Motion

Distance(d)- simply pertains to length (scalar) e.g. 5m, 5km

Displacement (d)- change in position specified by magnitude + direction (vector) e.g. 5m, north , 5m, 20˚ S of E.

Speed (v) – distance per unit time (scalar)Velocity (v) – displacement per unit time

(vector)-ve velocity is in the opposite direction

v= speed/velocity

d= distance/displacement

t= time

• Acceleration – time rate of change of velocity

if

Where: a= acceleration

t= time

vi= Initial velocity

vf= final velocity

Uniformly Accelerated Motion: the motion in a straight line in which the rate changes uniformly

𝑎=𝑣 𝑓 −𝑣 𝑖

𝑡

Equations

1.

NEWTON’S LAWS OF MOTION

First law: law of Inertiaa body at rest remains at rest and a body in motion continues to move in straight line at constant speed, unless an external unbalanced force acts on it.

A B

A=BExternal forces No movement

rotateA=B

Unbalanceforce

Second law: Law of Accelerationan external unbalance force acting on an object produces an acceleration in the direction of the net force, an acceleration that is directly proportional to the unbalanced force and inversely proportional to the mass of the body

Where:F=forcem=massa= acceleration

Where:w = weightm = massg = gravity

Third law: Interactionfor every force that a first body exerts upon second body, there is a force equal in magnitude but opposite in direction that a second body exerts upon the first body.

A 100N box is sliding down a frictionless plane inclined at an angle of 30° from the horizontal. Find the acceleration of the box.

a=?

Note: N is always ┴ to the plane

30°

N

F

W

FBD- free bodydiagram- showsall the forces actingon the object

Solution:

UNIFORM CIRCULAR MOTION

- motion of an object in a curved or circular motion

10 rev. in 5s

Frequency, F = the no. of revolution per unit time

Period, T = the time required for one complete revolution.

Where:v= linear velocityr= radius

Radian- angle subtended by the arc of a circle whose length is equal to theradius of the same circle.

r

x

Where: v= linear velocity

r= radius ω =angular velocity

Where: ω = angular velocity = angle turned throught = time elapse

Using linear velocity

if

Using angular velocity

Centripetal force

SAMPLE PROBLEM

If the radius of the circular path of the stone is 0.5m and its period is 0.5s. What is its constant speed?

Given: r= 0.5mT= 0.5s

Required : v=?Solution:

What is the angular velocity of a stone which makes 10 rev in 5 seconds? The radius of the circular path is 0.5m

Given: no. of revolution = 10T= 5sr=0.5mRequired: ω=?Solution:

A mass of 0.5kg is whirled in a horizontal circle of radius 2m. If it makes 5 rev in 5s

Find: a. speedb. accelerationc. centripetal force

Given: m=0.5 kgr=2mt=5srev=5T=1s

Solution:

End of presentation