2.1 - Kinematics - Mr.Niehaus

2.1 - Kinematics

Kinematics = the branch of mechanics that deals with pure motion, without reference to the masses or forces involved in it.

Average Speed and Average Velocity

Average speed describes how fast a particle is moving. It is calculated by:

Average velocity describes how fast the displacement is changing with respect to time:

always positivedistanceaverage speedelapsed time

=

avexvt

Δ=Δ sign gives direction

Note: Δ = “change in”

Average Acceleration

Average acceleration describes how fast the velocity is changing with respect to time. The equation is:

sign determines directionave

xv tat t

Δ⎛ ⎞Δ⎜ ⎟Δ Δ⎝ ⎠= =Δ Δ

Sample problem: A motorist drives north at 20 m/s for 20 km and then continues north at 30 m/s for another 20 km. What is his average velocity?

Sample problem: A motorist drives north at 20 m/s for 20 km and then continues north at 30 m/s for another 20 km. What is his average velocity?

Given: v1 = +20 m/s (Note positive sign!)x1 = +20 kmt1 = ?v2 = +30 m/s x2 = +20 kmt2 = ?vav = ?

avexvt

Δ=Δ

Sample problem: A motorist drives north at 20 m/s for 20 km and then continues north at 30 m/s for another 20 km. What is his average velocity?

Given: v1 = +20 m/s (Note positive sign!)x1 = +20 km = 20000 mt2 = ?v2 = +30 m/s x2 = +20 km = 20000 mt2 = ?vav = ?

V = x/t

t = x/v t1 = 20,000m / 20m/s = 1000 seconds

t2 = 20,000m / 30m/s = 666.6 seconds

vav = Δx / Δt = (20000+20000) / (1000+666.6)= 24 m/s

avexvt

Δ=Δ

Sample problem: A motorist drives north at 20 m/s for 20 km and then continues north at 30 m/s for another 20 km. What is his average velocity?

Given: v1 = +20 m/s (Note positive sign!)x1 = +20 kmt2 = ?v2 = +30 m/s x2 = +20 kmt2 = ?vav = ?

0 1000 1666 2000

Distance (m)

Time (s)

40,000

20,000

vav

*Note that the average velocity is the slope of the line

Sample problem: It takes the motorist one minute to change his speed from 20 m/s to 30 m/s. What is his average acceleration?

Sample problem: It takes the motorist one minute to change his speed from 20 m/s to 30 m/s. What is his average acceleration?

Given: t1 = 0 t2 = 1 min = 60 secv1 = 20 m/sv2 = 30 m/s

aav = Δv/Δt = (v2 - v1) / (t2 - t1)

= (30 - 20) / (60 - 0)

= 0.17 m/s2

Average Velocity from a Graph

t

xA

BΔx

Δt

avexvt

Δ=Δ

Average Velocity from a Graph

avexvt

Δ=Δ

t

x ABΔx

Δt

Average Acceleration from a Graph

t

vA

BΔv

Δt

avevat

Δ=Δ

• Sample problem: From the graph, determine the average velocity for the particle as it moves from point A to point B.

0

-1

-2

1

2

0 0.1 0.2 0.3 0.4 0.5-3

3

t(s)

x(m)

AB

• Sample problem: From the graph, determine the average speed for the particle as it moves from point A to point B.

0

-1

-2

1

2

0 0.1 0.2 0.3 0.4 0.5-3

3

t(s)

x(m)

AB

S = Δx/Δt = (1 - 0) / (0.5-0) = 2 m/s

2.2 - Instantaneous Speed, Velocity, and Acceleration

Average Velocity from a Graph

t

x

Remember that the average velocity between the time at A and the time at B is the slope of the connecting line.

AB

t

x

What happens if A and B become closer to each other?

AB

Instantaneous Velocity from a Graph

t

x

What happens if A and B become closer to each other?

A B

Instantaneous Velocity from a Graph

t

x

AB

What happens if A and B become closer to each other?

Instantaneous Velocity from a Graph

t

x

A

B

What happens if A and B become closer to each other?

