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CHAPTER11
Electindo.comPHYSICS FOR JUNIOR HIGH SCHOOL VII
STRAIGHT MOTION AT CONSTANT ACCELERATION (SMACA)
Electindo.comPHYSICS FOR JUNIOR HIGH SCHOOL VII
STRAIGHT MOTION AT CONSTANT ACCELERATION
DEFINITIONSMACA is a motion that has rectilinear track with constant acceleration and the velocity changes regularly. The example of SMACA can be found at the fruit falling down from its branch, or the motion of object throwing up.
OBJECT MOVES WITH INCONSTANT VELOCITY
ACCELARATED ( a + ) DECELERATED ( a - )
Electindo.com
PHYSICS FOR JUNIOR HIGH SCHOOL VII
GRAPHIC OF SMACAaccelerated ( a + )
space (s) velocity (v) acceleration (a)
Space (s) – Time (t) Velocity (v) – Time (t) Acceleration (a) – Time (t)
X = Vo.t + ½ at2
V = Vo + at
a = (V/Vo) : t
Vo2 = V2+ 2a.s
Electindo.com
PHYSICS FOR JUNIOR HIGH SCHOOL VII
GRAPHIC OF SMACA decelerated ( a — )
Jarak (s) kecepatan (v) Percepatan (a)
Jarak (s) – waktu (t) kecepatan (v) – waktu (t) percepatan (a) – waktu (t)
X = Vo.t - ½ at2
V = Vo - at
a = (V/Vo) : t
V2 = Vo2- 2a.s
Electindo.com
PHYSICS FOR JUNIOR HIGH SCHOOL VII
STRAIGHT MOTION AT CONSTANT ACCELERATION
Ball acceleration when leaving the player = -g.
Velocity in maximum-high V = 0
Answer :
t = (V-Vo)/g= (0 - 12) / (-9,8) = 1.2 s
V = Vo + gt
Time required to reach the maximum-high :
Maximum-high :
1. A BASEBALL PLAYER THROWS A BALL THROUGH Y AXIS WITH STARTING VELOCITY OF 12 M/S. HOW LONG THE TIME SPENT BY THE BALL TO REACH MAXIMUM-HIGH, AND HOW HIGH THE BALL WILL BE ?
( )( ) m3,7=
m/s 9.8-2m/s 12-0
=a2v-v
= y 2
22o
Y=0
Y = 7,3 m
Electindo.com
PHYSICS FOR JUNIOR HIGH SCHOOL VII
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