Velocity
Acceleration
Motion at Constant Acceleration
“Euclid taught me that without assumptions there is no proof.
Therefore, in any argument, examine the assumptions.”—
mathematician Eric Bell
2.2 Average Velocity
In contrast to loose usage in “everyday life,” speed and
velocity have precise definitions in physics.
Average speed is defined by
(average speed) = (distance traveled along a path
during elapsed time)/(elapsed time).
Average velocity is defined by
(average velocity) = (displacement during elapsed
time)/(elapsed time).
Do you see the difference?
A runner who travels once around an oval 400 meter track in
one minute has run with an average speed
Savg = 400 m / 60 s = 6.7 m/s.
The runner’s average velocity is
Vavg = 0 m / 60 s = 0 m/s.
This example makes average velocity look like a silly thing to
calculate, but we will find that velocity is generally much more
useful than speed.
Mathematically,
vx
x2 - x1
x
=
=
t2 - t1
t
The bar above the v means “average.” The subscript “x” on
the v is very important. It reminds us of the direction we have
chosen to be positive. It also reminds us that velocity is a
vector quantity.
Average velocity may also be written as vavg,x .
A note on notation and vector components: Giancoli follows
the usual Physics convention for 1-dimensional kinematics and
writes v for velocity.
When he goes to 2 dimensions, he uses subscripts x and y,
e.g., vx and vy, to denote the x- and y-components of velocity.
The neglect of the subscript for one dimension is potentially
confusing.
Velocity is a vector and has both magnitude and direction.
By convention, a “regular” (not bold, no arrow above) symbol
stands for a scalar.
Sometimes physicists use “v” to mean “the magnitude of the
velocity vector. Sometimes they use it to mean “the velocity
vector in one dimension.” Two different things !
You would think physicists would be more careful to say
precisely what they mean. But we physicists tend to think
alike, and take shortcuts that we believe everybody should
understand.
Remember: vx has both a numerical value and a sign “buried”
in it. The sign tells whether vx is directed along the +x or –x
direction.
This may seem trivial now, but more complex problems will
demand correctness in all details.
2.3 Instantaneous Velocity
Instantaneous velocity is defined as the average velocity over
an infinitesimal (very small) time interval:
vx
x
= lim
.
t  0  t
If you have taken calculus, you can see that the instantaneous
velocity is just the derivative of position with respect to time.
(Don’t worry, you don’t need to know that for this class.)
The OSE sheet contains the 3-dimensional version:
OSE
r
v = lim
.
t 0 t
Remember, this
component form of
the OSE can be used
as an OSE.
Speed and velocity (again).
Instantaneous speed is just the magnitude of the
instantaneous velocity vector:
v = speed = vx .
We’ll be doing plenty of problems involving velocity soon
enough, so I’ll hold off on examples for a bit.
2.4 Acceleration
Average acceleration is defined by
(average acceleration) = (change of velocity)
/(elapsed time).
Mathematically,
ax
v x2 - v x1
v x
=
=
t2 - t1
t
As you might suspect, instantaneous acceleration is defined by
ax
v x
= lim
, or OSE
t  0
t
v
a = lim
.
t 0 t
Conceptual example 2-4. (a) If the velocity of an object is
zero does it mean that the acceleration is zero? (b) If the
acceleration is zero, does it mean that the velocity is zero?
Give examples.
Well…? You don’t expect me to give all the answers, do you?
Negative acceleration means the direction of the acceleration
vector is in the negative x-direction.* A car could have
negative acceleration and be speeding up.
Deceleration means the object is slowing down, but does not
necessarily mean that acceleration is negative. It means that
the magnitude of the acceleration is decreasing. Again—sign
depends on your choice of axes.
*In one dimension, axis called “x-axis.”
A “better” way to think of acceleration:
v
a = lim
.
t 0 t
An acceleration gives rise to a change in velocity.
We’ll be doing plenty of problems involving acceleration soon
enough, so I’ll again hold off on examples for a bit.
2.5 Motion at Constant Acceleration
If acceleration is constant, the instantaneous and average
accelerations are equal, and motion is said to be “uniformly
accelerated motion.”
It is not too difficult to start with our definitions of velocity
and acceleration, and derive the following four equations,
which are valid if acceleration is constant:
v x = v0x + ax t
1
x = x0 + v0x t + ax t2
2
v2x = v20x + 2ax  x - x0 
vx
v x + v 0x
=
2
The three boxed equations are
“official” and are on your OSE
sheet. The other comes from the
definition of “average.”
By “official” I mean the equations on the previous slide are the
starting point for all your kinematics in one dimension
problems!
The equations describe the motion of an object which has an
acceleration ax.
These equations are only valid if acceleration is constant! Do
not use them unless you know that is true!
You can define any convenient coordinate axes, and you may
also use “y” (or some other symbol) instead of “x.”
A note on notation:
v2x = v20x + 2ax  x - x0 
acceleration of object
between x and x0
(must be constant)
velocity when
position is x
velocity when
position is x0
It might be “better” to write
2
v1x
= v20x + 2ax  x1 - x0 
but none of the major textbooks write it that way.
“Should I write v2x = v20x + 2ax  x - x0 
or v2x = v2x0 + 2ax  x - x0  ?"
I’ll accept either. But v0x is better than vx0 because the
subscript “x” identifies the symbol as being the x-component
of whatever comes before, and by v0x we actually mean the x
component of (v0).
It would be technically more correct for me to write
v
2
x
2
=  v0 x  + 2ax  x - x0 
but that’s lots of extra parentheses to keep track of.
Example: starting from a speed v1 at the beginning of a
freeway entrance ramp, you want to enter the freeway at a
speed v2 at the end of the ramp, which has length L. Calculate
the required acceleration. Assume a is constant.
If time permits, I’ll work this on the blackboard in class. The
following slides show the solution in all its detail.
Litany for Kinematics, step 1: draw a basic
representative sketch of the physical situation.
Step 2: draw and label vectors for the relevant
dynamical quantities.
Step 3: draw axes, indicating origin & + direction.
Example: starting from a speed v1 at the beginning of a
freeway entrance ramp, you want to enter the freeway at a
speed v2 at the end of the ramp, which has length L. Calculate
the required acceleration. Assume a is constant.
Step 4: indicate and label initial and final positions.
Step 5: OSE.
Step 6: Replace generics with specifics.
Step 7: Solve algebraically.
Step 8: Draw a box around your final answer.
Example: starting from a speed v1 at the beginning of a
freeway entrance ramp, you want to enter the freeway at a
speed v2 at the end of the ramp, which has length L. Calculate
the required acceleration. Assume a is constant.
Litany for Kinematics, step 1: draw a basic
representative sketch of the physical situation.
end of ramp
v1
L
v2
Step 2: draw and label vectors for the relevant
dynamical quantities.
v1
v1
There’s a reason for not “attaching”
the v’s to the object.
a=?
L
v2
v2
The labels v1 and v2 below the “car” didn’t help. Also are
misleading (vectors?). Wasted time! Dump them.
If you are given that v1 = 0, I do not require that you draw the
arrow if you write v1 = 0 next to the object.
You can indicate numerical values of vector magnitudes by the
arrow, or put them off to the side. If a value is unknown, you
should indicate that with a “?”
Step 3: draw axes, indicating origin & + direction.
v1
a=?
v2
x
L
Actually, there is nothing in the problem statement that
“demands” v1 pointing to the right. (I don’t see how you could
“enter” by going away from the entrance, so it makes sense
for v2 to be to the right.)
If you have information that tells you the direction of a vector,
use it. Otherwise, assume a reasonable direction. If your
assumed direction is wrong, your math will tell you.
Step 4: indicate and label initial and final positions.
v1
x1=0
a=?
v2
L
x2
x
What about time? No information given about time, maybe
we don’t need it; we can always add it later if necessary.
It would be a good idea to label the positions with x1=0 and
x2=L. You don’t need to if the diagram provides the
information. (I wrote x1=0 because it might not be clear the vertical line
there indicates x=0.)
Step 5: OSE.
a=?
v1
x1=0
v2
L
x2
Think… “I am given initial and final velocities and a total
distance; I want acceleration.”
“Aha!”
OSE:
v2x = v20x + 2ax  x - x0 
It never hurts to give me a clue that this is your intended OSE
by labeling it as such! In fact, on exams I will usually require
that you do so.
x
Step 6: Replace generics with specifics.
v1
a=?
x1=0
v2x = v20x + 2ax  x - x0 
v2
x2
L
You are not allowed to
alter this equation!
+(v22) = +(v
+(v
+ 12) + 2(+a)
2
(
(+
(+(+L)
- (0) )
The diagram says ax is directed to the right, which is why we put a in
with a + sign here.
The problem is worded so that ax could be directed to the left. Our final
answer would tell us if we “guessed” the direction wrong.
x
Step 7: Solve algebraically.
a=?
v1
x1=0
 