Instantaneous Velocity from a Graph

Instantaneous Velocity from a Graph

t

x

AB

The line “connecting” A and B is a tangent line to the curve. The velocity at that instant of time is represented by the slope of this tangent line. Calculus is required to find this tangent line = beyond the scope of this class. However it can be approximated using graphing techniques.

A and B are effectively the same point. The time difference is effectively zero.

• Sample problem: From the graph, determine the instantaneous speed and instantaneous velocity for the particle at point B.

0

-1

-2

1

2

0 0.1 0.2 0.3 0.4 0.5-3

3

t(s)

x(m)

AB

Draw a tangent line at point B. The velocity is the slope of the tangent.

(.4, 1)

(.6, -3)

Instantaneous Speed =|(-3-1))/(.6-.4)| = 20 m/s

Instantaneous Velocity = (-3-1)/(.6-.4) = -20 m/s

Average and Instantaneous Acceleration

t

v

Average acceleration is represented by the slope of a line connecting two points on a v/t graph.

Instantaneous acceleration is represented by the slope of a tangent to the curve on a v/t graph.

A

B

C

t

x

Instantaneous acceleration is negative where curve is concave down

Instantaneous acceleration is positive where curve is concave up

Instantaneous acceleration is zero where slope is constant

Average and Instantaneous Acceleration

Sample problem: Consider an object that is dropped from rest and reaches terminal velocity during its fall. What

would the v vs t graph look like?

t

v

Sample problem: Consider an object that is dropped from rest and reaches terminal velocity during its fall. What

would the x vs t graph look like?

t

x

Draw representative graphs for a particle which is stationary.

x

t

Positionvs

time

v

t

Velocityvs

time

a

t

Accelerationvs

time

Draw representative graphs for a particle which is stationary.

x

t

Positionvs

time

v

t

Velocityvs

time

a

t

Accelerationvs

time

Draw representative graphs for a particle which has constant non-zero velocity.

x

t

Positionvs

time

v

t

Velocityvs

time

a

t

Accelerationvs

time

Draw representative graphs for a particle which has constant non-zero velocity.

x

t

Positionvs

time

v

t

Velocityvs

time

a

t

Accelerationvs

time

x

t

Positionvs

time

v

t

Velocityvs

time

a

t

Accelerationvs

time

Draw representative graphs for a particle which has constant non-zero acceleration.

x

t

Positionvs

time

v

t

Velocityvs

time

a

t

Accelerationvs

time

Draw representative graphs for a particle which has constant non-zero acceleration.

2.3 - Kinematic Equations

212

2 20 2 ( )

o

o o

v v at

x x v t at

v v a x

= +

= + +

= + Δ

2.3 - Kinematic Equations

Sample problem: A body moving with uniform acceleration has a velocity of 12.0 cm/s in the positive x direction when its x coordinate is

3.0 cm. If the x coordinate 2.00 s later is -5.00 cm, what is the magnitude of the acceleration?

Sample problem: A jet plane lands with a speed of 100 m/s and can accelerate at a maximum rate of -5.00 m/s2 as it comes to a halt.

a) What is the minimum time it needs after it touches down before it comes to a rest?

b) Can this plane land at a small tropical island airport where the runway is 0.800 km long?

Air track demonstration

Kinematic graphs for uniformly accelerating object.Curve-fit and equation comparison

2.4 Freefall

Free Fall

Free fall is a term we use to indicate that an object is falling under the influence of gravity, with gravity being the only force on the object.

Gravity accelerates the object toward the earth the entire time it rises, and the entire time it falls.

The acceleration due to gravity near the surface of the earth has a magnitude of 9.8 m/s2. The direction of this acceleration is DOWN.

Air resistance is ignored.

Sample problem: A student tosses her keys vertically to a friend in a window 4.0 m above. The keys are caught 1.50 seconds later. a) With what initial velocity were the keys tossed?

b) What was the velocity of the keys just before they were caught?

Sample problem: A student tosses her keys vertically to a friend in a window 4.0 m above. The keys are caught 1.50 seconds later. a) With what initial velocity were the keys tossed?

Given:Note: up is positivexf = 4.0 mxo = 0 mt = 1.50 sa = -9.8 m/s2

vo = ?

b) What was the velocity of the keys just before they were caught?

