AP C UNIT 11-Dimension

•A quick review of basic kinematic variables

•Difference in math application towards physics

•Differentiation (Calculus)

Definitions

• Distance (x)• Displacement ( ) • Speed (v)• Velocity ( )• Avg = 2 pts in time, Instantaneous = 1pt in time

x

v

Definitions

• Acceleration ( ) • If v = constant, then a = ?• Velocity is maximum when acceleration is• Position is maximum when velocity is• Be careful of signs

a

zerozero

Calculus vs Algebra

• In HP, all problems were solved using a constant variable in the equations.

• Graphical analysis used lines of constant slope.• In APC, it requires utilizing the tool of differential

& integral calculus to further probe physics principles where variables might be changing.

• The “down & dirty” calculus skills presented here are not meant to replace those you learned/about to learn in your calculus class. Note a mixed ability group of Calc AB and BC.

Calculus – Differentiation & Slope

Last year we formulated linear regression lines of best fit, chose 2 points on the line, and determined the slope. No matter what 2 points were chosen, obviously, the slope was always the same since it was constant.

But what happens when we have a non-linear function? In the graph below, the slope varies from point to point. Trying to calculate Δy/Δx from A to B we would yield a negative slope, from B to C it would yield a positive slope, and from A to C, a zero slope.

To find the slope at ‘A’, we can examine a point just above and just below ‘A’. The closer those points are to ‘A’ (and to each other), the more accurately they would describe the slope at that point.

This idea is to compute the rate of change as the limiting value of the ratio of the differences Δy/Δx as Δx becomes infinitely small. It follows that this limit is the exact slope of the tangent.

Such a tiny or infinitesimal change in x and y is denoted by dx and dy, where we replace Δy/Δx with dy/dx.

• Differentiating or taking the derivative, f’(x), of a function, f (x), equates to the slope of the tangent line at a point when the run is reduced to such a small value…an instantaneous point

• To find instantaneous velocity at a point in time, we use the following notation (recall Δ was for 2pts in time)

limΔt 0 dt

dx

t

x

dx is a differential distance and dt is a differential time which is a fancy way of saying very, very small. In technical terms, dx is what happens to Δx in the limit when Δx approaches zero.

instv

• To find instantaneous acceleration at a point in time, we use the following notation

2

2

)(dt

xd

dt

dx

dt

d

dt

dvainst

1st derivative of velocity or 2nd derivative of displacement

Calculus Rules - Differentiation(down & dirty method…your calc teacher will fill in the details)

Rule #1: Derivative of a Constant

*Derivative of a constant is zero

If f(x) = 5, then f’(x) = 0.

Rule #2: Power Rule

If f(x) = xn , then f’(x) = nxn-1.

If f(x) = 5x2 + 3, then f’(x) = 2(5x1) + 0 = 10x…using derivative of constant & power rule.

NOTE that we are just finding slopes of a tangent using these rules

When a function reaches its maximum or minimum value, it must turn around where its slope = 0

Maximum and Minimum points (ie; when does max velocity occur?)

Relative maxima

relative minimum

y

x1 x2 x3 x

Thus, the derivative when evaluated at x1 , x2 ,x3 will equal zero and allows us to determine where max or min occurs.

Example: f(x) = x3 – 6x2 + 9x + 1

To find max or min values, set f’(x) = 0 and solve.f’ (x) = 3x2 – 12x + 9 = 0 = 3(x2 – 4x + 3) = 0 x = 1 and x = 3 (extreme points)

This is where the extreme points are….are they a max or min?

2nd Derivative Test: To find if it’s max or min (concavity), take 2nd derivative of functionf’’(x) = 6x-12Plug in extrema value. If f’’(x) < 0, then it’s a Max If f’’(x) > 0, then it’s a Minx = 1 is max since you get -6x = 3 is min since you get +6You could also plug in values above and below the extrema values (1 & 3) into the 1st derivative to determine if slope is incr or decreasing to determine concavity.Essentially, the 2nd derivative is looking at how the 1st derivative is changing.

ExampleThe position of a particle is given by:

x(t) = 7.8 + 9.2t – 2.1t3

a) Find the velocity of particle at t = 3.0s.

b) Find the maximum position of particle.

