Motion
The basic description of motion is how
location (position) changes with time. We
call this velocity.
Is velocity a vector? (Does it have magnitude
and direction?) YES!
v = (x,y) / t
where the  sign means “change in”, or “final
minus initial” .
Velocity
Now the question becomes: how do you
divide a vector by a scalar?
Since multiplication is simply multiple
additions (3*2 means 2+2+2), and since we
can add vectors nicely in rectangular form
(add the components), we should be able to
multiply a vector by a scalar by just
multiplying the vector’s components by the
scalar.
Velocity
And since division is simply the inverse of
multiplication, we can divide a vector by a
scalar by just dividing the components of
the vector by the scalar.
Hence v = (vx , vy) = (x,y) / t, and
vx = x / t and vy = y / t .
Velocity
• Note that the MKS units of velocity are m/s.
• This definition of velocity indicates that
position changes over time. This is really,
then, a calculation of an AVERAGE
VELOCITY.
• Is there such a thing as an
INSTANTANEOUS VELOCITY?
Velocity
According to the calculus,
in the limit as t approaches zero
(and so (x,y) also approaches zero),
the expression vx-avg = x / t becomes
vx = dx/dt where x is a function of t.
This is the mathematical way of saying we do
have a way of finding instantaneous
velocity!
Average versus Instantaneous,
Discrete versus Continuous,
Values versus Functions
When dealing with discrete values, we can find
values for averages. vx-avg = x / t .
In dealing with the continuous case, we use
functions. When you take the derivative of a
function, you get a function. This means that
x(t) [this means that x is a function of t] and so v(t) [v is
a function of t] in the instantaneous case.
Acceleration
But velocity is not the whole story of motion.
Sometimes (often) we are interested in how the
velocity changes with time!
This leads to the notion of ACCELERATION:
a = (ax , ay) = (vx,vy) / t
and
ax-avg = vx/ t , and ax = dvx/dt .
Note that acceleration is a vector (it has
components, it has magnitude and direction, we
have to work in rectangular components).
Note that the units of acceleration are
(m/s) / s or more commonly: m/s2 .
Question: what is a square s (s2) ?
Where do we stop?
• Is there a name for the change of
acceleration with time?
• Why haven’t most people heard of it, when
most people have heard of velocity and of
acceleration?
Jerk!
• To answer the first question, the change in
acceleration with respect to time is called
Jerk! jx = ax/t .
• To answer the second question, the reason
most people have not heard of jerk is
because it is not normally useful. This is
due to reasons we’ll see in Part II of the
course.
Signs (+ or -) for position
Position: Usually we have some reference
point that we call zero position.
For horizontal positions, plus usually means
to the right, and minus means to the left.
For vertical positions, plus usually means
above (up) and minus means below (down).
Warning: these are only the usual conventions; they can be
reversed if that is more convenient.
Signs (+ or -) for velocity
For horizontal motion, moving to the right usually
means a positive velocity component, and moving
to the left means a negative velocity component.
For vertical motion, moving up usually means a
positive velocity component, and moving down
means a negative velocity component.
Warning: if the usual conventions for position are switched, then the
sign conventions for the velocity will also be switched. For example,
if down is called a positive position, then moving down will be
considered a positive velocity.
Note: we can have a positive position with either a positive or negative
velocity, and we can have a negative position with either a positive or
negative velocity.
Signs (+ or -) for Acceleration
If the velocity is increasing in the positive direction, the
acceleration is positive, and if the velocity is
decreasing in the positive direction, the acceleration is
negative.
Warning: the case of negative velocities is more tricky & counter-intuitive!
If the velocity is becoming more negative (going faster in
the negative direction), the acceleration is negative,
and if the velocity is becoming less negative (getting
slower in the negative direction), the acceleration is
positive. Examples:
going faster in negative direction: ax = vxf – vxi = [(-7 m/s)–(-5 m/s)]/(1 s) = -2 m/s2
going slower in negative direction: ax = vxf – vxi = [(-3 m/s)–(-5 m/s)]/(1 s) = +2 m/s2
Signs (+ and -) for Acceleration
Language problems in the vertical
If we are going faster in the up direction, we say we are
speeding up (and going up). No problem. Is this
acceleration positive or negative?
