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 & VECTORS SCALARVECTOR Quantity that only has magnitude Does not have direction Example: mass, time, volume, area, temperature, distance, speed Quantity with magnitude and direction A or arrow notation, Commonly represent by boldface notation, A or arrow notation, Example: velocity, force, acceleration, momentum

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)

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 F1F1 F2F2 F x1 F y1 F x2 F y2 x x F = F 1 + F 2 F x = F x1 + F x2 F x2 F y = F y1 + F y2 F y1 F y2 F x1

(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 45 o 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, 45 o below the + x- axis (4 th quadrant) B x = (2.5m)x B y = (4.0m)y Find C such that A + B + C = D = (+6.0m) y x - componentsy - components A x = A cos 45 o = +4.24mA y = - A sin 45 o = m B x = + 2.5mB y = + 4.0m C x = ?C y = ? D x = 0D x = +6.0m

Calculate x – and y – components separately: x-components: Ax + Bx + Cx = Dx 4.24m + 2.5m + Cx = 0 ∴ Cx = m y-components: Ay + By + Cy = Dy m + 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:

EXAMPLE 2 For the vector shown in Figure above determine;

DISTANCEDISPLACEMENT 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 & VELOCITY SPEEDVELOCITY  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

EXAMPLE 3 A jogger jogs from one end to the other of a straight 300m track in 2.50 min and then jogs back to the starting point in 3.30 min. What was the jogger’s average 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

SOLUTION Given : Δx 1 = 300mΔt 1 = 2.50 min x 60 s = 150 s Δx 2 = -300m Δt 2 = 3.30 min x 60 s = 198 s a) b) c)

ACCELERATION - rate of change of velocity. SI Unit: meters per second squared (m/s2).

EXAMPLE 4 A couple of sport-utility vehicle (SUV) are traveling at 110km/h on a PLUS highway. The drives sees an 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 = m/s v 0 = 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 – v 0 )/t = (15.3m/s – 30.5 m/s)/10s = -15.2m/s 2  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 V o 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 at 2 v 2 = u 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;

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 car b)Time taken for the speed to decrease from 12m/s to 8m/s c)Time taken for the car to come to a halt from the speed of 12m/s d)Total distance travelled by the car during this time.