Practiceworksheet

Calculus - IntegrationIntegration is the reverse of differentiation where an integral is called the anti-derivative. Much like finding the slope of the tangent line at a point for differentiation, integration is finding the area under the curve or function. Like finding area for a v-t graph which equals displacement.

Last year we only looked area for linear functions such as:

However, if the function is not linear, we must use calculus. We could estimate the area by breaking the curve up into many rectangles and summing them up which can approximate the area. The more rectangles we use (smaller width), the better the estimation.

Consider the function at the left where we wish to find area from A to C. The problem is that the height keeps changing as we move from A to C along the function.

In order to minimize this problem (height changing), we focus on a very small region of the graph, where the height is relatively stable. Start by picking a point x along x-axis and another point extremely close but beyond it, (x+dx). Drawing vertical lines at these two points where they meet the function yields a shaded region.

If we treat this region as a rectangle, its area is equal to the height f(x) times the tiny width dx.

Obviously the shaded region isn't a rectangle, but as dx becomes smaller—as we bring the right side toward the left—the height change becomes less significant, and the region more closely resembles a rectangle.

We now just use an infinite amount of tiny rectangles and sum them up.

As dx approaches zero, this approximation becomes perfect: the height will equal f(x) on both left and right ends of rectangle…the area of the shaded region = f(x)*dx.

C

A

dxxf )(

‘S’ looking symbol means to sum up

Calculus Rules - IntegrationRule #1: Power Rule

Cxdxax nnan 1

1

where C is a constant and n ≠ -1

If f(x) = 6x2, then the integral of that function is:

DEFINITE INTEGRALS (these integrals are evaluated between 2 values)

Consider f(x) = 4x2 + 5x + 3. Evaluate the integral between the values of 1 and 2.

For each of the following expressions determine the indefinite integral with respect to time: (a) v(t) = 6t3 − 5t(b) a(t) = 3t2 − 4t + 7(c) a(t) = 10 + t−2

(d) v(t) = 2t5 − t4/3

Using the Constant, C.Acceleration of a bus is given by:

a(t) = 1.2t

a) If vbus = 5.0m/s at t = 1.0s, what is vbus at t = 2.0s?

b) If position of bus is 6.0m at t = 1.0s, find position at t = 2.0s?

Constant Acceleration Eqns

v

v

t

o

dvadtdt

dva

0

atvvvvat oo

x

x

t

o

dxvdtdt

dxv

0

x

x

tt

o

x

x

t

o

o

o

dxtdtadtv

dxdtatv

00

0)(

From above

2

21 attvx o

Use substitution to get other two equations

a

vvt o

t

vva o

tvv

x o )2

(

axvv o 222

Example: A car sits at a light. When light turns green, it accelerates at a constant rate of 2.5m/s2. At the moment the acceleration begins, a truck moves past the car with a uniform speed of 15m/s.

At what position beyond the starting point does the car overtake the truck?

A pair of identical balls are simultaneously released on a pair of equal length tracks, A and B, as shown. Friction is minimal. Both balls reach the ends of their tracks at the samea)timeb)speedc)Both of thesed)None of these

QUESTION

Graphical Analysis

i) Describes position of particle with respect to timeii) Slope (Δx/Δt) indicates the velocity of the particle (mag & dirn)

Position vs Time (x vs t):

Average Velocity (displacement/time) is slope of the secant line

Instantaneous Velocity is slope of the tangent line

Velocity vs Time (v vs t):i) Describes velocity of particle with respect to timeii) Slope (Δv/Δt) indicates the acceleration of the particle (mag &

dirn)

Average Acceleration (velocity/time) is slope of the secant line

Instantaneous Acceleration is slope of the tangent line

Note that particle can have negative velocity and positive acceleration

Area indicates the displacement of particle

Acceleration vs Time (a vs t):

i) Describes acceleration of particle with respect to time

ii) Area equates to change in velocity

Sketch the x vs t, v vs t, and a vs t graphs for the diagram assuming the ball stays in contact with the surface of ramp at all times.

x v a

*Graph worksheet

Freefall

Assume no air friction where acceleration due to gravity on Earth is given by a = ‘g’ = -9.8m/s2.

