Kinematics and One Dimensional Motion:

Non-Constant Acceleration

8.01W01D3

Today’s Reading Assignment: W02D1

Young and Freedman: Sections 2.1-2.6

Kinematics and One Dimensional Motion

Kinematics Vocabulary

• Kinema means movement in Greek

• Mathematical description of motion– Position– Time Interval– Displacement– Velocity; absolute value: speed– Acceleration

– Averages of the latter two quantities.

Coordinate System in One Dimension

Used to describe the position of a point in space

A coordinate system consists of:

1. An origin at a particular point in space2. A set of coordinate axes with scales and labels3. Choice of positive direction for each axis: unit vectors 4. Choice of type: Cartesian or Polar or Spherical

Example: Cartesian One-Dimensional Coordinate System

Position • A vector that points from origin to body.• Position is a function of time• In one dimension:

ˆ( ) ( )t x t=x ir

Displacement Vector

Change in position vector of the object during the time interval Δt=t2 −t1

Concept Question: Displacement

An object goes from one point in space to another. After the object arrives at its destination, the magnitude of its displacement is:

1) either greater than or equal to

2) always greater than

3) always equal to

4) either smaller than or equal to

5) always smaller than

6) either smaller or larger than

the distance it traveled.

Average Velocity The average velocity, , is the displacement divided by the time interval

The x-component of the average velocity is given by

rvave(t)

rvave ≡

Δrr

Δt=

ΔxΔt

i =vave,x(t)i

v

ave,x=

ΔxΔt

Δrr

Δt

Instantaneous Velocityand Differentiation

• For each time interval , calculate the x-component of the average velocity

• Take limit as sequence of the x-component average velocities

• The limiting value of this sequence is x-component of the instantaneous velocity at the time t .

tΔ

0tΔ → v

ave,x(t) =Δx / Δt

limΔt→ 0

ΔxΔt

= limΔt→ 0

x(t+ Δt) −x(t)Δt

≡dxdt

vx(t) =dx / dt

Instantaneous Velocity

x-component of the velocity is equal to the slope of the tangent line of the graph of x-component of position vs. time at time t

v

x(t) =

dxdt

Concept Question: One Dimensional Kinematics

The graph shows the position as a function of time for two trains running on parallel tracks. For times greater than t = 0, which of the following is true:

1. At time tB, both trains have the same velocity.

2. Both trains speed up all the time.3. Both trains have the same velocity

at some time before tB, . 4. Somewhere on the graph, both

trains have the same acceleration.

Average Acceleration

Change in instantaneous velocity divided by the time interval

The x-component of the average acceleration

ra

ave≡

Δrv

Δt=

Δvx

Δti =

(vx,2 −vx,1)

Δti =

Δvx

Δti =aave,xi

a

ave,x=

Δvx

Δt

Δt=t2 −t1

Instantaneous Accelerationand Differentiation

• For each time interval , calculate the x-component of the average acceleration

• Take limit as sequence of the x-component average accelerations

• The limiting value of this sequence is x-component of the instantaneous acceleration at the time t .

tΔ

0tΔ → a

ave,x(t) =Δvx / Δt

limΔt→ 0

Δvx

Δt= lim

Δt→ 0

vx(t+ Δt) −vx(t)Δt

≡dvx

dt

ax(t) =dvx / dt

Instantaneous Acceleration

The x-component of acceleration is equal to the slope of the tangent line of the graph of the x-component of the velocity vs. time at time t

a

x(t) =

dvx

dt

Table Problem: Model Rocket

A person launches a home-built model rocket straight up into the air at y = 0 from rest at time t = 0 . (The positive y-direction is upwards). The fuel burns out at t = t0. The position of the rocket is given by

 

 

with a0 and g are positive. Find the y-components of the velocity and acceleration of the rocket as a function of time. Graph ay vs t for 0 < t < t0.

 

 

2 6 400 0 0

1( ) / ; 0

2 30

ay a g t t t t t

⎧= − − < <⎨⎩

Non-Constant Accelerationand Integration

The area under the graph of the x-component of the acceleration vs. time is the change in velocity

Velocity as the Integral of Acceleration

010

( ) lim ( ) ( , )i

t t i N

x x i i xt

it

a t dt a t t Area a t′= =

Δ →=′=

′ ′≡ Δ =∑∫

ax( ′t )d ′t

′t =0

′t =t

∫ =dvx

d ′td ′t

′t =0

′t =t

∫ = d ′vx′vx =vx t=0( )

′vx =vx t( )

∫ =vx(t) −vx,0

Position as the Integral of Velocity

Area under the graph of x-component of the velocity vs. time is the displacement

( )x

dxv t

dt≡

0

0

( ) ( )t t

x

t

v t dt x t x′=

′=

′ ′= −∫

Ei is the error in the approximation for each interval

Summary: Time-Dependent Acceleration

• Acceleration is a non-constant function of time

• Change in velocity

• Change in position

,0

0

( ) ( )t t

x x x

t

v t v a t dt′=

′=

′ ′− = ∫

( )xa t

0

0

( ) ( )t t

x

t

x t x v t dt′=

′=

′ ′− = ∫

Table Problem: Sports CarAt t = 0 , a sports car starting at rest at x = 0 accelerates with an x-component of acceleration given by

and zero afterwards with

• Find expressions for the velocity and position vectors of the sports as functions of time for t >0.

(2) Sketch graphs of the x-component of the position, velocity and acceleration of the sports car as a function of time for

t >0

α , β > 0

Concept QuestionA particle, starting at rest at t = 0, experiences a non-constant acceleration ax(t) . It’s change of position can be found by

1. Differentiating ax(t) twice.

• Integrating ax(t) twice.

• (1/2) ax(t) times t2.

• None of the above.

• Two of the above.

Next Reading Assignment:

Young and Freedman: Sections 3.1-3.3, 3.5