### Unit 3

```Unit 3
Motion at Constant Acceleration
Giancoli, Sec 2- 5, 6, 8
Example 3-1 Consider the case of a car that accelerates from rest with a
constant acceleration of 15 m / s2 starting at t1 = 0.
a 
v 2  v1
t 2  t1

v
t

2
v 2  v1  a ( t 2  t1 )  0  15 m s  ( t 2  0 )   15 m s 2
t
2
We see that velocity increases by 15 m/s every second and is
thus a linear function of time.
t (s)
v ( m/s)
a ( m/ s2 )
0
0
15
1
15
15
2
30
15
3
45
15
4
60
15
5
75
15
Unit 3- 2
Derivation of Equations for
Motion at Constant Acceleration
•In the next 4 slides we will combine several known
equations under the assumption that the acceleration is
constant.
•This process is called a derivation.
•It will produce four equations connecting
displacement, velocity, acceleration and time.
•You will use these equations to solve most of the
problems in Units 3,4 and 7.
Skip Derivation
Unit 3- 3
Derivations
•Much research in Physics requires derivations.
•In more advanced courses students are required to be able to
perform derivations on tests and homework
•In this course you will need to know the initial assumptions, the
resultant equations and how to apply them.
•You do not need to memorize derivations.
•But I could ask you to derive an equation for a specific problem.
This is very similar to an ordinary problem without a numeric
Unit 3- 4
Motion at Constant Acceleration - Derivation
•Consider the special case acceleration equals a
constant:
a = constant
• Use the subscript “0” to refer to the initial
conditions
• Thus t0 refers to the initial time and we will set t0 =
0.
• At this time v0 is the initial velocity and x0 is the
initial displacement.
•At a later time t, v is the velocity and x is the
displacement
•In the equations t1t0 and t2  t
Unit 3- 5
Motion at Constant Acceleration - Derivation
The average velocity during this time is:
v
x  x0
t  t0

x  x0
(Eqn. 1)
t
v
The acceleration is assumed to be constant
a
v  v0
 Constant
t
( Eqn. 2 )
vo
t
From this we can write
v  v0  a t
( Eqn. 3 )
Unit 3- 6
Motion at Constant Acceleration - Derivation
Because the velocity increases at a uniform rate, the average
velocity is the average of the initial and final velocities
v
v
v0  v
( Eqn. 4 )
2
From the definition of average velocity
x  x0  v t
 x0  (
v0  v
)t

2
x0  (
vo
v 0  v 0  at
t
)t
2
And thus
x

x0  v0 t 
1
at
2
( Eqn. 5 )
2
Unit 3- 7
Equations for Motion at Constant Acceleration
• The book derives one more equation by eliminating time
• The notation in the equations has changed
• At t = 0, x0 is the displacement and v0 is the velocity
• At a later time t, x is the displacement and v is the
velocity
v  v0  a t
x  x0  v0 t 
1
at
2
2
v  v0  2 a ( x  x0 )
2
v 
2
v  v0
2
Unit 3- 8
Example 3-2. A world-class sprinter can burst out of the blocks to essentially
top speed (of about 11.5 m/s) in the first 15.0 m of the race. What is the
average acceleration of this sprinter, and how long does it take him to reach that
speed? (Note: we have to assume a=constant)
v  v  2ax  x
2
2
0
0

v0  0
x0  0
v  11 . 5 m s
x  15 . 0 m
v  v  at
0
v v
2
a 
2
t 
0
2( x  x )
2
2
(11 . 5 m )  0
s
a 
2 ( 15 . 0 m  0 )
s
0
a
0
a  4 . 41 m
vv
2
t
11 . 5 m
s
4 . 41 m
0
s
2
t  2 . 61 s
Unit 3-9
Graphical Analysis of Linear Motion
• Consider the graph of x vs. t. The graph is a straight line, which means the
slope is constant.
• The slope of the line is the rise (Δy) over the run (Δx).
slope 
 x
t
• If we compare this with the definition of velocity,
we see that the slope of the x vs. t graph is the
velocity.
v 
x
t
Unit 2 - 10
Graphical Analysis of Linear Motion
The graphs describe the motion
of a car whose velocity is
changing:
v
lim
x
 t  0 t
v is the slope of position vs.
time graph. Since the graph
is not linear, we draw a
tangent line at each point and
find slope of the tangent line.
a 
lim
v
t  0
t
Thus a is the slope of
velocity vs. time graph.
Unit 3- 11
Example 3-3: Calculate the acceleration between points A and B and B
and C.
a 
a
lim
v
t  0
t
v
( straight
t
aAB 
a BC 
v 2  v1
 15 . 0 m
s
s
20 . 0 s  15 . 0 s
15.0 m
t 2  t1 
v 2  v1
t 2  t1
line for A  B & B  C)

 15 . 0 m
s
s
25 . 0 s  20 . 0 s
5.0 m

0 .0 m

s
2
 2 .0 m
s
2
Unit 3- 12
Example 3-4. A truck going at a constant speed of 25 m/s passes a car at rest.
The instant the truck passes the car, the car begins to accelerate at a constant
1.00 m / s2. How long does it take for the car to catch up with the truck.
1
x  x v t
t
0t
0t
2
at
xc  x0c  v0ct 
2
t
x t  ( 25 m ) t
s
x

c
c
(1 . 0 m
2
1
s
2
2
)t
act
2
2
t
) t  ( 25 m ) t
s
s
(1 . 0 m
2
(1 . 0 m
2
x  x
When the car catches the truck:
1
1
1
2
2
s
2
) t  25 m
s
t  50 s
How far has the car traveled when it catches the truck?
1
x  (1 . 0 m )( 50 s )  1250 m
s
2
2
c
2
Unit 3- 13
Example 3-4: Graphical Interpretation
In order to understand the solution to example 4, we can graph the two
equations:
x
c

1
2
(1 . 0 m
s
2
)t
2
x t  ( 25 m ) t
s
•
•
•
•
•
The graph of the truck (red) is linear because the velocity is constant.
The car is accelerating so its graph (blue) is quadratic.
The two curves intersect at t = 50 s which agrees with the solution.
They intersect at x ~ 1250 m which also agrees.
The slopes of the two graphs at t = 50 s indicate that the car is traveling twice
as fast as the truck.
Unit 3- 14
Unit 3 Appendix
Photo and Clip Art Credits
Some figures electronically reproduced by permission of Pearson Education, Inc., Upper
Slide
3-9 Sprinter: Microsoft Clipart Collection, reproduced under general license for educational purposes.
4-5, 6, 7 Soccer Kicker: Microsoft Clipart Collection, reproduced under general license for educational
purposes.
4-8, 9 Photo of Leaning Tower of Pisa: Microsoft Clipart Collection, reproduced under general license
for educational purposes.
Unit 03- 15
A quadratic equation is a polynomial equation of the second degree (one term is
squared). The general form of a quadratic equation is
a x bxc  0
2
where a, b and c are constants and a is not equal to zero.
You will need to arrange your equation in this form in order to determine the
numeric value of a, b and c. Often this semester, the unknown variable will be t
A quadratic equation has two solutions or roots:
x
 b
b 4 ac
2
and
2a
x
 b
b 4 ac
2
2a
These solutions are usually combined using the ± symbol:
x
 b
b 4 ac
2
2a
• The solutions can be equal ( if b2 = 4 a c ).
• If they are not equal, you will often need to select the solution that makes
sense for the problem.
Unit 3- 16
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