12.3 Rectilinear Motion (cont’d)
12.3 Rectilinear Motion (cont’d)
READING QUIZ
1
In curvilinear motion, the direction of the instantaneous
velocity is always
(a)
(b)
(c)
(d)
2
GENERAL CURVILINEAR MOTION (Section 12.4)
tangent to the hodograph.
perpendicular to the hodograph.
tangent to the path.
perpendicular to the path.
ANS: (C)
• A particle moving along a curved path undergoes
In curvilinear motion, the direction of the instantaneous
acceleration is always
(a)
(b)
(c)
(d)
tangent to the hodograph.
perpendicular to the hodograph.
tangent to the path.
perpendicular to the path.
ANS: (A)
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12.3 Rectilinear Motion (cont’d)
curvilinear motion. Since the motion is often
three-dimensional, vectors are used to describe the
motion. A particle moves along a curve defined by the
path function, s.
• The position of the particle at any instant is designated by
the vector r = r(t). Both the magnitude and direction of r
may vary with time.
• If the particle moves a distance ∆s along the curve during
time interval ∆t, the displacement is determined by vector
subtraction: ∆r = r ′ − r
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12.3 Rectilinear Motion (cont’d)
APPLICATIONS
VELOCITY
• Velocity represents the rate of
• The path of motion of a
change in the position of a particle.
plane can be tracked with
radar and its x, y, and z
coordinates (relative to a
point on earth) recorded as
a function of time.
• The average velocity of the particle
during the time increment ∆t is
vavg = ∆r/∆t.
• The instantaneous velocity is the
time-derivative of position v = dr/dt.
• How can we determine the
• The velocity vector, v, is always tangent to the path of
velocity or acceleration of
the plane at any instant?
• The magnitude of v is called the speed. Since the arc
motion.
length ∆s approaches the magnitude of ∆r as t → 0, the
speed can be obtained by differentiating the path function
(v = ds/dt). Note that this is not a vector!
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12.3 Rectilinear Motion (cont’d)
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12.3 Rectilinear Motion (cont’d)
APPLICATIONS(continued)
ACCELERATION
• Acceleration represents the rate of change
in the velocity of a particle.
• If a particle’s velocity changes from v to v ′
over a time increment ∆t, the average
acceleration during that increment is :
aavg = ∆v/∆t = (v ′ − v)/∆t
• The instantaneous acceleration is the
time-derivative of velocity :
a = dv/dt = d2 r/dt2
• A roller coaster car travels down a fixed, helical path at a
constant speed.
• A plot of the locus of points defined by the
• How can we determine its position or acceleration at any
arrowhead of the velocity vector is called a
hodograph. The acceleration vector is
tangent to the hodograph, but not, in
general, tangent to the path function.
instant?
• If you are designing the track, why is it important to be able
to predict the acceleration of the car?
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12.3 Rectilinear Motion (cont’d)
12.3 Rectilinear Motion (cont’d)
CURVILINEAR MOTION: RECTANGULAR
COMPONENTS (Section 12.5)
RECTANGULAR COMPONENTS: ACCELERATION
• The acceleration vector is the time derivative of the velocity
vector (second derivative of the position vector):
• It is often convenient to describe
the motion of a particle in terms
of its x, y, z or rectangular
components, relative to a fixed
frame of reference.
dv
d2 r
= 2 = ax i + ay j + az k
dt
dt
where ax = v˙x = ẍ = dvx /dt, ay = v˙y = ÿ = dvy /dt, and
az = v˙z = z̈ = dvz /dt.
• The magnitude of the acceleration vector is
√
a = [(ax )2 + (ay )2 + (az )2 ]0.5 or a = a · a.
• The direction of a is usually not tangent to the path of the
particle.
a=
• The position of the particle can
be defined at any instant by the
position vector: r = xi + yj + zk.
• The x, y, z components may all be functions of time, i.e.,
x = x(t), y = y(t), and z = z(t) .
• The magnitude of the position vector is: r = (x2 + y 2 + z 2 )0.5
• The direction of r is defined by the unit vector: ur = r/r
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12.3 Rectilinear Motion (cont’d)
12.3 Rectilinear Motion (cont’d)
RECTANGULAR COMPONENTS: VELOCITY
EXAMPLE
• Given: The motion of two particles (A and B) is described
• The velocity vector is the time derivative of the position
by the position vectors
vector:
d(xi) d(yj) d(zk)
dr
=
+
+
v=
dt
dt
dt
dt
• Since the unit vectors i, j, k are constant in magnitude and
direction, this equation reduces to
rA = xA i + yA j = [3ti + 9t(2 − t)j] m
• Find: The point at which the particles collide and their
speeds just before the collision.
v = vx i + vy j + vz k
• Plan:
1
The particles will collide when their position vectors are
equal, or rA = rB .
2
Their speeds can be determined by differentiating the
position vectors.
where
vx = ẋ =
dx
dt
vy = ẏ =
dy
dt
vz = ż =
(5)
rB = xB i + yB j = [3(t2 − 2t + 2)i + 3(t − 2)j] m (6)
dz
dt
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12.3 Rectilinear Motion (cont’d)
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12.3 Rectilinear Motion (cont’d)
RECTANGULAR COMPONENTS: VELOCITY
EXAMPLE (continued)
Solution:
1
The point of collision requires that rA = rB , so xA = xB
and yA = yB .
