kinematics 2d

8/13/2019 Kinematics 2D

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Kinematics in Two-Dimensions

1. Vectors1.1. Vector Addition/Subtraction

1.2. Vector Multiplication:

Cross Product

Dot Product

2. Motion in 2-Dimensions

Projectil es, maximum height, time, range

3. Relative velocity


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Lesson Outcomes

Students should be able to:

1. define the componentsof displacement, velocityand

accelerationin both dimensions

2. define projecti le motion

3. derive the projectile equations of motion4. apply the projectile equations of motion to determine

the maximum heightand rangeand the total time of

motion

5. define the relative velocity


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Displacement, velocity, acceleration,momentumand forceare examples of vector

quantities.

Vectors

A vector quantity is any quantity with

magnitudeand direction.


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a

b

Lets say you have 2 vectors a & b


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ab

a + b

Adding Vectors

Triangle Method

a

b

a + bParallelogram Method


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Adding Vectors

a

Method 3: Resolving the vectors

ax= a cos

ay= a sin

2

y

2

x aaa

x

y

a

a1tan


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Adding VectorsMethod 3: Resolving the vectors

ax= a cos a

ay= a sin a

c = a + b cx= ax+ bxcy= ay+ by

x

y

2

y

2

x

c

ctan

ccc

1

c

bx= b cos bby= b sin b


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a

b

c = a + b

cx

ay

bx

cyax

ax bx

ay

by

by


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8 N + 6 N = ? 14 N? 2 N? 10 N?


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There are two type of vector multiplication:-

Vector Mul tipl ication

dot multiplication scalar

and

cross multiplication vector


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Scalar quantity

Dot multiplication: a b = abcos

a

b


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a b = a b cos 0= a b

Special cases:

Case 1:

= 0o

Case 2:

= 90o

a b = a b cos 90= 0a

b

a

b


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a b = a b sin 0= 0

Special cases:

Case 1:

= 0o

Case 2:

= 90o

a b = a b sin 90= a b

ab

a

b


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x

y

z

a

i

j

k

ax

ay

az

a =axi +

ayj +

azk

i i = 1 i j = 0j j = 1 j k = 0k k = 1 k i = 0i j = k i i = 0

j k = i j j = 0

k i = j k k = 0


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Displacement, velocity and acceleration

Displacement

0rrr

Average velocity

t

r

tt

rr

v0

0av

Instantaneous velocity

dt

dr

t

r

v 0t

lim


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Vector components of v are vxand vy

sinvv,cosvv yx and

Average acceleration

t

v

tt

vvaav

0

0

dt

dv

t

vlimat

0

Instantaneous acceleration

The acceleration has a vector components axand ayinx any-directions.


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ExampleA spacecraft has initial velocity component of

v0x= +22m/s and acceleration component ax = +24m/s2. In

they-direction it has v0y = +14m/s and ay = +12m/s2. The

direction to the right and upward have been chosen as

positive directions.

Find (a) x and vx, (b) y and vy and (c) the final velocity

(magnitude and direction) of the spacecraft at

time, t = 7.0 s.


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The spacecraft motion is twodimensional motion

X-part of the motion is independent of they-part. Similarlythey-part is independent ofx-part of the motion.


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Solution

mss/mss/m 74272421722 2

m/ss.s/ms/mtavv xxx 1900724220

2

1 20 tatvy yy

m/sss/ms/mtavv yyy 98712140

a)x and vx

b) y and vy

v0x= +22m/s ax = +24m/s2

v0y =+14m/s ay = +12m/s2

21 2

0 tatvx xx

mss/mss/m 3927122

1714

2


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The magnitude and the direction of spacecraft are

s/m)s/m()s/m(vvv yx 21098190

2222

27

190

98tanor 1-

x

y

v

vtan

+x


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Projectile Motion Projectile motion is a motion of object in the 2-D plane

under the inf luence of gravity, as shown in Fig. 2.

To analyze a projectile motion we need to consider the

components of the motion in the x- and y- directions

separately.

We note that the x-component of the acceleration is zero

(ax= 0), and they-component is constant and equal tog or

g, (ay= g), depending on whether we take upwards to be the

+ve y-direction orvey-direction, respectively.

Fig. 2: The trajectory of a body projected with an initial velocity v at an angle above the


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Fig. 2:The trajectory of a body projected with an initial velocity voat an angle above the

horizontal. The distance Ris the horizontal range, and his the maximum height to which

the particle rises.

vy= voy

vx= vox

vy= 0

R

h

O

vy

vx

vo

vx= vox

vy= - voy

vx

y

x

sinandcos 0y00x0 vvvv


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In projectile motion we can express all the vector

relationships in terms of separate equations for thehorizontal and vertical components. The components of

the acceleration are:

ax= 0 and ay = g

Therefore we can still directly use the previous

equations of motion with constant acceleration.

