AP Physics C

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)

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.

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.

Time (sec)

Δt

Δx

slopet

xvavg

tan0

lim)( slopetv tx

tinst

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.

Tangent line

Time (sec)

Δt

Δv

slopet

vaavg

tan0

lim)( slopeta tv

tinst

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.

Tangent line

tv

t

0lim

This limit statement is read as “the limit of ∆v/∆t as ∆t approaches zero”

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”

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

if

ifavg tt

txtx

t

xv

)()(

Average velocity

Average acceleration

if

ifavg tt

tvtv

t

va

)()(

)()( txdt

dtvinst

Instantaneous velocity

Instantaneous velocity

)()( tvdt

dtainst

x(t) = 5t3-3t2+8t+9, where x is given in meters

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.

Since…if

ifavg tt

txtx

t

xv

)()( we must first find values of position

for time t=1 sec and t=3 sec.

612

19141

13

)1()3(

xx

t

xvavg

So… x(3) = 5(3)3-3(3)2+8(3)+9 = 141x(1) = 5(1)3-3(1)2+8(1)+9 = 19

Then plug into the formula

The answer is: 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.

86158*13*25*3)()( 22 tttttxdt

dtv

Now that we have velocity as a function of time, we can simply evaluate the function at t=2 .

56826215)2( 2 instv

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…if

ifavg tt

tvtv

t

va

)()( we must first find values of velocity

for time t=1 sec and t=3 sec.

So… v(3) = 15(3)2 - 6(3) + 8 = 125v(1) = 15(1)2 - 6(1) + 8 = 17

Then plug into the formula 542

17125

13

)1()3(

vv

t

vaavg

The answer is: aavg = 54 m/s2

Given position as a function of time as shown below, find the instantaneous velocity during at t=2 sec.

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.

Now that we have acceleration as a function of time, we can simply evaluate the function at t=2 .

So the answer is… vinst = 54 m/s

v(t) = 15t2 - 6t + 8 , where v is in m/s

630)6*1()15*2()()( tttvdt

dta

546)2(30)2( insta

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 1-D 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 + CYou 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.

10

Time (s)

A1A2 A3

2 10 7

A1 = ½ bh = ½ (2sec)(10m/s) = 10 m A2 = l*w = (5 sec)(10 m/s) = 50 mA3 = ½ bh = ½(3 sec)(10 m/s) = 15 m

Total Area = A1+A2+A3

Total Displacement =75 m

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.

Time (s)

2

v(t) = function of the line

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) dtThis 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…

Time (sec)

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 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

ttvtxAreat

*)(lim)(0

∆t ∆t∆t∆t ∆t

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…

1

1)(

)(

n

n

tn

adttv

attvGiven:

Then:

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) dtThen, apply the power rule to each term of the expression and don’t forget about the constant…

CtttCtttdttttx

52526)( 23)10(

105)11(

112)12(

1262

Now, plug in the values of x and t that you have been given to find C

252)(

2,

)0(50)0(22

23

23

ttttx

Cthen

C

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

v(t) = ∫ a(t) dt

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.

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