that

rPB = rPA + rAB (27)Position of P Position of P Position of A

w.r.t.B w.r.t. A w.r.t. BDifferentiating the above equation, we get

vPB= vPA+ vAB (28)

Fig.(29) A point P is located with respect to

two framesA andB

P

yB

yA

xA

xB

rAB

rPA

rPB

OA

OB

Note that equation (28) is a vector sum we worry about signs only when we take

components. Let us consider the case where a man walks across a train with velocity vo.

The ground forms a stationary frame (x,y) while the train forms a moving frame (x ,y ).We want to find the velocity of the man relative to the ground (v).Using equation (28)

v = vrel+ v0





23/29

An observer on the train will see the man walking along the x axis. An observer on the

ground will see him moving in a direction given by tan =o

rel

v

vto thexaxis.

Fig.(30)(a) The diagram shows the position of man (M) at three instants as he walks across a train

with velocity ov relative to the ground.(b) His velocity w.r.t. ground is shown by the velocity vector diagram.

orel vvv +=

O

x

y

vovrel

vrel

vrel

M

M

M

(a) (b)

v

vo

vrel

x

y

Example 16

A boat can travel at 10 m/s relative to the water. It starts at one bank of river that is 100 m

wide and flows with a velocity u = 7.5 m/s. If the boat points directly across, find

(a) Its velocity relative to the bank

(b) How far downstream it travels.

SolutionIn this problem the train is replaced by the

river and the man is replaced by the boat(a) From the velocity vector diagram, the

magnitude of the velocity of boat w.r.t.

ground is

v=

( ) smuvrel /5.125.7)10(2222 =+=+

The direction is indicated by the angle with theyaxis.

tan =4

3

10

5.7==

relv

u

b = 100 m

x

yu

vrel

Fig.(31 a) Example 16

or = tan1

(3/4) = 37(b) The distance travelled along the bank is

x = utwhere t is the time taken to cross the river.

t=relv

b

Thus,

x = mv

ub

rel

75)100(10

5.7=

=

vrel v

u

Fig. 31 b





24/29





25/29

Example 17

In which direction the boat must point so as to cross the river in a direction perpendicular

to river currents. Also, find out the time taken to cross the river. Use the information given

in the previousExample 16.

Solution

x

u

Fig.(32 ) Example 17

y

vrel

vrelv

u

To get directly across the river, the

boat must be pointed upstream

(a)(b)

From the velocity diagram

The magnitude of the velocity w.r.t. bank is

v = ( ) 61.6)5.7(10 2222

==uvrel m/s

The direction of motion of the boat w.r.t. the river current is represented by the angle measured anticlockwise from theyaxis.

tan = 13.161.65.7 ==

vu or = tan

1 (1.13) = 48.6

and time taken t = 13.1561.6

100 ==v

bs





26/29

Objective Solved Examples

1. A particle moves along thexaxis in such a way that its coordinate (x) varies with time (t)

according to the expression

x = 2 5t+ 6t2 m. The time tis in second.

The initial velocity of the particle is

(a) 5 m/s (b) 3 m/s

(c) 6 m/s (d) 3 m/s

Solution x = 2 5t+ 6t2

v =dt

dx= -5 + 12 t

Initial velocity means velocity at t= 0

vinitial = -5 m/s

(a)

2. In the arrangement shown in the

figure, the ends P and Q of an

unstretchable string move downward

with uniform speed u. PulleysA and

B are fixed. At the instant shown,the mass M moves upward with a

speed of

(a) 2u cos (b) u / cos (c) 2u / cos (d) u cos

A

M

B

P Q

Solution

In time t, ends P and Q move downwardthrough a distance u t and the mass Mmoves upward to M . During this time, thestrings AM and BMare reduced to AM and

BM andMC= MD = u t.

Now

u

MCMM'

coscos==

Speed of M=

u

t

MM'

cos=

(b)

A

M

P

utC D

M





27/29

3. A ball is released from a height h above the ground. It takes a time Tto reach the ground.

Where is the ball at the time T/2(a) at a height h/4 from the ground

(b) at a height h/2 from the ground

(c) at a height 3h/4 from the ground(d) none of these

Solution

h =2

2

1gT

s =422

12

hTg =

height from the ground = h -4

3

4

hh=

(c)

4. Two bodies, one held 30 cm directly above the other, are released simultaneously and fall

freely under gravity. After 2 s their separation will be

(a) 10 cm (b) 20 cm

(c) 30 cm (d) zero

SolutionMotion parameters i.e. the initial velocity and acceleration are same for both the bodies.

Therefore, they will fall by the same distance in a given time, maintaining their initial

separation.

(c)

5. A boat which has a speed of 5 km/hr in still water crosses a river of width 1 km along the

shortest possible path in 15 minutes. The velocity of the water in km/hr is

(a) 1 (b) 3

(c) 4 (d) 14

SolutionShortest possible path isAB.

the boat must move in the directionshown in the figure. For this, we must have

5 cos = minutes15km1

= 4 km/hr

and 5 sin = v v = 3 km/hr (b)

A

B

1 km

v

5





28/29

6. An object is dropped from rest. Its velocity versus displacement graph isv

(a) (b) (c) (d)s

v

s

v

s

v

s

SolutionFor free fall

v2 = u2 + 2gs

Here u = 0 so v2 = 2gs

graph between v ands will be a parabola symmetric abouts axis. (c)

7. A particle moves in a circle of radius 25 cm at 2 revolutions per second. The accelerationof the particle in m/s2 is

(a) 2 (b) 8 2(c) 4 2 (d) 2 2

Solution

The angular velocity =1

22 = 4 rad/s

a = 2R = (4 )2 (1/4) = 4 2 m/s2 (c)

8. A simple pendulum is oscillating without damping. When the displacement of the bob is

less than maximum, its acceleration vector a is correctly shown in:

(a)

a

(b)

a

(c)

a

(d)

a





29/29

SolutionAs displacement of the bob is less than

maximum, the acceleration will have

two components, radial and tangential.

So resultant will be inclined at someangle with string.

(c)

a

ta

ra

9. A particle starts from rest. Its

acceleration ( ) versus time (t) is asshown in the figure. The maximum

speed of the particle will be

(a) 110 m/s (b) 55 m/s

(c) 550 m/s (d) 660 m/s

10 m/s2

11 t(s)

SolutionThe area under the acceleration time graph gives change in velocity.

(b)

10. A small block slides without friction down an inclined plane starting from rest. Let Snbe

the distance travelled from time t= n - 1 to t= n. Then1+n

n

S

Sis

(a) n

n

2

12 (b) 12

12

+n

n

(c)12

12

+

n

n(d)

12

2

+nn

Solution

( )122

1+= tauSt

(c)


Al i i i d i


031 kinematics final

Documents