If we are going slower in the up direction, we say we are
slowing down (but going up). See the language
problem? Is this acceleration positive or negative?
If we are going faster in the down direction, we say we are
speeding up (but going down). See the language
problem here? Is this acceleration + or - ?
If we are going slower in the down direction, we say we
are slowing down (and going down). Is this acceleration
+ or - ?
Signs (+ and -) for Acceleration
Language problems in the vertical
If we are going faster in the up direction, we say we are
speeding up (and going up). No problem. This
acceleration is positive.
If we are going slower in the up direction, we say we are
slowing down (but going up). See the language
problem? This acceleration is negative.
Warning: these next two are counter-intuitive:
If we are going faster in the down direction, we say we are
speeding up (but going down). See the language
problem here? This acceleration is negative.
If we are going slower in the down direction, we say we
are slowing down (and going down). This acceleration
is positive .
Motion
We now have two useful definitions
(relations):
vx-avg = x / t , or vx(t) = dx(t)/dt
ax-avg = vx/ t , or ax(t) = dvx(t)/dt .
If we know position and time, we can
calculate velocity; if we know velocity and
time, we can calculate acceleration.
Discrete Case - an example
Given the following data, find vx and ax:
x (in meters) at t (in seconds)
-2
0
+1
0.5
+6
1
+4
1.5
0
2
can you picture this?
Discrete case: a picture
Note: the time (Δt) for each arrow (which is
the change in position, Δx) is 0.5 seconds.
-2 m
1m
6m
4m
0m
-2
0 sec
0.5 sec
1.0 sec
1.5 sec
2.0 sec
0
2
4
6
x
Discrete Case - an example
Since we know the position at 0 sec and 1 sec, we can
find the average velocity in this interval:
vx-average = x / t
-2 m 0 sec
vx-avg (between 0 and 1 sec)) =
1 m 0.5 sec
(+1 m - -2 m) / (0.5 sec - 0 sec)
6 m 1.0 sec
= +6 m/s.
4 m 1.5 sec
Since this is the velocity between
0 m 2.0 sec
0 and 0.5 seconds, we can say
that this probably is close to the
speed at 0.25 seconds.
Discrete Case - an example
Can you figure out the value for the other times:
x (in meters) at t (in sec.) vx (in m/s)
-2
+1
+6
+4
0
0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
+6
Discrete Case - an example
Doing similar calculations for the other times:
x (in meters) at t (in sec.) vx (in m/s)
-2
+1
+6
+4
0
0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
+6
+10
-4
-8
Discrete Case - an example
For acceleration we do the same thing:
Since we know the approximate velocity at 0.25 sec
and 0.75 sec, we can find an approx. average
acceleration in this interval:
ax-average = vx / t
ax-avg (between 0.25 and 0.75 sec) =
(+10 m/s - +6 m/s) / (.75 sec - .25 sec) = +8 m/s2.
Since this is the acceleration between .25 and .75
seconds, we can say that this probably is close to
the acceleration at 0.5 seconds.
Discrete Case - an example
Doing similar calculations for the other times:
x (in meters) at t (in sec.) vx (in m/s) ax (in m/s2)
-2
+1
+6
+4
0
0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
+6
+8
+10
-28
-4
-8
-8
Motion
Often we do NOT know position and time,
but rather something else and we wish to
predict what the position versus time will
be! Can we go backwards as well as
forwards in these relations? (That is,
knowing acceleration and time, can we
figure out what the velocity will be?)
Going backwards:
the discrete case
vx-avg = x / t and ax-avg = vx/ t
Since the above definitions involve division,
the inverse of division is multiplication. In
the calculus (for functions), the inverse of
the derivative is the integral.
Knowing the AVERAGE velocity and the
time interval, we can find the CHANGE IN
position: x = vx-avg * t .
