Chapter 2 Motion in One Dimension
Particle model: a particle is a point-like object, that is, an object that has mass but is
of infinitesimal size
2.1 Positon, Velocity, and Speed
A vector quantity requires the specifications of both direction and magnitude
r r
r
vector: displacement: ∆x = x f − xi
nonsymmetry nature
x
0
10
5
xi
15
xf
r
r
r r
r
xi = 10iˆ , x f = 8iˆ --> ∆x = x f − xi = 8iˆ − 10iˆ = −2iˆ
xi
xf
r
r
r r
r
xi = 8iˆ , x f = 10iˆ --> ∆x = x f − xi = 10iˆ − 8iˆ = 2iˆ
r r
r
scalar: distance: ∆x = x f − xi (no negative sign)
x
0
5
10
15
r
r
r
r r
r
xi = 10iˆ , x f = 8iˆ --> ∆x = x f − xi = 8iˆ − 10iˆ = −2iˆ --> ∆x = 2
r
r
r
r r
r
xi = 8iˆ , x f = 10iˆ --> ∆x = x f − xi = 10iˆ − 8iˆ = 2iˆ --> ∆x = 2
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Active Figure\AF_0201.swf
r
r
∆x
vector: average velocity: v avg =
∆t
r
∆x
total _ dis tan ce
scalar: average seed: v avg =
, Average _ Speed =
∆t
total _ time
Example: A particle moving along the x axis is located at xi = 12 m at ti = 1s and xf = 4
m at tf = 3 s. Find its displacement and average velocity during this time interval.
r
r
Displacement: ∆x = 4 − 12 = −8 m, distance: ∆x = ∆x = 4 − 12 = 8 m
x f − xi 4 − 12
r
Average velocity: vavg =
=
= −4 m/s, Average speed: 4 m/s
t f − ti
3 −1
What is position?
2.2 Instantaneous Velocity and Speed
See active figure AF0203.
r
r
∆x d r
vector: velocity: v = lim
= x
∆t − > 0 ∆t
dt
r
∆x
r
d r
scalar: speed: v = lim
=
x
∆t − > 0 ∆t
dt
Example: The position of a particle moving along x axis varies in time according to
r
the expression x = 3t 2 iˆ , where x is in meters and t is in seconds. Find the velocity
in terms of t at any time.
Find the average velocity in the intervals t = 0 s to t = 2 s.
2
r r
r
r
xi = x (t ) = 3t 2iˆ , x f = x (t + ∆t ) = 3(t + ∆t ) 2 iˆ
r r
x f − xi
r
3t 2 + 6t∆t + 3(∆t ) 2 − 3t 2 ˆ
v = lim
= lim
i = 6tiˆ
∆t − > 0
∆t − > 0
∆t
∆t
r
r
x (t 2 ) − x (t1 ) 3 ∗ 2 2 − 3 ∗ 0 2 ˆ
vavg =
=
i = 6 m/s
t 2 − t1
2−0
2.3 Analysis Model: The Particle Under
Constant Velocity
r
d r r
x = v = v ⋅ iˆ , dx = vdt ⋅ iˆ ,
dt
r
xf
t
r
xi
0
r
r r
r
∫ dx = v ∫ dt ⋅ iˆ , x f = xi + v t
r
Example: A particle move at a constant velocity v = 5 ⋅ iˆ(m / s ) , the initial position xi
= 10 m, find the final position after a time interval of t = 10 s.
r
r r
r
x f = xi + v t --> x f = (10 + 5 ∗10 )iˆ = 60iˆ(m)
2.4 Acceleration
a: Position, b: velocity, c: acceleration
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r
r
∆v
vector: average acceleration: a avg =
∆t
r
r
∆v d r d  d r  d 2 r
vector: instantaneous acceleration: a = lim
= v =  x = 2 x
∆t − > 0 ∆t
dt
dt  dt  dt
Example: A particle's position on the x axis is given by x = 4 − 27t + t 3 with x in
meters and t in seconds. (a) Find the particle's velocity function v(t) and
acceleration function a(t).
dx(t )
dv(t )
v(t ) =
= 0 − 27 + 3t 2 a (t ) =
= 6t
dt
dt
,
2.5 Motion Diagrams
See active figure AF0209.
