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Problem 13.29 A car is traveling at 30 mi/hr when
a traffic light 295 f ahead turns yellow. The light will
remain yellow for 5 s before turing red.
(a)
30 mi/hr
What constant acceleration will cause the car to
reach the light at the instant it turns red, and what
will the velocity of the car be when it reaches the
light?
If the driver decides not to try to make the light,
what constant rate of acceleration will cause the car
to come to a stop just as it reaches the light?
(b)
Solution: The displacement for straight line motion under constant
acceleration is
a 2
t + v(0)t + s(0).
2
s(t) =
(b)
Two conditions are to be met: the car comes to a stop as it reaches
the light, therefore the distance traveled is 295 ft and the velocity
is zero at the end of that travel. The distance traveled is
s(t) = 295 =
Assume that s(0) = 0.
5280 ft
v(0) = 30 mi/hr = 30
1 mi
a 2
t + 44t,
2
and the velocity is v(t) = 0 = at + 44. The time required is
The velocity at t = 0 is given:
(a)
295 ft
1 hr
3600 s
= 44 ft/s.
t =−
44
.
a
Substitute into the displacement equation to obtain
The distance traveled in 5 seconds is s(5) = 295 ft, from which
a=−
a
295 = (5)2 + 44(5).
2
(44)2
= −3.28 ft/s2 .
2(295)
The time required is t = 13.4 s.
Solve: a = 6 ft/s2
The velocity for straight line motion under constant acceleration
is
ds(t)
= v(t) = at + v(0).
dt
At t = 5 seconds
v(5) = 6(5) + 44 = 74 ft/s = 50.45 mi/hr.
Problem 13.52 A car’s acceleration is related to its
position by a = 0.01s m/s2 . When s = 100 m, the car
is moving at 12 m/s. How fast is the car moving when
s = 420 m?
Solution:
a=v
vf
dv
= 0.01s m/s2
ds
v dv = 0.01
12
v2
2
420
s ds
100
vf
= 0.01
12 m/s
vf2
z
=
s2
2
420
m
100 m
122
(4202 − 1002 )
+ 0.01
z
z
vf = 42.5 m/s
Problem 13.98 A gas turbine starts rotating from rest
and has angular acceleration α = 6t rad/s2 for 3 seconds.
It then slows down with constant angular deceleration
α = −3 rad/s2 until it stops. (a) What maximum angular
velocity does it attain? (b) Through what total angle does
it turn?
Solution:
(a)
(b)
α dt + C = 3t 2 + C.
ω(t ≤ 3) =
At t = 0, ω(0) = 0, from which C = 0. The maximum is
[ω(t)]t=3 = 27 rad/s .
The angular travel is
θ(t ≤ 3) =
ω dt + C = t 3 + C.
At t = 0, θ(0) = 0, from which C = 0. The total travel during
the first interval is [θ(t)]t=3 = 33 = 27 rad.
Problem 13.109 A powerboat being tested for maneuverability is started from rest at t = 0 and driven in
a circular path of 12 m radius. The tangential component of the boat’s acceleration as a function of time is
at = 0.4t m/s2 .
(a) What are the boat’s velocity and acceleration in
terms of normal and tangential components at t =
4 s?
(b) What distance does the boat move along its circular
path from t = 0 to t = 4 s?
Solution:
(a)
at = 0.4t m/s2
an = +v 2 /r
v = 0.2t 2 m/s
At t = 4 s,
a = 0.4tet + v 2 /ren
a = 1.6et + 0.853en
v = 3.2et m/s
(b)
For t > 3, the angular velocity is
For 0 ≤ t ≤ 3 the angular velocity is
s = 0.2t 3 /3
s|4s = 4.27 m
ω(t > 3) =
α dt + C = −3t + C.
At t = 3, ω(3) = 27 rad/s, from which C = 36. The angular
travel is
θ(t > 3) =
3
ω dt + C = − t 2 + 36t + C.