Going backwards:
the discrete case
x = vx-avg * t , or
xfinal = xinitial + vx-avg*t
Note that the velocity in this formula is the AVERAGE
velocity. If the velocity is constant, then this equation
works exactly. However, if the velocity changes, then
we need to know the real average velocity. The real
average velocity is not necessarily the sum of the
initial and final divided by 2!
final = 4
Using just the endpoints,
avg = (2+4)/2 = 3.
3
initial = 2
avg = 3
avg < 3
avg > 3
Going Backwards
Knowing the AVERAGE acceleration and the
time interval, we can find the CHANGE IN
velocity: vx = ax-avg * t .
If the acceleration is constant, so that the
average acceleration is equal to the
acceleration at all times, then we have the
formula:
vxfinal = vxinitial + ax* t
Example - discrete case
x (m)
t(sec)
0
v (m/s)
a (m/s2)
0.8
+3
1.6
-1
2.4
-2
3.2
Example - discrete case
Knowing the acceleration at t=0.8 sec, we can use the
definition of acceleration: a = v/t to get: v =
a*t . Since the accelerations are given in 0.8
second intervals, let’s choose t = 0.8 sec. This
leads to:
v(t=1.2 sec) - v(t=0.4 sec) = (3 m/s2) * 0.8 sec, or
v(t=1.2 sec) = v(t=0.4 sec) + (3 m/s2) *0.8 sec
However, unless we know one of these two v’s, we
can’t solve this. Let’s say that we do know the
velocity at t=0.4 sec is v(t=0.4 sec) = +5 m/s.
v(t=1.2 sec) = +5 m/s + (3 m/s2) * 0.8 sec = +7.4 m/s.
Example - discrete case
x (m)
t(sec)
0
0.4
0.8
1.2
1.6
2.0
2.4
2.8
3.2
v (m/s)
a (m/s2)
+5.0
+3
+7.4
-1
-2
Example - discrete case
We now proceed as before to get the next
velocities:
v(t=2.0 sec) = v(t=1.2 sec) + (-1 m/s2) * 0.8 sec ;
from the previous calculation, we know v(t=1.2 sec) =
7.4 m/s, so
v(t=2.0 sec) = 7.4 m/s + (-1 m/s2) * 0.8 sec = 6.6 m/s.
Proceeding:
v(t=2.8 sec) = v(t=2.0 sec) + (-2 m/s2) * 0.8 sec gives
v(t=2.8 sec) = 6.6 m/s + (-2 m/s2) * 0.8 sec = 5 m/s.
Example - discrete case
x (m)
t(sec)
0
0.4
0.8
1.2
1.6
2.0
2.4
2.8
3.2
v (m/s)
a (m/s2)
+5.0
+3
+7.4
-1
+6.6
-2
+5.0
Example - discrete case
To get position, we now go backwards from
velocity: Knowing the velocity at t=0.4 sec, we can
use the definition of velocity: v = x/t to get: x =
v*t . Since the velocities are given in 0.8 second
intervals, let’s choose t = 0.8 sec. This leads to:
x(t=0.8 sec) - x(t=0 sec) = (5 m/s) * 0.8 sec , or
x(t=0.8 sec) = x(t=0 sec) + (5 m/s) * 0.8 sec
However, unless we know one of these x’s, we can’t
solve this. Let’s say that we do know the position at
t=0 sec is x(t=0 sec) = -2 m.
x(t=0.8 sec) = -2 m + (5 m/s) * 0.8 sec = +2.0 m.
Example - discrete case
x (m)
-2 .0
+2.0
t(sec)
0
0.4
0.8
1.2
1.6
2.0
2.4
2.8
3.2
v (m/s)
a (m/s2)
+5.0
+3
+7.4
-1
+6.6
-2
+5.0
Example - discrete case
We now proceed as before to get the next
positions:
x(t=1.6 sec) = x(t=0.8 sec) + (+7.4 m/s) * 0.8 sec ;
from the previous calculation, we know x(t=1 sec) = 2.0
m, so
x(t=1.6 sec) = 2.0 m + (+7.4 m/s) * 0.8 sec = 7.92 m.
