Kinematics and One Dimensional Motion 8.01t
Sept 10, 2004
Kinematics
• Kinema means movement
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Mathematical description of motion
Position
Displacement
Velocity
Acceleration
Coordinate System in One Dimension
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Choice of origin
Choice of coordinate axis
Choice of positive direction for the axis
Choice of unit vectors at each point in space
Position
• Vector from origin to body G
x(t ) = x(t ) î
Displacement
• change in position coordinate of the object
between the times t1 and t2
G
∆x ≡ ( x(t2) − x(t1 ) ) î ≡ ∆x(
t)î
Average Velocity
• component of the average velocity, vx ,
is the displacement ∆x divided by the time interval ∆t
G
∆x
v(t ) ≡
î = vx
(t )î
∆t
Instantaneous velocity
• For time interval ∆t , we calculate the average
velocity. As ∆t → 0 , we generate a sequence of
average velocities. The limiting value of this
sequence is defined to be the x-component of
the instantaneous velocity at the time t .
x ( t + ∆t ) − x ( t ) dx
∆x
vx (t) ≡ lim vx = lim
= lim
≡
∆t →0
∆t→0 ∆t
∆t →0
∆t
dt
Instantaneous velocity
Average Acceleration
• Change in velocity divided by the time
interval
vx,2 − vx,1 )
(
G
∆vx
∆vx
ˆ
a = ax î ≡
î =
i=
î
∆t
∆t
∆t
Instantaneous acceleration
• For time interval ∆t , we calculate the average
acceleration. As ∆t → 0 , we generate a
sequence of average accelerations. The limiting
value of this sequence is defined to be the xcomponent of the instantaneous velocity at the
time t.
v ( t + ∆t ) − v (t )
G
∆v
dv
a(t ) = ax (t )î ≡ lim ax î = lim
î = lim î ≡ î
∆t →0
∆t→0 ∆t
∆t →0
∆t
dt Instantaneous Acceleration
Constant acceleration: area under the acceleration vs. time graph
∆vx vx (t ) − vx,0
ax = ax =
=
∆t
t
vx (t ) = vx,0 + a x t
Constant acceleration: Area
under the velocity vs. time graph
1
Area(vx , t ) = vx,0t + (vx (t ) − vx,0 )t
2
vx (t ) = vx,0 + ax t
1
1 2
Area(vx
, t ) = vx,0t + (vx,0 + ax t − vx,0 )t = vx,0t + ax
t
2
2
Constant acceleration: Average velocity
• When the
acceleration is
constant, the velocity
is a linear function of
time. Therefore the
average velocity is
1
vx = ( vx (t ) + vx,0 )
2
(
)
1
1
1
vx = ( vx (t ) + vx,0 ) = ( vx,0 + ax t ) + vx,0 = vx,0 + ax t
2
2
2
Constant acceleration: Area under the velocity vs. time graph
• displacement is equal to the area under
the graph of the x-component of the
velocity vs. time
1 2
∆x ≡ x(t ) − x0 =vxt = vx,0t + ax t = Area(vx ,t)
2
1 2
x(t ) = x0 + vx,0t + ax t
2
Summary: constant acceleration
• Position
• velocity
1 2
x(t ) = x0 + vx,0t + ax t
2
vx (t ) = vx,0 + ax t
Velocity as the integral of the acceleration
Velocity as the integral of the acceleration
• the area under the graph of the
acceleration vs. time is the change in
velocity
t ′ =t i=N
∫ a (t ′)dt ′ ≡ lim ∑ a (t )∆t
x
t ′ =0
t ′ =t
∆ti →0
t ′ =t x
i
i
= Area(ax , t )
i=1
v′x = vx ( t )
dvx
∫t′=0 ax (t ′)dt′ = t′∫=0 dt dt ′ = v′ =v∫ t =0 dv′x = vx
(t ) −
vx,0
)
x
x(
Position as the integral of velocity
• area under the graph of velocity vs. time is
the displacement
dx
vx (t ) ≡
dt
t ′=t
∫ v (t ′)dt
′ = x(t ) − x
x
t ′ =0
0
Example:
• A runner accelerates from rest for an
interval of time and then travels at a
constant velocity. How far did the runner
travel?
I. Understand – get a conceptual
grasp of the problem
• stage 1: constant
acceleration
• stage 2: constant
velocity.
Tools:
Coordinate system
Kinematic equations
Devise a Plan:
Stage 1: constant acceleration
Initial conditions:
Kinematic Equations:
x0 = 0
vx,0 =
0
1
2
x ( t ) = ax
t
2
Final Conditions: end acceleration at
position:
1
xa ≡ x ( t = ta ) =
ax ta 2
2
velocity
vx ,a ≡
vx (
t =
ta )
=
a
x ta
vx
(t ) =
ax t
t = ta Devise a Plan
Stage 2: constant velocity, time interval
• runs at a constant velocity for the time
• final position
[ta ,
tb ]
tb − t a
xb ≡ x ( t = tb ) = xa + vx,a ( tb − ta )
III. Solve
• three independent equations
1
2
xa = ax ta
2
vx ,a = ax t
a
xb = xa + vx,
a ( tb −
ta ) xb
• Six unknowns:
xa
vx , a
a x ta t
b
• Need three extra facts: for example:
ax
ta
tb
III. Solve
• solve for distance the runner has traveled 1
1
2
xb ≡ x ( t = tb ) = ax ta + ax ta ( tb − ta ) = ax ta tb − ax ta 2
2
2
IV. Look Back : Choose Values
ta = 3.0 s
runner accelerated for
ax = 2.0 m ⋅ s -2
initial acceleration:
runs at a constant velocity for
tb − ta = 6.0 s
tb = ta + 6.0 s = 3.0 s + 6.0 s = 9.0 s
Total time of running
Total distance running
1
1
2
xtotal = ax ta tb − ax ta 2 = ( 2.0 m ⋅ s -2 ) ( 3.0 s )( 9.0s ) − ( 2.0 m ⋅ s -2 ) ( 3.0 s ) = 4.5×101 m
2
2
Final velocity
vx ,a = a x ta = ( 2.0 m ⋅ s -2 ) ( 3.0 s ) = 6.0 m ⋅ s -1