Objects in freefall are always accelerating downwards towards center of Earth.

Can change based on location.

In 1D, what is the value of velocity at apex of flight?

What is the value of acceleration at apex?

A balloonist, riding in the basket of a hot air balloon that is rising at 10.0m/s, releases a sandbag when the balloon is 40.8m above the ground. Determine the velocity of the bag when it hits the ground.

Sketch a graph of x, v, and a vs t for motion of sandbag

A model rocket is fired from rest vertically and ascends with a constant vertical acceleration of 4.0m/s2 for 6.0s. Its fuel is then exhausted and it continues as a free-fall particle. What is the total time elapsed from takeoff to striking the Earth?

A. TERMS DEFINED

*SCALAR *VECTOR

22 bac

B. ADDITION / SUBTRACTION METHODS

*2 OR MORE VECTORS COMBINED YIELDS RESULTANT

*GRAPHICAL

HEAD 2 TAIL, PARALLELOGRAM

*ANALYTICAL

VECTORS

*SUBTRACTION

C. COMPONENTS

aX= acosθ ay= asinθ

)( baba

22yx aaa

x

y

a

atan

i

j

k

E. UNIT VECTOR

•Vector with magnitude = 1 w/ direction•Lacks dimension & unit•Purpose is to ‘point’

LABELED

ˆ = hat, replaces

jaiaa yxˆˆ

a

bthese are vector

components of and jbibb yxˆˆ

ax and ay are scalar components of a

EXAMPLE

Find the vector sum,

r in unit vector notation.

a = (4.2m)

i - (1.5m)

j

b = (-1.6m)

i + (2.9m)

j

c = (-3.7m)

j

jyixr ˆˆ

jyyixxrrr ˆ)(ˆ)( 121212

Find displacement of particle as it moves from

jir ˆ2ˆ31 to jir ˆ4ˆ92

.

The position of a particle that moves in both the x & y plane at the same time can be described by:

< position vector >

2-D Motion

t

rvavg

Similarly, it’s velocity is described as:

jdt

dyi

dt

dx

dt

rdvinst ˆˆ

instv has a direction that is always tangent to path of

particle

2

2

dt

rd

dt

vda

Example:

jtittr ˆ)2

11(ˆ)43( 23

find instv

and insta

for arbitrary times.

It’s acceleration is described as:

Projectiles

Neglecting air resistance,

Horizontal velocity = constant

Vertical velocity changes

Horizontal acceleration = zero

Vertical acceleration = ‘g’

tvtvxrange oox cos

gtvgtvv ooyy sin

22

2

1sin

2

1gttvygttvyy oooyo

a) Rank the paths (use only the symbols > or = , for example, a>b=c) according to time of flight, greatest first.

b) Rank the paths (use only the symbols > or = , for example, a>b=c) according to initial speed, greatest first.

y

x

Derive an expression to solve for minimum initial v to just clear gap (ignoring bike width)

θ

Relative VelocityUsed whenever you see “velocity relative to” or “velocity with respect to”, addition and subtraction of velocities is done so by utilizing a subscript method.

vAC = vAB + vBC

Label each object with its F.O.R.

You want the inner subscripts to match up

Switching the order of subscripts is like multiplying by negative 1 where vAB = -vBA

vAC refers to velocity of A relative to C

Example: A car is moving north at 88km/h when a truck approaches it from the other direction moving at 104km/h.a) What is the truck’s velocity relative to the car?

b) What is the car’s velocity relative to truck?

c) How do relative velocities change after the pass each other?

2D example:A supersonic aircraft is moving in still air with a constant velocity of . jhkmihkm ˆ)/20(ˆ)/200(

Suddenly, at t = 0 a wind gusts with a velocity of . Assuming the pilot makes no attempt to correct for wind, find plane’s displacement after 1hr relative to ground.

jthkmithkm ˆ)/30(ˆ)/20( 232

A canoe has a velocity of 0.40m/s SE relative to the Earth. The canoe is on a river flowing 0.50m/s East relative to the Earth. Find velocity of canoe relative to the river.