Proceeding:
x(t=2.4 sec) = 7.92 m + (+6.6 m/s) * 0.8 sec = 13.2 m
x(t=3.2 sec) = 13.2 m/s + (+5 m/s) * 0.8 sec = 17.2 m.
Example - discrete case
x (m)
- 2.0
+ 2.0
+ 7.92
+13.2
+17.2
t(sec)
0
0.4
0.8
1.2
1.6
2.0
2.4
2.8
3.2
v (m/s)
a (m/s2)
+5.0
+3
+7.4
-1
+6.6
-2
+5.0
Example - discrete case
Note: In going backwards, we needed to
know the acceleration, but we also needed
to know where to start, both for the velocity
and for the position. These starting points
are called “initial conditions”.
In going forward, we had no need for such
initial conditions.
Continuous Case
(derivations based on calculus)
In dealing with the continuous case (equations instead
of values), we again start with the definitions:
v(t) = dx(t)/dt
and
a(t) = dv(t)/dt
where x, v, and a are all functions of time.
We can go “forward” if we know the function of
position, x, with time: x(t) – we simply differentiate.
We can go “backward” if we know the function of
acceleration, a, with time: a(t) – do the inverse of
differentiation: integration:
v
t
x
t
vo∫ dv = 0∫ a(t) dt and
xo∫ dx = 0∫ v(t) dt .
Special Case:
Constant Acceleration
If the acceleration is constant, then we get two
equations from our two starting definitions:
vxfinal = vx-initial + ax*t .
and
xfinal = xinitial + vx-initial *t + (1/2)*ax*t2 .
Since there is acceleration, the velocity does not remain
constant and so the formula for x with constant
velocity does not hold.
Falling (without air resistance)
In the case of something falling, the
acceleration due to gravity near the earth’s
surface is approximately constant, if we can
also neglect the effects of air resistance. In
this special case, we can use the equations
for constant acceleration. [In part two we will
investigate what it means to be “near” the earth.]
Falling (without air resistance)
If we treat up as +y, then we have these two
equations:
y = yo + vyo *t + (1/2)*g*t2 and
vy = vyo + g*t where g = -9.8 m/s2 .
Here, we have simply used y for yfinal and we
have used yo for yinitial. The same notation is
also used for v.
Solving Problems
Note that when we have identified a problem
as being one of constant acceleration, we
have two equations:
• y = yo + vyo*t + (1/2)*a*t2 and
• vy = vyo + a*t .
Note that in these two equations we have six
quantities: y, yo, v, vo, a, and t. This
means we have to identify four of the six
in order to use the two equations to solve
for the other two quantities.
Solving Problems
Reading the description of a problem involves
several steps:
• Identify the problem type: does this
problem have constant acceleration? If so,
we know we have the two equations to work
with.
• Identify what you know: does this problem
involve falling under the influence of
gravity? If so, we know a = g = -9.8 m/s2.
Solving Problems
(list continued from previous slide)
• We can usually pick out where to start
from (if gravity, the ground is usually where
y=0 is). This is important for identifying y
and yo. Sometimes we are given
information about yo, sometimes about y.
• Special words: The word “stop” or
“stationary” means that at this time v=0.
This may apply to either v or vo.
Solving Problems
(list continued from previous slide)
• Make sure you know what negative signs
mean. For y, positive usually means above
ground, negative will mean below ground.
For v, positive usually means going up (or
forward), negative will mean going down
(or backwards).
Solving Problems
List continued from previous page
• Note that in the y equation for constant
acceleration, there is a t2 term:
y = yo + vyo*t + (1/2)*a*t2 .
That means that, when solving for time, there
may be two solutions. Can you identify in
the problem what the two solutions would
be for? The computer homework
assignment on Quadratic Equations
should provide a review in this area.
Example of a falling problem
To find the height of a tree, a person throws a
baseball up so that it just reaches the height
of the tree. The person then uses a
stopwatch to time the fall of the ball from
the highest point (the height of the tree) to
the ground. If the time on the stopwatch is
3.4 seconds, how high is the tree?
