Lecture 4
Constant acceleration in 1D
x
(t)
↔
v
(t)
↔
a
(t)
In general, for 1D motion along a straight line:
0
0
( )
t
dx
v
x
x
v t dt
dt
=
⇔
=
+
∫
0
0
( )
t
dv
a
v
v
a t dt
dt
=
⇔
=
+
∫
0
0
0
0
( )
( )
t
t
dv
a
v
v
a t dt
dt
dx
v
x
x
v t dt
dt
=
⇔
=
+
=
⇔
=
+
∫
∫
One Dimensional Constant acceleration.
1
Basic math:
1
n
n
x
x dx
n
+
=
+
∫
When
a
= constant, the equations are simple:
0
0
0
2
0
0
(
)
1
2
t
x
x
v
at dt
x
x
v t
at
=
+
+
=
+
+
∫
0
0
0
t
v
v
a
dt
v
v
at
=
+
=
+
∫
One Dimensional Constant acceleration.
From these we can derive a couple more useful equations:
2
0
0
0
1
2
v
v
at
x
x
v t
at
=
+
=
+
+
2
2
0
2
v
v
a
x
−
=
∆
0
2
v
v
v
+
=
2
0
0
0
0
0
0
1
1
1
2
(
)
0
2
2
2
v t
at
x
x
v
v
v
v
at
v
v
v
t
t
+
−
+
=
=
=
+
=
+
−
=
−
2
0
0
0
0
0
2
2
2
0
0
0
0
2
2
0
1
2
2
2
2
v
v
v
v
v
v
t
x
x
v
a
a
a
a
vv
v
v
v
vv
a
a
v
v
a
−
−
−
=
⇒
−
=
+
−
+
−
=
+
−
=
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
One Dimensional Constant acceleration.
0
v
v
at
=
+
2
2
0
2
v
v
a
x
−
=
∆
0
2
v
v
v
+
=
2
0
0
1
2
x
x
v t
at
=
+
+
A car is traveling with
v
0
= 10 m/s. At
t
= 0, the driver puts on
the brakes, which slows the car to a stop in 2 seconds.
a.
What is the acceleration produced by the brakes?
•
“Translate” the problem –understand it: Draw a figure.
Identify and include initial (
t
= 0,
v
0
= 10 m/s) and final situation
(
t
= 2s; car stopped)
Start of braking
t
= 0,
v
0
= 10 m/s
t
= 2 s
,
stopped
v
= 0
EXAMPLE: Braking Car
EXAMPLE: Braking Car
A car is traveling with
v
0
= 10 m/s At
t
= 0, the driver puts on
the brakes, which slows the car to a stop in 2 seconds.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '06
 Johnson
 Physics, Acceleration, Velocity, Automobile, Brake

Click to edit the document details