When
t
=
2.11
{\displaystyle t=2.11}
θ
=
6.28
;
ω
=
2
−
1.34
=
0.66
rad./sec.
{\displaystyle \theta =6.28;\quad \omega =2-1.34=0.66\,{\text{ rad./sec.}}}
α
=
−
1.27
rad./sec
2
{\displaystyle \alpha =-1.27\,{\text{ rad./sec}}^{2}}
t
=
3.03
{\displaystyle t=3.03}
θ
=
6.28
;
ω
=
2
−
2.754
=
−
0.754
rad./sec.
{\displaystyle \theta =6.28;\quad \omega =2-2.754=-0.754{\text{ rad./sec.}}}
α
=
−
1.82
rad./sec
2
{\displaystyle \alpha =-1.82{\text{ rad./sec}}^{2}}
The velocity is reversed. The wheel is evidently at rest between these two instants; it is at rest when
ω
=
0
{\displaystyle \omega =0}
0
=
2
−
0.3
t
3
{\displaystyle 0=2-0.3t^{3}}
t
=
2.58
{\displaystyle t=2.58}
θ
2
π
=
3
+
2
×
2.58
−
0.1
×
2.58
3
6.28
=
1.025
{\displaystyle {\frac {\theta }{2\pi }}={\frac {3+2\times {2.58}-0.1\times 2.58^{3}}{6.28}}=1.025}
Exercises V (See page 256 for Answers)
(1) If
y
=
a
+
b
t
2
+
c
t
4
{\displaystyle y=a+bt^{2}+ct^{4}}
d
y
d
t
{\displaystyle {\dfrac {dy}{dt}}}
d
2
y
d
t
2
{\displaystyle {\dfrac {d^{2}y}{dt^{2}}}}
Ans.
d
y
d
t
=
2
b
t
+
4
c
t
3
{\displaystyle {\dfrac {dy}{dt}}=2bt+4ct^{3}}
d
2
y
d
t
2
=
2
b
+
12
c
t
2
{\displaystyle {\dfrac {d^{2}y}{dt^{2}}}=2b+12ct^{2}}
(2) A body falling freely in space describes in
t
{\displaystyle t}
s
{\displaystyle s}
s
=
16
t
2
{\displaystyle s=16t^{2}}
s
{\displaystyle s}
t
{\displaystyle t}
t
=
2
{\displaystyle t=2}
t
=
4.6
{\displaystyle t=4.6}
t
=
0.01
{\displaystyle t=0.01}
(3) If
x
=
a
t
−
1
2
g
t
2
{\displaystyle x=at-{\tfrac {1}{2}}gt^{2}}
x
˙
{\displaystyle {\dot {x}}}
x
¨
{\displaystyle {\ddot {x}}}
(4) If a body move according to the law
s
=
12
−
4.5
t
+
6.2
t
2
{\displaystyle s=12-4.5t+6.2t^{2}}
find its velocity when
t
=
4
{\displaystyle t=4}
s
{\displaystyle s}
