Download Motion Derivatives and Anti-derivatives

AP Physics C
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A vector quantity
The straight line distance and direction from the
starting point to the ending point
Usually graphed on a position vs. time graph
Can also be calculated from the area under a velocity
time graph
Measured in length units such as meters, centimeters,
kilometers, miles, yards, feet, etc.
Position variables may be
x (along the x axis),
y (along the y axis),
z (along the z axis),
or r (for position in 2-D or 3D)
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A vector quantity
Speed in a specific direction
“how fast” and “which way”
average velocity or instantaneous velocity must be
distinguished
Can be found from the slope of the position vs time
graph
Average velocity is the slope of the line connecting 2
points on a position curve
Instantaneous velocity is the slope of the tangent line
at a specific point on the position curve.
Measured in units of speed such as meters per
second, kilometers per hour, miles per hour, etc.
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A vector quantity
Rate of change of velocity
Physics 1 assumed constant acceleration and all of the
formulas from Physics 1 only work for uniform
(constant) acceleration
Acceleration is NOT always constant, sometimes time
dependant.
Can be calculated from the slope of a velocity vs. time
graph
Average acceleration is the slope of the line connecting
two points on a velocity vs. time graph
Instantaneous acceleration is the slope of the tangent at
a given point in time on the velocity graph.
Measured in units of meters per second per second or
meters per second squared.
Position (m)
vavg
Δx
Δt
Time (sec)
Tangent line
x

 slope
t
Average velocity is the
slope of the line connecting 2
points on a position curve.
Instantaneous velocity is
the slope of the tangent
line at a specific point on
the position curve.
vinst (t )  lim
t 0
x
t
 slopetan
Velocity (m/s)
aavg
Δv
Δt
Time (sec)
Tangent line
v

 slope
t
Average acceleration is the
slope of the line connecting 2
points on a velocity curve.
Instantaneous acceleration
is the slope of the tangent
line at a specific point on the
velocity curve.
ainst (t )  lim
t 0
v
t
 slope tan
This limit statement is read as “the limit
of ∆v/∆t as ∆t approaches zero”
lim
t 0
v
t
As we make the time interval smaller
and smaller, the average value moves
toward a value at a specific point in
time…thus an instantaneous value.
Taking the limit of a function as the time
interval approaches zero allows you to
ultimately find the slope of the tangent
line at that specific point in time.
We know from Physics 1 that the slope
of a velocity time graph represents
acceleration.
These limits can be replaced by a
calculus function called a “derivative”
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We looked at the limit as Δt approaches zero
The derivative is the calculus operation that
replaces the limit process
It describes the slope of a function at any given
point.
The general form of the power rule (simplest
one) is as follows if
x(t) = a * tn , where a is constant, t is a variable, and n is its power
then the derivative, dx/dt = n*a* tn-1
read as “the derivative of the function “x”
(position function) with respect to time”
Since derivative gives us the slope of a tangent
line, then:


The derivative of a position function (position – time
graph) gives velocity as a function of time
The derivative of a velocity function (velocity –time
graph) gives acceleration as a function of time
These functions (derivatives) can then be evaluated at
the instance in time of interest to calculate an
instantaneous velocity or acceleration value.
define a 1-D position function in the x –
direction as a function of time, for example
x(t) = 5t3-3t2+8t+9 , where x is in meters , t is in seconds
Then the velocity function is the derivative,
so…
dx/dt = v(t) = 15t2-6t+8 , where v is in m/s
To get acceleration, take the derivative of the
velocity function or the second derivative of
position
dv/dt = a(t) = 30t–6, where a is in m/s2
AVERAGE
INSTANTANEOUS
Average velocity
Instantaneous velocity
vavg
x x(t f )  x(ti )


t
t f  ti
Average acceleration
aavg
v v(t f )  v(ti )


t
t f  ti
d
vinst (t )  x(t )
dt
Instantaneous velocity
d
ainst (t )  v(t )
dt
Given position as a function of time as shown below, find
the average velocity during the interval of time from
t=1sec to t=3sec.
x(t) = 5t3-3t2+8t+9, where x is given in meters
Since…
So…
vavg 
x x(t f )  x(ti )

t
t f  ti
we must first find values of position
for time t=1 sec and t=3 sec.
x(3) = 5(3)3-3(3)2+8(3)+9 = 141
x(1) = 5(1)3-3(1)2+8(1)+9 = 19
Then plug into the formula
The answer is:
vavg
x x(3)  x(1) 141  19