Projectile Motion

Th f i i i l l i


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x-direction y-direction

Acceleration

Velocity

Displacement

The components of initial velocity, v0are

and 0000 sinvvcosvv yx

xx vv 0 gtvv yy 0

tvxSxx 0

202

1gttvyS yy

ay = gax= 0


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Projectile Motion

Using the information given in the table above and theequations of motion, you can solve any problem dealing

with motion in a plane, provided you make an assumption

that there is no air resistance.

And always remember that there is no accelerationin the

x-direction.


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Time taken for a projectile to reach the

maximum height.

You can see that the object in the projectile changed its

direction from going upwards to coming downwards.

That clearly indicates that it reached a point whereby the

vertical velocity was zero.

The maximum height occurs when vyis equals to zero.

But the horizontal velocity remained constant.

P j til M ti


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Projectile Motion

Time taken to reach maximum height, when vy=0,

The equation to be used is

0 gtvv yy

g

v

g

v

t

y sin00

and vy=0

0 gtv y

M i h i ht f j til


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Maximum height of projectile

Maximum height when vy=0

The equation to be used is

ygvv yy 22

0

2

g

v

g

vhy

y

2

sin

2

22

0

2

0

and vy=0

Thus,

P j til T t l Fli ht ti


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Projectile Total Flight time

When the projectile reach the ground, the y-component of

the displacement is zero.

02

1 20 gttvy y

021

0

gtvt y 02

1or,0 0 gtvt y

g

v

g

vt

y sin22 00

Compare to time taken to reach maximum height

P j til T t l Fli ht ti


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Projectile Total Flight time

The total flight time will be 2t, i.e. twice time

taken to reach maximum height.

g

v22T 0

sin

t

The time taken to reach the maximum height is

the same as the time taken to reach the ground

after achieving the maximum height.

Note: only true IF the projectile return to same height

Examples of Projectile Motion


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Examples of Projectile Motion


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NOTE

You are advised not to memorize the general

equations above. Instead you must train

yourself to derive the equations to be used

whenever it is necessary.

Example 1


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Example 1A handball is thrown with an initial vertical velocity component

of 18.0 m/s and a horizontal velocity component of 25.0 m/s.

(a) How much time is required for the handball to reach the

highest point of the trajectory?

(b) How high is this point?

(c) How much time (after being thrown) is required for thehandball to return to its original level? How does this compare

with the time calculated in part (a)?

(d) How far has it traveled horizontally during this time?

Solution to Example 1


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Solution to Example 1

a) The time required to reach the maximum height is obtained from

s81m/s819

m/s1800Thus,

2

0.

.g

vt

y

m516m/s8192-

m/s)(18-0

2

0yThus

2

20

..g

v y

s635

18Thus518

2

10 220 .tttgttvy y

m.90s63m/s250 .tvxR x

b) At max height, vy= 0,

c) The time taken to reach its original level y=0,

(which is twice with that calculated in (a).

d) It traveled horizontally during this time a distance of

00

,gtvvyy

0220

2,gyvv y

Projectile Range


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Projectile Range Range, R, is actually x-displacement just as the maximum

height is the vertical displacement.

The equation used for calculating the range is

tvRxxx x00

g

v

g

vvR

2sinsin2cos

2

00

0

cossin22sin

Range R, here represents the maximum horizontal

displacement, the time will be the total flight time 2t.

Using the trigonometric identity

Which launch angle 30 45 and 60 gives


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Which launch angle, 30, 45 and 60 gives

greatest range?

This equation showsRvaries with angle as sin2.

2sin2

0

g

vR

g

v

R

2

0

max

ThusRis largest when sin 2is largest, that is when

sin 2=1. Since sin 90 = 1, its follows that 2= 90 , thus =45

gives the maximum range.

Projectile Motion


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Projectile Motion

The distance rof the projectile from the origin at any time (the

magnitude of the position vector r) is given by:

The projectiles speed at any time is

The direction of the velocity is given by

The velocity vector vis tangent to the trajectory at each point.

22 yxr 2y

2x vvv

x

y

v

vtan

C t l Q ti


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Conceptual Question

A wrench is accidentally dropped from the top ofthe mast on a sailboat. Will the wrench hit at the

same place on the deck whether the sailboat is at

the rest or moving with a constant velocity? Justifyyour answer.

REASONING AND SOLUTION


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The wrench will hit at the same place on the deck of the ship

regardless of whether the sailboat is at rest or moving with aconstant velocity.

If the sailboat is at rest, the wrench will fall straight downhitting the deck at some pointP.

If the sailboat is moving with a constant velocity, the motion

of the wrench will be two dimensions. However, thehorizontal component of the velocity of the wrench will be thesame as the velocity of the sailboat.

Therefore, the wrench will always remain above the samepointPas it is falling.


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Discussion

Suppose you are driving in convertible with the topdown. The car is moving to the right at a constant

velocity. You point a gun straight upward and fire it. In

the absence of air resistance, where would the bullet

land- behind you, ahead of you, or in the barrel of thegun?

Conceptual Question


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Conceptual Question

A stone is thrown horizontally from the top of a cliffand eventually hits the ground below. A second stoneis dropped from rest from the same cliff, fallsthrough the same height, and also hits the groundbelow. Ignore air resistance. Discuss whether each ofthe following quantities is different or the same inthe two cases; if there is difference, describe thedifference: (a) displacement, (b) speed just beforeimpact with the ground and (c ) time of flight.



