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Kinematics in Two-Dimensions

1. Vectors

1.1. Vector Addition/Subtraction

1.2. Vector Multiplication: Cross Product

• Dot Product

2. Motion in 2-Dimensions

• Projectiles, maximum height, time, range

3. Relative velocity

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Lesson ObjectivesStudents should be able to:1. define the components of displacement, velocity and

acceleration in both dimensions2. define projectile motion3. derive the projectile equations of motion4. apply the projectile equations of motion to determine the

maximum height and range and the total time of motion5. define the relative velocity

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• Displacement, velocity, acceleration, momentum and force are examples of vector quantities.

Vectors

• A vector quantity is any quantity with magnitude and direction.

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a

b

Let’s say you have 2 vectors a & b

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a b

a + b

Adding Vectors

Triangle Method

a

b

a + bParallelogram Method

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Subtracting Vectors

a b

a - b = a + (-b)

- b

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Adding Vectors

a

Method 3: Resolving the vectors

ax = a cos

ay = a sin

2y

2x aaa

x

y

a

a1tanθ

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Adding VectorsMethod 3: Resolving the vectors

ax = a cos a

ay = a sin a

c = a + b cx = ax + bx

cy = ay + by

x

y

2y

2x

c

ctan

ccc

1c

bx = b cos b

by = 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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There are two type of vector multiplication:-

Vector Multiplication

• dot multiplication scalar

and

× cross multiplication vector

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Scalar quantity

Dot multiplication: a b = ab cos

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 = 0

ab

a

b

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Vector quantity

Cross multiplication: a b = ab sin

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

ai

j

k

ax

ay

az

a = ax i + ay j + az k

i i = 1 i j = 0

j j = 1 j k = 0

k k = 1 k i = 0

i 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

rrv

0

0av

Instantaneous velocity

dt

dr

t

rv

0t

lim

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Vector components of v are vx and vy

sin and ,cos vvvv yx

Average acceleration

t

v

tt

vva

0

0av

dt

dv

t

va

0t

lim

Instantaneous acceleration

The acceleration has a vector components ax and ay in x an y-directions.

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Example

A spacecraft has initial velocity component of

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

In the y-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 two–dimensional motion

X-part of the motion is independent of the y-part. Similarly the y-part is independent of x-part of the motion.

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Solution

m/s190s07sm24sm22tavv xx0x .//

m/s98s7sm12sm14tavv yy0y //

a) x and vx

b) y and vy

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

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

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The magnitude and the direction of spacecraft are

sm210sm98sm190vvv 222y

2x /)/()/(

27190

98tanor tan 1-

x

y

v

v

+x

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

under the influence 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 the y-component is constant and equal to – g or g, (ay= g), depending on whether we take upwards to be the +ve y-direction or –ve y-direction, respectively.

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Fig. 2: The trajectory of a body projected with an initial velocity vo at an angle above the horizontal. The distance R is the horizontal range, and h is 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

sin and cos 0y00x0 vvvv

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In projectile motion we can express all the vector relationships in terms of separate equations for the horizontal 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

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x-direction y-direction

Acceleration

Velocity

Displacement

The components of initial velocity, v0 are

θandθ 0000 sinv v cosvv yx

xx vv 0 gtvv yy 0

tvx x0 20 2

1gttvy y

ay = -gax = 0

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Projectile Motion

• Using the information given in the table above and the equations 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 acceleration in the x-direction.

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Time taken for a projectile to reach the maximum height.

• The maximum height occurs when vy is equals to zero.

• 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.

• But the horizontal velocity remained constant.

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Velocity Vectors for a Projectile Launched at the Origin

h

vv

v

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Projectile Motion• Maximum height when vy=0

• The equation to be used is

yg2vv 2y0

2y

g2

v

g2

vhy

220

2y0

sin

and vy=0

Thus,

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Projectile Motion

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

• The equation to be used is

gtvv y0y

g

v

g

vt 0y0

sin

and vy=0

gtv y0

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Projectile Total Flight time

• The time taken to reach the maximum height is the same as the time taken to reach the ground after achieving the maximum height.

• The total flight time will be 2t.

g

v22T 0

sin

t

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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.

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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 the handball 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?

