1-dimensional KinematicsKinematics

Here are our old friends, the kinematic equations ☺

or

221

o

oo

t

vaatvv

attvxx

∆=+=

++=

( )

2

2

or

22

o

o

o

vvv

xavv

taatvv

+=

∆+=

=+=

Note: these equations are valid for constant acceleration.

Average Speed, Velocity, and Acceleration

Average Speed and Average Velocity

§ Average speed: how fast a particle is moving.

always positivedistance

average speedelapsed time

=

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

always positiveelapsed time

sign gives directiont

xv

∆∆

=

Velocity 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?

xv

∆=

ssm

mt 1000

/2020000

1 ==∆

v

xt

t

xv

∆=∆

∆∆

=s

sm

mt

sm

667/30

20000/20

2 ==∆

smss

m

t

xv /0.24

667100040000

=+

=∆∆

=

Average Acceleration

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

sign determines directionx∆ ∆ sign determines direction

ave

v ta

t t

∆ ∆ ∆ = =∆ ∆

Acceleration ProblemIt takes the motorist one minute to change his speed from 20 m/s to 30 m/s. What is his average acceleration?

t

vv

t

va o

∆−

=∆∆

=

2/17.060

/20/30

sm

s

smsm

tt

=

−=

∆∆

Average Velocity from a Graph

x AB∆x

∆t

ave

xv

t

∆=∆

t∆t

Average Acceleration from a Graph

vA

B∆v

∆tt

∆t

ave

va

t

∆=∆

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

0

1

2

3x(m)

A-1

-2

0 0.1 0.2 0.3 0.4 0.5-3

t(s)

A

B

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

0

1

2

3x(m)

A0

-1

-2

0 0.1 0.2 0.3 0.4 0.5-3

t(s)

A

B

Instantaneous Speed, Velocity, and Acceleration

Velocity from a Graph

x AB

t

The average velocity between A and B is the slope of the connecting line.

Velocity from a Graph

xB

t

The instantaneous velocity is the slope of the line tangent to the curve at the point of interest.

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

0

1

2

3x(m)

A0

-1

-2

0 0.1 0.2 0.3 0.4 0.5-3

t(s)

A

B

Average and Instantaneous Accelerationv

Average

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

A

t

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

the curve on a v/t graph.

B

C

x Instantaneous acceleration is positive where curve is concave

Instantaneous acceleration is zero where slope is constant

Average and Instantaneous Acceleration

t

Instantaneous acceleration is negative where curve is concave down

positive where curve is concave up

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?

v

t

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?

x

t

Estimate the net displacement from 0 s to 4.0 s

v (m/s)

2.0

t (s)2.0 4.0

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

a (m/s2)

1.0

t (s)2.0 4.0 t (s)2.0 4.0

-1.0

Derivatives

Sample problem. From the position-time graph draw the corresponding velocity-time graph

x

t

v

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) m§ x = (-4 + 3t) m§ x = (1.0 + 2.0t – ½ 3 t2) m§ x = -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

ave

xv

t

∆=∆

( )0 0

lim liminst avet t

x dxv v

t dt→ →

∆ = = = ∆

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

Instantaneous Acceleration

ave

va

t

∆=∆

( )0 0

lim limavet t

t

v dva a

t dt→ →

∆∆ = = = ∆

§ 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.

x∆ 2

20limt

x

d xta

t dt→

∆ ∆ ∆ = =∆

§ Just don’t let the new notation scare you; think of the d as a baby ∆, 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 At

dxv nAt

dt−

=

= =

Kinematics ProblemA particle travels from A to B following the function x(t) = 3.0 – 6t + 3t2.