• The magnitude of the velocity vector is
0.5
v = (vx )2 + (vy )2 + (vz )2
• Set the x-components equal: xA = xB , 3t = 3(t2 − 2t + 2)
• Simplifying: t2 − 3t + 2 = 0
• Solving: t = 3 ± [32 − 4(1)(2)]0.5 /2(1) so that t = 2 or 1 s.
• The direction of v is tangent to the path of motion.
• Set the y-components equal: yA = yB , 9t(2 − t) = 3(t − 2)
• Simplifying: 3t2 − 5t − 2 = 0
• Solving: t = 5 ± [52 − 4(3)(−2)]0.5 /2(3)
so that t = 2 or −1/3 s.
• So, the particles collide when t = 2 s (only common time).
Substituting this value into rA or rB yields xA = xB = 6 m
and yA = yB = 0.
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12.3 Rectilinear Motion (cont’d)
12.3 Rectilinear Motion (cont’d)
EXAMPLE (continued)
From
GROUP PROBLEM SOLVING (continued)
Solution:
rA = xA i + yA j = [3ti + 9t(2 − t)j] m
2
rB = xB i + yB j = [3(t − 2t + 2)i + 3(t − 2)j] m
1
1
x-components:
dx
vx = ẋ =
= (16t2 )
dt
Z x
Z t
16 3 2
dx = x =
16t2 dt =
= 42.7
t
3
0
0
0
Differentiate rA and rB to get the velocity vectors.
vA =
vB =
drA
= ẋA i + ẏA j = [3i + (18 − 18t)j] m/s (7)
dt
drB
(8)
= ẋB i + ẏB j = [(6t − 6)i + 3j] m/s
dt
2
• Speed is the magnitude of the velocity vector.
|vB |
=
=
(32 + (−18)2 )0.5 = 18.2 m/s
(62 + 32 )0.5 = 6.71 m/s
(14)
⇒ x = (16/3)t3 = 42.7 meter at t = 2 s
2)
= 32t = 64 m/s2
Acceleration: ax = ẍ = v¨x = d(16t
dt
• At t = 2 s: vA = [3i − 18j] m/s and
• At t = 2 s: vB = [6i + 3j] m/s
|vA |
(13)
(9)
(10)
y-components:
= 4t3 m/s
Velocity known as : vy = ẏ = dy
dt 2
Ry
Rt 3
Position: 0 dy = 0 4t dt ⇒ y = t4 0 = 16 meter at t = 2s
3)
2
2
Acceleration: ay = ÿ = v¨y = d(4t
dt = 12t t=2 = 48m/s
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12.3 Rectilinear Motion (cont’d)
12.3 Rectilinear Motion (cont’d)
CHECK YOUR UNDERSTANDING QUIZ
1
2
GROUP PROBLEM SOLVING (continued)
Solution:
If the position of a particle is defined by
r = [(1.5t2 + 1)i + (4t − 1)j] (m), its speed at t = 1 s is
(a)
(b)
(c)
(d)
3. z-components:
Velocity known as : vz = ż = dz
dt = (5t + 2)m/s Rz
Rt
2
Position: 0 dz = 0 (5t + 2) dt ⇒ z = 52 t2 + 2t 0 = 14 m at
t = 2s
Acceleration: az = z̈ = v¨z = d(5t+2)
= 5 m/s2
dt
2 m/s
3 m/s
5 m/s
7 m/s
ANS: (c)
The path of a particle is defined by y = 0.5x2 . If the
component of its velocity along the x-axis at x = 2 m is
vx = 1 m/s, its velocity component along the y-axis at this
position is
(a)
(b)
(c)
(d)
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4. The position vector and magnitude of the acceleration
vector are written using the component information found
above.
Position vector: r = [42.7i + 16j + 14k] m.
Acceleration vector: a = [64i + 48j + 5k] m/s2
Magnitude: a = (642 + 482 + 52 )0.5 = 80.2 m/s2
0.25 m/s
0.5 m/s
1 m/s
2 m/s
ANS: (d)
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12.3 Rectilinear Motion (cont’d)
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12.3 Rectilinear Motion (cont’d)
GROUP PROBLEM SOLVING
ATTENTION QUIZ
y
• The velocity of the particle is
2
3
v(t) = [16t i + 4t j + (5t + 2)k] m/s
= vx i + vy j + vz k
1. If a particle has moved from A to B
along the circular path in 4s, what is the
average velocity of the particle?
(11)
(12)
(a)
(b)
(c)
(d)
and x = y = z = 0 at t = 0.
• Find: The particle’s coordinate position and the magnitude
2
x
B
2.5i m/s
2.5i + 1.25j m/s
1.25πi m/s
1.25πj m/s
ANS: (a)
of its acceleration when t = 2 s.
• Plan: Note that velocity vector is given as a function of
time.
1
R=5m
A
2. The position of a particle is given as r = (4t2 i − 2xj) m.
Determine the particle’s acceleration.
Determine the position and acceleration by integrating and
differentiating v, respectively, using the initial conditions.
Determine the magnitude of the acceleration vector using
t = 2 s.
(a)
(b)
(c)
(d)
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(4i + 8j) m/s2
(8i − 16j) m/s2
(8i) m/s2
(8j) m/s2
ANS: (b)
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