Example of a falling problem
Draw a diagram to help define the situation:
yo = ? vo = ? (to=0 sec.)
a = g = -9.8 m/s2
y = ?, v = ?, t = 3.4 seconds
Example of a falling problem
We will assume that air resistance is
negligible, and that the tree is not too high
so we can consider gravity constant. In this
case we then have the constant acceleration
situation and so can use the two equations:
• y = yo + vyo*t + (1/2)*a*t2 and
• vy = vyo + a*t .
Example of a falling problem
We can assume that the ground is where
y=0. Then from the statement of the
problem, we are looking for yo (which
would correspond to the height of the tree),
and we know the time for y=0. Thus we
know three quantities and have one
unknown so far:
Example of a falling problem
• yo = ?
• yfinal = y = 0 m
• a = -9.8 m/s2
• t = 3.4 seconds
That leaves the initial and final velocity. To
solve the problem, we need to know four
things and can have two unknowns (since
we have two equations).
Example of a falling problem
From the statement of the problem, the ball
falls from the highest point, so vo = 0. We
do NOT know the final velocity. Note that
the ball will HIT the ground, but that does
NOT make the final velocity zero - just
before it hits it is travelling rather fast! The
act of hitting destroys our assumption of
constant acceleration due only to gravity.
Example of a falling problem
Draw a diagram to help define the situation:
yo = ? vo = 0 (highest point)
a = g = -9.8 m/s2
y = 0, v = ?, t = 3.4 seconds
Example of a falling problem
•
•
•
•
•
•
y = yo + vyo*t + (1/2)*a*t2 and
vy = vyo + a*t .
yo = ?
y=0m
a = -9.8 m/s2
t = 3.4 seconds
vo = 0 m/s
v=?
Example of a falling problem
Putting the knowns into the two equations gives:
• 0 m = yo + (0 m/s)*(3.4 s) + (1/2)*(-9.8 m/s2)*(3.4 s)2
which we see is one equation in one unknown
and can be directly solved
yo = 56.64 m .
• v = 0 m/s + (-9.8 m/s2)*(3.4 s) = -33.3 m/s.
2nd Example: accelerating car
A car accelerates (assume constant acceleration)
from rest up to a speed of 65 mph in a time of
7 seconds.
What is the average acceleration of the car?
How far does the car go during the 7 seconds
while it is accelerating?
Accelerating Car
We recognize this as a constant acceleration problem, so we
have our two equations:
x = xo + vo*t + (1/2)*a*t2 and v = vo + a*t
and six quantities:
• xo =
• x=
car
car
• a=
• t=
x=
xo =
a=
• vo =
v=
vo =
t =
to = 0 s
• v=
Accelerating Car
From the statement of the problem, we see the
following being given:
“from rest up to a speed of 65 mph in a time
of 7 seconds”
Can we determine which symbols go with
which values?
Accelerating Car
“from rest” means vo = 0;
“up to a speed of 65 mph” means v = 65 mph, but mph
is not the MKS unit. We need to convert it to m/s
(2.24 mph = 1 m/s), so
v = 65 mph * (1 m/s / 2.24 mph) = 29 m/s.
“in a time of 7 seconds” means t = 7 s.
This gives 3 of the six quantities, and we have two
equations, so we need one more. This one is “hidden”
in the problem – since we don’t have a definite starting
position, we can assume xo = 0.
Accelerating Car
x = xo + vo*t + ½*a*t2 and v = vo + a*t
and six quantities:
• xo = 0 m
x = xo + vo*t + ½*a*t2
• x= ?
x = 0 m + 0 m/s*(7 s) + ½ a*(7 s)2
• a= ?
• t= 7s
v = vo + a*t
• vo = 0 m/s
29 m/s = 0 m/s + a*(7 s)
• v = 29 m/s
Accelerating Car
The first (x) equation has two unknowns (x and a).
However, the second (v) equation has only one
unknown (a). Therefore, we can solve the second
equation for a, and then use the first equation to solve
for x.