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a) The displacement is greater for the stone that is thrown

horizontally, because it has the same vertical component as the

dropped stone and, in addition, has a horizontal component.b) The impact speed is greater for the stone that is thrown

horizontally. The reason is that it has the same vertical

velocity component as the dropped stone but, in addition, also

has a horizontal component that equals the throwing velocity.c) The time of flight is the same in each case, because the vertical

part of the motion for each stone is the same. That is, each

stone has an initial vertical velocity component of zero and

falls through the same height.

Conceptual Example

Two youngsters dive off an overhang into a lake. Diver 1 drops straight down,


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y g g p g ,

diver 2 runs off the cliff with an initial horizontal speed v0. Is the splashdown

speed of diver 2 (a) greater than,

(b) less than, or (c) equal to the splashdown speed of diver 1?

Reasoning and Discussion

Note that neither diver has an initial y component of velocity, and that they both


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fall with the same vertical acceleration the acceleration of gravity. Therefore, the

two divers fall for the same amount of time, and their y components of velocity are

the same at splashdown. Since diver 2 also has a nonzero x component of

velocity, unlike diver 1, the speed of diver 2 is greater.

Answer:(a) The splashdown speed of diver 2 is greater than that of

diver 1.

P bl


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Problem:

A plane travelling horizontally at 115 m/sis to drop a package from a height of

1050 m above the ground.

How far before the plane is above thetarget point should the package bedropped?

The package falling from the plane is a projectile motion.


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Determine the range of the package (the x-displacement)before it reaches the ground.Ans: 1683 m

Exercise


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A projectile is shot from the edge of a high cliff with initial velocity 25 m/s at an angle of

30above the horizontal. It reaches the ground after 3.5 seconds.

a) What is the height of the cliff?

b) How far from the edge of the cliff does the projectile land?

c) Determine the velocity of the projectile as it strikes the ground.

Relative Velocity


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Relative Velocity

Relative velocityis the velocity of an object relative to the

observer who is making the measurement.

The velocity of object A relative to object B is written vAB, and

velocity of object B relative to C is written as vBC.The velocity

of A relative C is (note the ordering of subscript)

While the velocity of object A relative to object B is vAB, the

velocity of B relative A is vBA= vAB

BCABAC vvv

Example


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TGPTPG VVV Vector sum of the two- velocity-vectors.

VPT= velocity of the Passengerrelative to the Train.

VTG= velocity of the Train relative to the Ground.

VPG

= velocity of the Passengerrelative to the Ground.

Relative Velocity


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y

VPT= velocity of the Passengerrelative to the Train.

VTG= velocity of the Train relative to the Ground.

VPG= velocity of the Passengerrelative to the Ground. Each velocity symbol contains two-letter subscript, the

1st for the moving body, the 2nd indicates the objective

relative to it the velocity is measured.

TGPTPG VVV

Example 1


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On a pleasure cruise a boat is traveling to the water at

a speed of 5.0 m/s due south. Relative to the boat, apassenger walks toward the back of the boat at a speed

of 1.5 m/s. (a) What is the magnitude and direction of

the passengersvelocity relative to the water? (b) How

long does it take for the passenger to walk a distance of27 m on the boat? (c) How long does it take for the

passenger to cover a distance of 27 m on the water?

REASONING


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The time it takes for the passenger to walk the distance on the

boat is the distance divided by the passengers speed vPBrelative to the boat.

The time it takes for the passenger to cover the distance on the

water is the distance divided by the passengers speed vPW

relative to the water. The passengers velocity relative to the boat is given.

However, we need to determine the passengers velocity

relative to the water.

ANSWERS


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b) The time it takes for the passenger to walk a distance of 27 m

on the boat iss18

m/s5.1

m27m27

PBvt

PW

27 m 27 m7.7 s

3.5 m/s

t

v

BWPBPW vvv a) passengers velocity relative to the water, vpw

c) The time it takes for the passenger to cover a distance of 27 m

on the water is

soutm/s,3.5south5.0m/s,northm/s,51 .

Example 2


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p

The engine of a boat drives it across a river that is

1800 m wide. The velocity vBW of the boat relative to

the water is 4.0 m/s, directed perpendicular to the

current, as shown in the Fig. The velocity vws of the

water relative to the shore is 2.0 m/s. a) What is the

velocity vBS of the boat relative to the shore? b) How

long does it take for the boat to cross the river?

WSBwBS vvv


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m/s5.4m/s0.2m/s0.4 2222 WSBwBS vvv

m/s04.vBW

m/s02.vWS

632tanThus

02

04

1-

,.

.tan

a)

s450/04

m1800Widthtime )b

shore


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m/s04vBW .

s450m/s54

m0202

v

traveledDistancetime

BS

.

)b

1800 m

shore

shore

63

d

m202063sin

1800

63sin1800

d

d

kinematics 2d

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