Sample problem: A ball is thrown directly downward with an initial speed of 8.00 m/s from a height of 30.0 m. How many seconds later does the ball strike the ground?

Picket Fence Lab

Use a laptop, Science Workshop 500, accessory photogate, and picket fence to determine the acceleration due to gravity. You must have position versus time and acceleration versus time data, which you will then fit to the appropriate function. Gravitational acceleration obtained in this experiment must be compared to the standard value of 9.81 m/s2.

Estimate the net change in velocity from 0 s to 4.0 s

a (m/s2)

1.0

t (s)2.0 4.0

-1.0

Estimate the net displacement from 0 s to 4.0 s

v (m/s)

2.0

t (s)2.0 4.0

Lesson

Derivatives

Sample problem. From this position-time graph

x

t

Draw the corresponding velocity-time graph

x

t

Suppose we need instantaneous velocity, but don’t have a graph?

Suppose instead, we have a function for the motion of the particle.Suppose the particle follows motion described by something like

x = (-4 + 3t) mx = (1.0 + 2.0t – ½ 3 t2) mx = -12t3

We could graph the function and take tangent lines to determine the velocity at various points, or…We can use differential calculus.

Instantaneous Velocity

avexvt

Δ=Δ

( )0 0

lim liminst avet t

x dxv vt dt→ →

Δ⎛ ⎞= = =⎜ ⎟Δ⎝ ⎠

Mathematically, velocity is referred to as the derivative of position with respect to time.

Instantaneous Acceleration

( )0 0

lim lim

ave

avet t

vat

v dva at dt→ →

Δ=Δ

Δ⎛ ⎞= = =⎜ ⎟Δ⎝ ⎠Mathematically, acceleration is referred to as the derivative of velocity with respect to time

Instantaneous Acceleration

Acceleration can also be referred to as the second derivative of position with respect to time.

2

20limt

xd xta

t dt→

Δ⎛ ⎞Δ⎜ ⎟Δ⎝ ⎠= =Δ

Just don’t let the new notation scare you; think of the d as a baby Δ, indicating a very tiny change!

Evaluating Polynomial Derivatives

It’s actually pretty easy to take a derivative of a polynomial function. Let’s consider a general function for position, dependent on time.

1

n

n

x Atdxv nAtdt

−

=

= =

Sample problem: A particle travels from A to B following the function x(t) = 2.0 – 4t + 3t2 – 0.2t3.

a) What are the functions for velocity and acceleration as a function of time?

b) What is the instantaneous acceleration at 6 seconds?

Sample problem: A particle follows the function2

4.21.5 5x tt

= − +

a) Find the velocity and acceleration functions.

b) Find the instantaneous velocity and acceleration at 2.0 seconds.

Lesson

Kinematic Graphs --Laboratory

Kinematic Graphs Laboratory• Purpose: to use a motion sensor to collect graphs of position-

versus-time and velocity-versus-time for an accelerating object, and to use these graphs to clearly show the following:A. That the slopes of tangent lines to the position versus time curve yield

instantaneous velocity values at the corresponding times.B. That the area under the curve of a velocity versus time graph yields

displacement during that time period.• Experiments: You’ll do two experiments. The first must

involve constant non-zero acceleration. You can use the cart tracks for this one. The second must involve an object that is accelerating with non-constant non-zero acceleration. Each experiment must have a position-versus-time and velocity-versus-time graph that will be analyzed as shown above.

• Report: A partial lab report in your lab notebook must be done. The report will show a sketch of your experimental apparatus. Computer-generated graphs of position versus time and velocity versus time must be taped into your lab notebook. An analysis of the graphical data must be provided that clearly addresses A and B above, and a brief conclusion must be written.

Setting up your Lab Book• Put your name inside the front cover. Write it on

the bottom on the white page edges.• For each lab:

– Record title at top where indicated– Record lab partners and date at bottom.– Record purpose under title.– List equipment.– Note computer number, if relevant

• For these labs, you will insert printouts of graphs in your lab books, and do the analysis on the graphs and in the book.