OBJECTIVES• Ability to understand and define scalar and vector

quantity.• Ability to understand the concept of vector

addition, subtraction & components and applying the analytical component method.

• Ability to understand and distinguish between speed, velocity and acceleration

• Ability to apply motion equation based on physical situations.

• Ability to understand the Newton’s Law and its application.

SUBTOPICS

• Scalars & Vectors• Speed, Velocity & Acceleration• Motion Equation• Newton’s Law• Force From Newton’s Law

SCALARS & VECTORSSCALAR VECTOR

• Quantity that only has magnitude• Does not have direction• Example: mass, time, volume, area, temperature, distance, speed

• Quantity with magnitude and direction• Commonly represent by boldface notation, A or arrow notation, • Example: velocity, force, acceleration, momentum

A

VECTORS

2 vectors are the same if :

(a) magnitude a = magnitude b |a| = |b|

(b) a and b parallel or same direction

VECTOR ADDITION

R = A + B ?

1)

2)

3)

ABBAC

the sum of vectors is independent of the order in which the vectors are added, as long as we maintain their length and direction

VECTORS SUBSTRACTION

R = A – B = A + (–B)The –ve of a vector is represented by an arrow of the same length as the original vector, but pointing in the opposite direction

VECTOR COMPONENTS

Vector components

Phase

Magnitudes of components

• component of a vector is the influence of that vector in a given direction• component method: Most widely used analytical method for adding multiple vectors

UNIT VECTOR•Unit vector has a magnitude of unity, or one, and thereby simply indicates a vector’s direction.

VECTORS ADDITION BY COMPONENTS

• resolve the vectors into rectangular vector components and adding the components for each axis independently

y y

F1

F2

Fx1

Fy1

Fx2

Fy2

x x

F = F1 + F2

F = F1 + F2

Fx = Fx1 + Fx2

Fx2

Fy = Fy1 + Fy2

Fy1

Fy2

Fx1

(a) Resolve the vectors into their x- and y-components. (b) Add all of the x-components and all of the y-components together vectorally to obtain the x- and y-components Cx and Cy respectively

EXAMPLE 1

You are given two displacement vectors: 1) A with magnitude of 6.0m in the direction of 45o below the + x-axis, and 2) B, which has an x – component of +2.5m and a y-component of +4.0m.

Find a vector C so that A + B + C equals a vector D that has magnitude of 6.0m in the + y-direction.

SOLUTION• A = 6.0m, 45o below the + x-

axis (4th quadrant)• Bx = (2.5m)x

• By = (4.0m)y• Find C such that A + B + C = D =

(+6.0m) y

x - components y - components

Ax = A cos 45o = +4.24m Ay = - A sin 45o = - 4.24m

Bx = + 2.5m By = + 4.0m

Cx = ? Cy = ?

Dx = 0 Dx = +6.0m

• Calculate x – and y – components separately: x-components:

Ax + Bx + Cx = Dx4.24m + 2.5m + Cx = 0

∴Cx = - 6.74m

y-components: Ay + By + Cy = Dy

- 4.24m + 4.0m + Cy = 6.0m ∴ Cy = +6.24m

• So, C = (-6.74m) x + (6.24m) y• We may also express the results in magnitude-angle form: Magnitude:

Phase:

2.4724.6

74.6tantan 11

m

m

C

C

x

y

mmmCCC yx 18.9)24.6()74.6( 2222

EXAMPLE 2

For the vector shown in Figure above determine;

CBAandCBA

DISTANCE DISPLACEMENT

• scalar quantity• Total path length traversed in moving from one location to another• Only +ve value

• vector quantity• straight line distance between 2 points along with the direction from the starting point to another• Can have +ve or –ve values (indicate the direction)

DISTANCE & DISPLACEMENT

SPEED & VELOCITYSPEED VELOCITY

refers to how fast an object is moving scalar quantity defined as the rate of motion, or rate of change in position.fast moving = high speed, slow moving = slow speed zero speed = no movement

Refers to how fast something is moving and in which direction it is moving. vector quantity defined as the rate of change of displacement or the rate of displacement. velocity = 0, does not mean the object is not moving.

SPEED

SI Unit: m/s

VELOCITY

SI Unit: m/s

12 tt

d

t

ds

distance that travel to timetotal

traveleddistances speed, average

12

12

t time, traveltotal

x nt,displacemev velocity,average

tt

xx

EXAMPLE 3

A jogger jogs from one end to the other of a straight300m track in 2.50 min and then jogs back to thestarting point in 3.30 min. What was the jogger’saverage velocity (a) in jogging to the far end of the track(b) coming back to the starting point, and(c) for total jog

300m

2.5 minutes

3.3 minutes

SOLUTIONGiven : Δx1= 300m Δt1= 2.50 min x 60 s = 150 s

Δx2 = -300m Δt2 = 3.30 min x 60 s = 198 s

a)

b)

c)

sms

m

t

x/2

150

300v velocity,average

1

11

sms

m

t

x/51.1

198

300v velocity,average

2

22

smss

mm

t

x/0

198150

)300(300v velocity,average

3

33

ACCELERATION

0

0

12

12

change themake totime

yin velocit change on,accelerati average

tt

vv

tt

vvt

va

a

- rate of change of velocity.

SI Unit: meters per second squared (m/s2).

EXAMPLE 4

A couple of sport-utility vehicle (SUV) are travelingat 110km/h on a PLUS highway. The drives seesan accident in the distance and slows down to 55km/h in 10s. What is the average acceleration of the SUV?

SOLUTION• Change velocities to SI unit.

1km/h = 0.278 m/s

• v0 = 110kmh-1 x (0.278ms-1/1kmh-1) = 30.5m/s

v = 55kmh-1 x (0.278ms-1/1kmh-1) = 15.3m/s

t = 10s

Therefore, average acceleration:a = (v – v0)/t

= (15.3m/s – 30.5 m/s)/10s = -15.2m/s2 decelaration

• Equation that describe the behavior of system (e.g the motion of a particle under an influence of a force) as a function of time

• Sometimes the term refers to the differential equations that the system satisfies and sometimes to the solutions to those equations.

MOTION EQUATION

When an object moves along the straight line and velocity increase uniformly from Vo to v in time t.

constant acceleration: a = change in velocity / time taken

= (v-u )/ t v = u+at

Motion With Constant Acceleration

derivation of motion equation:

• v = u + at

• v= ½(u+v)t

• s = ut + ½ at2

• v2 = u2 + 2 as

FREE FALL

• Objects in motion solely under the influence of gravity.

• Expressing a=-g in the kinematics equation for constant acceleration in the y-direction yields the following;

ov v gt 21

o o 2y y v t gt 2 2

o o2 ( )v v g y y 2 2

o o2 ( )v v g y y

EXERCISE• The speed of a car travelling along a straight

road decreases uniformly from 12m/s to 8 m/s over 88 m. Calculate the

a)Decelaration of the carb)Time taken for the speed to decrease from

12m/s to 8m/sc) Time taken for the car to come to a halt

from the speed of 12m/sd)Total distance travelled by the car during

this time.