Kinematics and One Dimensional Motion: Non-Constant Acceleration 8.01 W01D3.
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Transcript of Kinematics and One Dimensional Motion: Non-Constant Acceleration 8.01 W01D3.
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