2
At t = 3, θ(3) = 27 rad, from which C = −67.5. The time of
travel is obtained from ω(ts ) = 0 = −3ts + 36, from which the
36
stopping time is ts =
= 12 seconds. The total travel is
3
3
θ(t = 12) = − t 2 + 36t − 67.5
= 149 rad
2
t=12
Problem 13.141 The radial line rotates with a constant
angular velocity of 2 rad/s. Point P moves along the line
at a constant speed of 4 m/s. Determine the magnitude
of the velocity and acceleration of P when r = 2 m.
y
2 rad/s
4 m/s
P
r
Solution: The angular velocity of the line is
dθ
= ω = 2 rad/s,
dt
from which
d2θ
= 0.
dt 2
The acceleration is
a = [−2(4)]er + [2(4)(2)]eθ = −8er + 16eθ (m/s2 ).
The radial velocity of the point is
dr
= 4 m/s,
dt
from which
x
O
The magnitude is
|a| =
√
82 + 162 = 17.89 m/s2
y
d2r
= 0.
dt 2
2 rad/s
4 m/s
P
The vector velocity is
v=
dr
dt
er + r
dθ
dt
eθ = 4er + 4eθ (m/s).
r
The magnitude is
|v| =
√
x
0
42 + 42 = 5.66 m/s .
Problem 14.4 The 2-kg collar a is initially at rest on
the smooth horizontal bar. At t = 0 it is subjected to a
constant horizontal force F = 4 N.
F
A
(a) What is the resulting horizontal acceleration of the
collar?
(b) How fast is it moving at t = 1 s?
(c) What distance has it moved at t = 1 s?
Strategy: Use a Cartesian coordinate system with the
x-axis parallel to the bar. Draw the free-body diagram of
the collar, and apply Eqs. (3.5) to determine the collar’s
acceleration.
Solution:
F
m
m = 2 kg
F =4N
Fx :
(b)
F = ma
(a)
s = (2)(1)2 /2 m
a = 2 m/s2
v = at + v0
At t = 1 s,
v = 2 m/s
s = at 2 /2 + s0 ,
4=2a
s=1m
v0 = 0
x
s0 = 0
Problem 14.80 An airplane of weight W = 200,000 lb
makes a turn at constant altitude and at constant velocity
v = 600 ft/s. The bank angle is 15◦ . (a) Determine the
lift force L. (b) What is the radius of curvature of the
plane’s path?
15°
L
W
Solution: The weightis W = −jW = −j(2 × 105 ) lb. The normal
v2
. The lift is L = |L|(i cos 105◦ +
ρ
j sin 105◦ ) = |L|(−0.2588i + 0.9659j).
acceleration is an = i
(a)
From Newton’s second law,
F = L + Wn = man , from
which, substituting values and separating the j components:
|L|(0.9659) = 2 × 105 , |L| =
(b)
2 × 105
= 207055 lb.
0.9659
The radius of curvature
2 is obtained from Newton’s law:
v
|L|(−0.2588) = −m
, from which
ρ
ρ=
W
g
v2
|L|(0.2588)
= 41763.7 ft.
15°
L
W
Problem 14.101 The 1-lb block A is given an initial
velocity v0 = 14 ft/s to the right when it is the position
θ = 0, causing it to slide up the smooth circular surface.
By using Newton’s second law in terms of polar coordinates, determine the magnitude of the velocity of the
block when θ = 60◦ .
4 ft
θ
A
Solution: For this problem, Newton’s second law in polar coordi-
eθ
nates states
4 ft
Fr = mg cos θ − N = m(d 2 r/dt 2 − rω2 ) and
Fθ = −mg sin θ = m(rα + 2ω(dr/dt)).
equations reduce to N = mrω2 + mg cos θ and rα = −g sin θ. The
first equation gives us a way to evaluate the normal force while the second can be integrated to give ω(θ). We rewrite the second equation as
g dω
dω dθ
dω
sin θ
=
=ω
=−
dt
dθ dt
dθ
r
and then integrate
ω60
ω0
ωdω = −
g r
60
0
◦
θ
A
W
In this problem, r is constant. Thus (dr/dt) = (d 2 r/dt 2 ) = 0, and the
α=
θ
er
sin θdθ. Carrying out the
integration, we get
2
g
g
ω2
◦
ω60
− 0 =−
(− cos θ)|60
(1 − cos 60◦ ).
0 =−
2
2
r
r
Noting that ω0 = v0 /R = 3.5 rad/s, we can solve for ω60 = 2.05 rad/s
and v60 = Rω60 = 8.20 ft/s.
θ
N