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

ttdt

dxtv 66)3(260)( +−=+−== 660)( =+==

dt

dvta

b) What is the instantaneous velocity at 6 seconds?

c) What is the initial velocity?

ttdt

tv 66)3(260)( +−=+−== 660)( =+==dt

ta

30)6(66)6(

66)(

=+−=

+−=

v

txv

6)0(66)0(

66)(

−=+−=

+−=

v

txv

Kinematics Problem

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

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

22 364)(3)3(240)( ttttdx

tv −+−=−+−==

b) What is the instantaneous acceleration at 6 seconds?

22 364)(3)3(240)( ttttdt

dxtv −+−=−+−==

ttdt

dvta 66)3(260)( −=−+==

30)6(66)6( −=−=a

A particle follows the function:

• Find the velocity and acceleration functions.

Kinematics Problem

tttx 52.45.1)( 2 +−= −

2

4.21.5 5x t

t= − +

33 4.85)2.4)(2(0)( −− =+−−== ttdt

dxtv

• Find the instantaneous velocity and acceleration at 2.0 seconds.

44 2.25)4.8)(3()( −− −=−== ttdt

dvta

05.124.8

)2(4.8)2( 33 === −v 575.1

22.25

)2(2.25)2( 44 =−=−= −a

Another problem

A dog wanders around with a displacement (in m) that follows the equation

What is the dog’s average velocity between 0s and 20s?

33001204.02 ttx +−=

and 20s?

( ) ( )( ) ( )

sms

mm

t

xv

st

mx

mx

/535.020

27.12

20

7.12202004.02)20(

20004.02)0(3

30012

330012

=−

=∆∆

=

=∆

=+−=

=+−=

Integrals

anti-derivatives (integrals)

• But how to go from a à v à x?

– Graphs: determine the area under the curve– Equations: integrate!– Equations: integrate!

Evaluating Polynomial Integrals

Consider a general polynomial function for acceleration, dependent on time.

dtAtadtv

Ata

n

n

==

=

∫∫

consttn

A

dttA

dtAtadtv

n

n

n

+

+

=

=

==

+

∫∫∫

1

11

Relationships between x, v, and a

Derivatives Integrals

dxv

x

= ∫= adtv

a

const. = v

2

2

dt

xd

dt

dva

dt

dxv

==

=

∫∫

=

=

vdtx

adtv const. = vo

const. = xo

Note: these equations are also valid for changing acceleration!

Integral ProblemA particle accelerates at 4 m/s2 .

Determine equations for the velocity and displacement of the particle, if at time t=0 it had a velocity of 3m/s and a displacement of 4m.

m/sin 3444 +=+=== ∫∫ tctdtadtv

min 43234)34( 2221 ++=++=+== ∫∫ ttcttdttvdtx

Integral Problem

A particle moves according to the equationin m/s

Determine its displacement from 2s to 5s.36 −= tv

( )dttvdtx 365

−== ∫∫ ( )

[ ][ ]( ) ( )

m

cc

ctt

ctt

dttvdtx

54

660

)2(3)2(3)5(3)5(3

33

36

36

22

5

22

5

22

21

2

=

−=

+−−+−=

+−=

+−=

−== ∫∫

Final note: Free Fall

§ Free fall means that an object is moving 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.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.

Practice Problems

Draw representative graphs for a particle which is stationary.

x v a

t

Positionvstime

t

Velocityvstime

t

Accelerationvstime

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

x v a

t

Positionvstime

t

Velocityvstime

t

Accelerationvstime

x v a

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

acceleration.

t

Positionvstime

t

Velocityvstime

t

Accelerationvstime

Sample problem: A body moving with uniform acceleration has a velocity of 12.0 m/s in the positive x direction when its x coordinate is 3.0 m. If the x coordinate 2.00 s later is -5.00 m, what is the magnitude

of the acceleration?

smv

mx

mxo

/12

5

3

=

−=

=2

21

221

tvxx

attvxx

attvxx

oo

oo

−−

=−−

++=

st

smvo

2

/12

=

=

( )2

221

221

/16

2)2)(/12(35

sma

s

ssmmma

at

tvxx oo

=

−−−=

=−−

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?

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 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?