+ v22
v2
L
x2
 
= + v12 + 2  a+L  - 0  
v22 - v12 = 2 a L
2 a L = v22 - v12
v22 - v12
a=
2L
I don’t need all the extra ()’s any more.
Do one plodding step at a time! (I didn’t!)
I assumed that acceleration is to the right.
Caution! I could give you a problem in
which the car slows down. We’ll soon
learn techniques for handling this.
x
Step 8: Draw a box around your final answer.
v1
a=?
x1=0
v22 - v12
a=
2L
v2
x2
L
Done!
“I want numbers! Give me
numbers to plug in!” No you
don’t. Really.
x
Because you really think you want numbers, I’ll give you some.
Starting from a speed of 10 mph at the beginning of a
freeway entrance ramp, you want to enter the freeway at a
speed of 65 mph at the end of the ramp, which has length of
400 yards. Calculate the required acceleration. Assume
constant acceleration.
v22 - v12
a=
2L
I inverted these when I first did the calculation,
but the units showed me my error.
2
2
1000 meters
hour   10 miles 1.61 km
1000 meters
hour 
 65 miles 1.61 km







 -

hour
mile
km
3600 s   hour
mile
km
3600 s 

a=

3 feet
1 meter 
2  400 yards 

yard
3.28 feet 

m
a = 1.13 2
s
Do you have any clue if this is correct or not? I
think the ramp length is quite a bit longer than
is typical in “reality.”
Do your remember the quote at the start of this lecture about
examining assumptions?
“Euclid taught me that without assumptions there is no proof.
Therefore, in any argument, examine the assumptions.”—
mathematician Eric Bell
In the solution in these notes, I assumed v1x, v2x and ax are
directed towards positive x. I also assumed ax is constant.
If you don’t know the direction a vector component points,
leave it in the form vx, ax, etc. in your solution and do not
include a sign for assumed direction.
Example: Use the data from the previous example to show
that it took 21.8 seconds to accelerate from 10 to 65 mph.
You learn relatively little just by watching me work problems.
If you wished, you could work the problem and see if my
calculation of time is correct.