29 m/s = 0 m/s + a*(7 s) or a = 29 m/s / 7s = 4.14 m/s2
and then x = 0 m + 0 m/s*(7 s) + ½ (4.14 m/s2)*(7 s)2 =
101.5 m .
Note that vavg = x /t , but x = vavg*t ≠ (29 m/s)*(7 s) = 203 m because v is not
a constant 29 m/s over the 7 second time interval. However, because the
acceleration is constant, we can use vavg = ½*(vo+vf) = ½*(0 m/s + 29 m/s) =
14.5 m/s, so that x = (14.5 m/s)*(7 s) = 101.5 m. Note that vavg = ½*(vo+vf)
does not always work – it only works if a = constant, which it is here.
Graphical Representations
From the definition of velocity (working in
rectangular components):
vx = x / t
we can recognize that the SLOPE of the x vs t
curve at any time = VALUE of the velocity at
that time.
Note that this means the VALUE of x has
NOTHING to do with the VALUE of v. It is
only how x CHANGES that affects the velocity.
Graphical Representations
Given the graph of x vs t, can you figure out
the graph of v vs t?
x
v
t
t
Graphical Representations
At t=0, the SLOPE of x is a small positive
amount, so the VALUE of v is a small
positive. [recall: v = x/t = slope of x vs t]
x slope
v value
t
t
Graphical Representations
A little before t=0, the x vs t curve is flat
SLOPE=0, so at this time the VALUE of
v=0.
x slope
v value
t
t
Graphical Representations
At an even earlier time, the SLOPE of x is
slightly negative, so the VALUE of v is also
slightly negative at that earlier time.
x slope
v value
t
t
Graphical Representations
At a later time, the SLOPE of x is slightly
more positive than it was at t=0, so the
VALUE of v is also slightly more positive.
x slope
v value
t
t
Graphical Representations
At the latest time, the SLOPE of x is about the
same, so the VALUE of v is also about the
same.
slope
value
x
v
t
t
Graphical Representations
Now we just connect the dots to get a graph of
v vs t based on the graph of x vs t.
x
v
t
t
Graphical Representations
• To go in reverse, that is, knowing the graph
of v and trying to find the graph of x, work
with the idea that the VALUE of v gives the
SLOPE of x.
• One thing to note: since the value of v
gives no information about the value of x,
only about the slope of x, we need to be
given one value of x to begin. This is
usually xo .
Graphical Representations
Since the definition of acceleration is
ax = vx/ t
we see that the SLOPE of v gives the VALUE of
a. Thus we can use the same procedure to get
the graph of a vs t from the graph of v vs t as
we did to get the graph of v vs t from x vs t.
Graphical Representations
Given this curve for v(t), can you sketch
(roughly) the curve for x(t)?
v
x
t
t
Graphical Representations
For the most negative time on the graph, the velocity is zero,
which means the slope of the x curve at this time is zero.
[recall: v = x/t = slope of x vs t]
slope
v
x
1
t
1
value
t
But this doesn’t tell us where to draw a flat curve. We need
to know where to start. Let’s assume that x is negative at
this time.
Graphical Representations
Now a little later, the velocity is a small positive value, so the
slope of the x curve should be a little positive (going up)
and getting a little more positive (steeper going up).
slope
v
x
value
3
2
3
2
t
t
Graphical Representations
At t=0, the velocity value is still positive, but not as large as it
was a little before that. This means the slope of the x curve
is still positive, but not as steep as it was a little earlier. A
little after t=0 the velocity is zero so the x curve is flat at
that point.
slope
value
v
x
4
5
4
5
t
t
Graphical Representations
for t>0, the velocity value turns negative. This means the slope
of the x curve will be negative (heading down), first getting
steeper and then getting flatter until it is flat when the
velocity reaches zero.
slope
v
x
value
t
6
t
7
6
7
Graphical Representations
Computer homework programs #7 (on
Motion Graphs) and #8 (on Acceleration
Due to Gravity) provide some information
that you can get by graphing the
information.
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Motion