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An object starts at position xi and travels to position xf
in a time interval t
xi
xf
ti tf
We represent this motion with a
position-time graph, with position on the
vertical axis and time on the horizontal.
Define the object’s average velocity during
the interval t: f ix x
vt
It should be clear that this average
velocity is also the slope of the object’s position-time graph.
xi
xf
ti tf
Note that average velocity over an interval is signed
f ix xv
t
Positive velocity is going forwards. xi
xf
ti tf
And the sign of velocity is relative to the position
coordinate systemf ix x
vt
Negative velocity is going
backwards! xf
xi
ti tf
Displacement varies in sign, but distance traveled
does not
f ix x x v t
xf
xi
ti tf
Displacement:
Distance traveled: | | | |f ix x v t
Distance traveled: Same same
This distinction is especially important for a
round trip
xi
xf
ti tf
Distance traveled is 2|xf - xi|
Displacement = 0
Speed is the magnitude of velocity, which cannot be
negative
xi
xf
ti tf
Round trip speed is 2|xf - xi| (tf – ti)
Displacement = 0
The average velocity during a time interval
t:
2
21
2
f
i
f
i
i
ii
v
But
So
v
v v a t
v av
v a t
v
v t
Substitute this form of the average velocity back into
the expression for xf
21( )
2f i ix x v t a t
1
2i
f i
v v a t
x x v t
Thus: Three Equations of Motion for constant acceleration during a time interval t
2
1( )
2
1
2
f i i f
f i
f i i
x x vt v v t
v v at
x x v t at
For constant acceleration, distance traveled is a quadratic function of time:
21( )
2f i ix x v t a t
Another way: Velocity-time graphs
vi
vf
ti tf
How far does an object traveling at constant vi go in
time t?
Ans: x = vi t
x = v t
vi
vf
ti tf
Both distances are numerically equal to the area of the
rectangle of height v and width t.
Suppose velocity changes by constant acceleration:
vi
vf
ti tf
How far does an object accelerating from vi to vf go in
time t?
Ans: x = vit + 1/2 (vf - vi)t
vi
vf
ti tf
How far does an object accelerating from vi to
vf go in time t?
But vf - vi = at!
vi
vf
ti tf
So: x = vit + 1/2 at2
Note that the velocity graph tells
us nothing about the initial xi
These relationships between a function, its slope and the
area below its graph
vi
vf
ti tf
are the key ties between the Physics of
Motion and the Calculus
One more equation:
2 2:
2
2
f i i f
f i
f i
v v and v v
Multiply v v
But
a t
a t
v
v
t x x x
v
2 2 2f iS v v xao
The 4 Equations of Motion for Constant Acceleration
2 2
2
1( )
2
2
1
2
f i i f
f i
f i
f i i
x x vt v v t
v v at
v v ax
x x v t at