 61
t
3 1
2
vavg = 61 m/s
Given position as a function of time as shown below, find the
instantaneous velocity during at t=2 sec.
x(t) = 5t3-3t2+8t+9, where x is given in meters
Since velocity is the slope of the position function, we need to take the
derivative of the position function with respect to time. Use the power
rule for each individual term.
d
v(t )  x(t )  3 * 5t 2  2 * 3t  1* 8  15t 2  6t  8
dt
Now that we have velocity as a function of time, we can simply evaluate
the function at t=2 .
2
inst
v (2)  152  62  8  56
So the answer is…
vinst = 56 m/s
Given velocity as a function of time as shown below, find
the average acceleration during the interval of time from
t=1sec to t=3sec.
v(t) = 15t2 - 6t + 8 , where v is in m/s
Since…
So…
aavg
v v(t f )  v(ti )


t
t f  ti
we must first find values of velocity
for time t=1 sec and t=3 sec.
v(3) = 15(3)2 - 6(3) + 8 = 125
v(1) = 15(1)2 - 6(1) + 8 = 17
Then plug into the formula
The answer is:
aavg
v v(3)  v(1) 125  17



 54
t
3 1
2
aavg = 54 m/s2
Given position as a function of time as shown below, find the
instantaneous velocity during at t=2 sec.
v(t) = 15t2 - 6t + 8 , where v is in m/s
Since acceleration is the slope of the velocity function, we need to take
the derivative of the velocity function with respect to time. Use the
power rule for each individual term.
d
a(t )  v(t )  (2 *15)t  (1* 6)  30t  6
dt
Now that we have acceleration as a function of time, we can simply evaluate
the function at t=2 .
ainst (2)  30(2)  6  54
So the answer is…
vinst = 54 m/s
To find average and instantaneous values
for velocity and acceleration in two
dimensional motion, given position as a
function of time in the vector format
using the unit vectors i and j, simply treat
each direction independently as if it were
a 1-D problem like the previous
examples. Be sure to put the results back
together to express your answer in vector
(i, j) form.
To find average and instantaneous values
for velocity and acceleration in three
dimensions, given position as a function
of time in the vector format using the unit
vectors i, j, and k, simply treat each
direction independently as if it were a 1D problem like the previous examples.
Be sure to put the results back together to
express your answer in vector (i, j,k)
format.
So…if the derivative of position with respect to time, x(t), or the slope of the
position-time graph gives velocity as a function of time…
…How do we go from velocity as a function of time to find the position as a
function of time?...
…We would need to find a function of time that, when its derivative is taken,
would give us the velocity function we are given.
…Could it be called an “Anti-derivative”? Try it!
Given a velocity function
v(t) = 2t + 5
find the function whose derivative will give you this answer…
What did you get? … see the next page for the answer.
Did you get something like this?
x(t) = t2 + 5t
+C
You should have gotten this…but …What is that C all about?, you ask.
Think back to the power rule of taking a derivative…
If x(t) = a * tn ,
then the derivative is dx/dt = n*a* tn-1
If you take the derivative of a constant, for example 3, we could write that in
the form given by the power rule as 3t0, since t0 = 1. Then apply the power
rule…0*3t-1 which equals zero. So the derivative of a constant is always
zero! When we take the anti-derivative we have to be sure to put the
constant back in because chances are it is not zero.
So how do we solve for that constant (C)? If you knew the value of the
position function at some time, such as x = 3 when t = 0, then you can
substitute the numbers into the function and solve for C.
3 = (0)2 + 5 (0) + C,
so C = 3
Now our function is complete!
x(t) = t2 + 5t + 3
Remember, from Physics 1, that the “displacement” can be