2.6 Motion Under Constant Acceleration
r r
v − v0
r r
a = a avg =
t −0
r r r
v = v0 + at
r r r
d r r r
x = v0 + a t , dx = (v0 + a t )dt ,
dt
- The 1st formula
r
x
t
r
x0
t0
r
r r
∫ dx = ∫ (v0 + at )dt
4
r r
r
r1
r r r
1r
x − x0 = v0 (t − 0) + a (t 2 − 0) , x = x0 + v 0 t + at 2
2
2
t =
r
r
v − v
r
a
0
- The 2nd formula
r r
r r 2
r r
r v − v0 1 r v − v0 
, x − x0 = v0 r + a  r 
a
2  a 
r
r
r r r
r r r
r r r
r r 1r
1r
a ( x − x0 ) = v0 v − v02 + v 2 − v0 v + v02 , v 2 = v02 + 2a ( x − x0 ) - The 3rd formula
2
2
Example: Spotting a police car, you brake a Porsche from a speed of 100 km/h to a
speed of 80.0 km/h during a displacement of 88.0 m, at a constant acceleration. (a)
What is that acceleration? (b) How much time is required for the given decrease in
speed?
r
r r
r r
r r r
1r
v f = v0 + at , x = x0 + v0t + at 2 , v 2 = v02 + 2a ⋅ ∆x
2
v f = 80
km
m 1 h
1000
1000
= 22.2 m/s, vi = 100
= 27.8 m/s
h
km 3600 s
3600
r
∆x = 88 m -> choose the 3rd formula, 22.2 2 = 27.82 + 2 ∗ a ∗ 88 , a=-1.6 m/s2
Example: Accelerating an Electron
An electron in the cathode-ray tube of a television set enters a region in which it
accelerates uniformly in a straight line from a speed of 3 x 104 m/s to a speed of 5 x
106 m/s in a distance of 2 cm.
(
For what length of time is the electron accelerating?
Choose the 3rd formula: 5 × 10 6
Choose the 1st formula: ∆t =
) = (3 × 10 )
2
4 2
+ 2a ∗
2
, a=6.2*1014 m/s2
100
5 × 10 6 − 3 × 10 4
= 8.0 × 10 −9 s
6.2 × 1014
Example: A car traveling at a constant speed of 45.0 m/s passes a trooper on a
mortorcycle hidden behind a billboard. One second after the speeding car passes the
billboard, the trooper sets out from the billboard to catch the car, accelerating at a
constant rate of 3.00 m/s2. How long does it take her to overtake the car?
1
45 + 45t = 3t 2
2
5
2.7 Freely Falling Objects
up: +y
r
m
a = − g = −9.8 2
s
t
a = a0
v = v0 + at = a0t
1
1
x = x0 + v0t + at 2 = a0t 2
2
2
Example: A stone thrown from the top of a building is
given an initial velocity of 20.0 m/s straight upward.
The building is 50.0 m high, and the stone just misses
the edge of the roof on its way down, as shown in Fig.
2.14.
Using tA = 0 as the time the stone leaves the
thrower’s hand at position A, determine (A) the time
at which the stone reaches its maximum height, (B)
the maximum height, (C) the time at which the stone
returns to the height from which it was thrown, (D)
the velocity of the stone at this instant, and (E) the
velocity and position of the stone at t = 5.00 s.
See active figure AF0213.
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2.8 Kinematic Equations Derived from
Calculus
tf
∆x = ∑ v xn ∆t n = lim ∑ v xn ∆t n = ∫ v x (t )dt (What’s definite & indefinite integral ? )
∆t − > 0
n
dx
vx =
-> dx = vx dt ->
dt
ax =
dv x
dt
ti
n
x2
t2
x1
t1
∫ dx = ∫ v dt
x
t
->
v xf − v xi = ∫ a x dt
0
For the case of constant acceleration, we can derive the kinematic equations:
dv
= a0 = const. -> v = v0 + at
dt
v=
dx
1
= v0 + at -> x = x0 + v0t + at 2
dt
2
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