Acceleration
Graphs to Functions


A simple graph of constant
velocity corresponds to a
position graph that is a
straight line.
The functional form of the
position is
x  v0t  x0

This is a straight line and
only applies to straight lines.
x
t
x0
v
v0
t
Changing Velocity

In more complicated motion the velocity is not
constant.

We can express a time rate of change for velocity just
as for position, v = v2 - v1.

The acceleration is the time rate of change of
velocity: a = v / t.
Average Acceleration
Example problem
 A jet plane has a takeoff speed of 250 km/h. If the
plane starts from rest, and lifts off in 1.2 min what is
the average acceleration?
a = v / t = [(250 km/h) / (1.2 min)] * (60 min/h)
a = 1.25 x 104 km/h2


Why is this so large? Is it reasonable?
Does the jet accelerate for an hour?
Instantaneous Acceleration

Instantaneous velocity is
defined by a derivative.

Instantaneous acceleration
is also defined by a
derivative.
v dv
a  lim

t 0 t
dt
x dx
v  lim

t  0 t
dt
v
P2
P1
P3
t
P4
Second Derivative



The acceleration is the derivative of velocity with
respect to time.
The velocity is the derivative of position with respect
to time.
This makes the acceleration the second derivative of
position with respect to time.
dv d dx
d 2x
a
 ( ) 2
dt dt dt
dt
Constant Acceleration



Constant velocity gives a
straight line position graph.
Constant acceleration gives
a straight line velocity graph.
The functional form of the
velocity is
v  a0t  v0
v
t
v0
a
a0
t
Acceleration and Position

For constant acceleration the
average acceleration equals
the instantaneous
acceleration.
v
a0(½t) + v0
v0

Since the average of a line
of constant slope is the
midpoint:
1
1 2
x  ( a0t  v0 )t  x0  a0t  v0t  x0
2
2
½t
t
Acceleration Relationships

Algebra can be used to
eliminate time from the
equation.

This gives a relation
between acceleration,
velocity and position.

For an initial or final velocity
of zero. This becomes
• x = v2 / 2a
• v2 = 2 a x
v  v0
from
a
1
x  at 2  v0t  x0
2
v  at  v0
t
1  v  v0 
 v  v0 
x  a

v

  x0
0
2  a 
 a 
2
2

v  v0   x
v  v0 
x
v
2a
0
a
0
Take Off
Example problem
 A jet plane has a takeoff speed of 250 km/h. If the
plane starts from rest, and has a constant
acceleration of 1.25 x 104 km/h2, what is the length of
the runway?
x = v2 / 2a = (250 km/h)2 / (2.5 x 104 km/h2)
x = 2.5 km

Is this reasonable?
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