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Solution to Example 1a) The time required to reach the maximum height is obtained from

s 81m/s 819

m/s 1800 Thus,

2

0 ..g

vt y

m 516m/s 8192-

m/s) (18-0

2

0y Thus

2

20 .

.g

v y

s 635

18 Thus 518

2

10 22

0 .tttgttvy y

m. 90s 63m/s 250 .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 ,gtvv yy

0220

2 ,gyvv y

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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

2v

g

v2vR

200

0

sinsincos

cossin22sin

• Range R, here represents the maximum horizontal displacement, the time will be the total flight time 2t.

• Using the trigonometric identity

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Which launch angle, 30, 45 and 60 gives greatest range?

• This equation shows R varies with angle as sin2.

2g

vR

20 sin

g

vR

20max

• Thus R is largest when sin 2 is largest, that is when sin 2=1.

• Since sin 90 = 1, its follows that 2 = 90 , thus =45 gives the maximum range.

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Projectile MotionThe distance r of 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 v is tangent to the trajectory at each point.

22 yxr 2y

2x vvv

x

y

v

vtan

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Conceptual Question

• A wrench is accidentally dropped from the top of the 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? Justify your answer.

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REASONING AND SOLUTION

• 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 a constant velocity.

• If the sailboat is at rest, the wrench will fall straight down hitting the deck at some point P.

• If the sailboat is moving with a constant velocity, the motion of the wrench will be two dimensions. However, the horizontal component of the velocity of the wrench will be the same as the velocity of the sailboat.

• Therefore, the wrench will always remain above the same point P as it is falling.

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Discussion

• Suppose you are driving in convertible with the top down. 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 the gun?

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Conceptual Question

• A stone is thrown horizontally from the top of a cliff and eventually hits the ground below. A second stone is dropped from rest from the same cliff, falls through the same height, and also hits the ground below. Ignore air resistance. Discuss whether each of the following quantities is different or the same in the two cases; if there is difference, describe the difference: (a) displacement, (b) speed just before impact with the ground and (c ) time of flight.

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REASONING AND SOLUTION

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.

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Conceptual Example

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The package falling from the plane is a projectile motion

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Relative Velocity

• Relative velocity is 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

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

VPT= velocity of the Passenger relative to the Train.

VTG= velocity of the Train relative to the Ground.

VPG= velocity of the Passenger relative to the Ground.

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Relative Velocity

• VPT= velocity of the Passenger relative to the Train.

• VTG= velocity of the Train relative to the Ground.

• VPG= velocity of the Passenger relative 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

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Example 1

• 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, a passenger walks toward the back of the boat at a speed of 1.5 m/s. (a) What is the magnitude and direction of the passenger’s velocity relative to the water? (b) How long does it take for the passenger to walk a distance of 27 m on the boat? (c) How long does it take for the passenger to cover a distance of 27 m on the water?

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

boat is the distance divided by the passenger’s speed vPB relative to the boat.

• The time it takes for the passenger to cover the distance on the water is the distance divided by the passenger’s speed vPW relative to the water.

• The passenger’s velocity relative to the boat is given. However, we need to determine the passenger’s velocity relative to the water.

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

s 18m/s 5.1

m 27m 27

PBvt

PW

27 m 27 m7.7 s

3.5 m/st

v

south m/s, 3.5south 5.0m/s,north m/s, 51

.

vvv BWPBPW

ANSWERSa) passenger’s 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

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Example 2

• 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?

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WSBwBS vvv

m/s 5.4m/s 0.2m/s 0.4 2222 WSBwBS vvv

m/s 0.4sin BWBS vv

m/s 0.2cos WSBS vv

632tan Thus 02

04 1- ,.

.tan

s 450m/s 0.4

m 1800

sinv

Width time)

BS

b

a)

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Summary

X-Direction Y-Direction

Acceleration

Velocity

Displacement

The components of initial velocity, v0 are θsinv vθ cosvv yx 0000 and

X-Direction Y-Direction

Acceleration

Velocity

Displacement

xx vv 0 gtvv yy 0

tvx x0 20 2

1gttvy y

ay = -gax = 0

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0rrr

r

r0

r0

r +( r0 )