motion. equations n ewton ’ s laws of motion first law: law of inertia a body at rest remains at...
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Motion
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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
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• 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
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Equations
1.
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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
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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
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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.
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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
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Solution:
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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
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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
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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
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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:
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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:
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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
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Solution:
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End of presentation