By Abdul Khan
Period 2
 Change
in position;
 Can be positive or negative, depending
on direction;
 ∆x = Xf – Xi change in position equals
final position minus initial position;
 Displacement and distance traveled are
not the same thing;
∆x = f-i
:-10-0 = -10
f= -10
∆x=f-i
:10-0=0
i=0
 Average Velocity
= Change in
position/change in direction;
 V-avg = ∆x/ ∆t = (xf-xi)/(tf-ti);
 Velocity is not the same as speed,
velocity has direction;
 Average velocity and instantaneous
velocity are not the same thing either.
 Put
time on the x-axis;
 Put displacement on the y-axis;
 The slope of the line is ∆x/ ∆t so the
slope is velocity:

x
∆x


∆t
y
 Acceleration
is the rate of change in
velocity with respect to time
 Aavg = ∆v/∆t = (v -v )/(t -t )
 Notice how this form looks similar to that
of velocity (∆x/∆t)
 Just as the slope of x vs. t is velocity, the
slope of v vs. t is acceleration.
f
i
f
i
d
= displacement (∆x)
 t = time of travel (∆t)
 a = rate of constant acceleration
 v = initial velocity
 v = final velocity
i
f
ā
= ∆v / ∆t = (Vf-Vi)/(Tf-Ti);
 a/1=(Vf-Vi)/t;
 at=Vf-Vi;
 atVvi=Vf;
 Vf=Vi+at
(equation # 1)
v
= ∆x/ ∆t;
 ∆x=d, ∆t=t, V =1/2(vi+vf);
 ½(Vi+Vf)=d/t;
 t/2(Vi+Vf)=d or ½((Vi+Vf)t (equation #
2).
Vf=Vi+at #1 into ½((Vi+Vf)t
d = ½(Vi + at + Vi)t
d = ½(2Vi + at)t
d = (Vi + ½at)t (equation # 3)
 Vf= Vi
+at;
 Vf-Vi=at;
 (Vf-Vi)/a=t. (i)
 D=1/2(Vi+Vf)/[(Vf-Vi)/a];
 D=[(Vi+Vf)(Vf-Vi)]2a;
 D=(VfVi+Vf^2-Vi^2VfVi)/2a
 D=(Vf^2-Vi^2)/2a
 2ad=Vf^2-Vi^2
 Vf^2=2ad+Vi^2 (equation # 4).