Motion

Group III

GE2-1

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

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

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

where v= speed

s= distance

t= time

Velocity ( v )- displacement per unit timewhere: v = velocity

s = displacement

t = time

• Acceleration – time rate of change of velocity

if

Where: a= acceleration

t= time

v0= Initial velocity

vi= final velocity

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

Equations

1.2.3.4.

5.

V=average velocity/speed

Sample problems:

1.An airplane taking off from a landing field has a run of 1200ft. If it starts from rest, moves with a constant acceleration and makes he run in 30s, with what velocity in m/s did it take off?

Illustration:

1ft=0.3048m

(1)

From we can derive

(2) From we can derive

2. A car comes to a stop in 6s from a velocity of 30m/s.

a.) what is its acceleration?b.) at the same acceleration,

how long would it take the car to come to a stop from a velocity of 40 m/s?

Given: Vo=30m/s

t=6sVI=0

Sol’n:A.)

B.)

3. a runner “A” can run the mile race in 4.25min. Another runner “B” requires 4.55mins to finish this distance. If they start out together and maintain their normal speed, how far apart will they be at the end of the race?

Solution:

Freely falling body- a body which is acted on by no force of appreciable magnitude other than its weight.system

sS T V G G

Mks m s m/s m/s2 9.8m/s2

Cgs cm s cm/s cm/s2 980cm/s2

Fps ft s ft/s ft/s2 32ft/s2

equations

1.2.3.4.5.

Cases object dropped from a height

An object thrown vertically downward

S max g=t

S max g=t

An object thrown vertically upward

S max +up

0

+down

g=- g =+

+up=+down

Sample Problem

A stone is dropped from the edge of a cliffa.) what is its velocity 3s later

b.) how far does it fall in this time?c.) how far will it fall in the next second?

B.

A.

C.

A girl throws the ball 60ft vertically into the aira. how long does she has to wait to catch it on the way down?b. what is its initial velocityc. what is its final velocity

given:

solution

Projectile MotionProjectile motion is an object which is given an

initial velocity and after some time allowed to move under the action of gravity

S maxg=- g=+

t up t down

(R)

Trajectory- path followed by projectileRange- horizontal distancet'= total time of flightt'=2 t up -2 t down

Note: horizontal comp. of velocity is constant

Sample problemA canon is elevated at an angle of 45˚. It fires a

ball with a speed of 300m/sA.) what height does the ball reach?B.) how long is the ball in the air?C.) what is the range?

a.) S max (S)b.) t'=?c.) range (R)Solution:

a.0

b.

.

c.

A stone is thrown from a window with an initial horizontal velocity of 10m/s. if the window is 20m high, and the ground is level,a. in how many seconds will the stone reach the ground?b. how far in the ground will it reach?

Required:a. t=?b. R=?Solution:a. b.

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.

System m(mass) a(acceleration)

F(force) w (weight)

Mks kg m/s2 N N

Cgs G cm/s2 dyne dyne

Fps Slug ft/s2 lb lb

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

Group III

Angelo James AmbionCamille PonienteJunelle Christian ObiceJumel HernandezJohn Pierre PereñaMarvin FeranilMarvin MaañoRyan SanchezWilfredo Peña