calculated using the “area under the curve” and most graphs were
given so that they could be divided up to calculate the area of
simple geometric shapes.
A1 = ½ bh = ½ (2sec)(10m/s) = 10 m
A2 = l*w = (5 sec)(10 m/s) = 50 m
A3 = ½ bh = ½(3 sec)(10 m/s) = 15 m
Velocity (m/s)
10
Total Area = A1+A2+A3
A2
A1
2
A3
7
10
Time (s)
Total Displacement =75 m
Velocity (m/s)
But…What do we do if the velocity graph is actually curved? What we may
not have mentioned in Physics 1 is that finding the area under a curve is one
of the main functions for Calculus. In essence, when you find the area, you
are then doing calculus…you have just been doing it with geometry formulas
instead of actual calculus operations.
Now we are going to introduce the
calculus method of finding the area.
It is called “finding the integral” and
the notation for finding the integral of
a function like v(t) is
v(t) = function of the line
∫ v(t) dt
2
Time (s)
This is read as “the integral of v as a function of
time with respect to time” (the variable in this case)
which finds the area under the curve…which just
happens to represent the displacement during that
time interval of the object that followed that motion.
Let’s look at what this process of finding the area under the curve of
any function really involves…first let’s have a graph to start with.
In general,
v(t) = 2t2 + 3
which looks kind of like this…
Velocity (m/s)
We can approximate the area by using a
series of rectangles of equal width (time
interval) and adding up their areas.
Area = length * width = v(t) * ∆t
∆t
∆t
∆t
∆t
∆t
Time (sec)
Area  x(t )  lim v(t ) * t
t 0
As you can see, there is plenty of error in this
method. We can reduce the amount of error
(empty space) by decreasing the width of the
rectangles. This method is called “finding the
Riemann Sum”. As we decrease the width
(∆t), theoretically to zero, we find that we are
taking the limit.
This limit becomes the definition of the
Integration operation…so
x(t)=∫ v(t) dt
So…we have determined that the integral of the velocity function will give
us time. We can find an “anti-derivative” or “integral” for simple
functions by simply thinking backwards…what if it gets more complicated
than simple numbers?
The simple rule for power functions is like this…
Given:
v(t )  at
Then:
a n1
v
(
t
)
dt

t

n 1
n
Given the velocity function
v(t) = 6t2 – 2t + 5
find the position function x(t)
Note: x = 2 at t = 0
First, write the integral expression…x(t)=∫(6t2 – 2t + 5) dt
Then, apply the power rule to each term of the expression and
don’t forget about the constant…


x(t )   6t 2  2t  5 dt 
6 ( 2 1)
2 1
t
 121 t (11)  051 t ( 01)  C  2t 3  t 2  5t  C
Now, plug in the values of x and t that you have been given to find C
2  2(0) 3  0 2  5(0)  C
then, C  2
x (t )  2t 3  t 2  5t  2
So…If the area under the curve is the displacement
and finding the integral calculates the area under
the curve, then we know that
x(t) = ∫ v(t) dt
To find the position, integrate the velocity
It just so happens that finding the velocity from
acceleration works the same way. Since the velocity
is the area under the curve of a velocity-time graph,
so finding the integral of the acceleration function will
give us the velocity function.
v(t) = ∫ a(t) dt
To find the velocity, integrate the acceleration
Note: a(t) means that acceleration can no longer be assumed to be constant, instead it may change as a
function of time! Also, all quantities can be calculated using this method for function in 2 and 3 dimensions.
Okay. I give up! It is after 1 am for the 4th
night in a row that I have worked on this thing.
My alarm goes off at 6 am, so I am tired and I
don’t even know if any of this makes sense. I
promise, it did in my head while I was
typing…but who knows. I hope it helps! Good
luck!
S.Ingle