Problem 1. Answers: 1. 216v i j =+
; 8a j = ; 7.13︒.(cos a v av θ⋅= ) 2. 1/3(3/)f t v k = 3. a-e, b-d, c-f. 4. [d]: 222x y L +=, 0dx dy x
y dt dt
+= dx v dt =, B dy v dt =, 0B xv yv +=, cot B x
v v v y
θ== 5. (a)32(102)3
t r i t t j =+-
, (Answer)
(b) 912r i j =+
, (3)(0)343
avg r r v i j -=
=+ , (Answer) (3)(0)343
avg v v a i j -==- (Answer)
(c) 92v i j =-
2tan 9
y x v v θ==-, 12.5θ=- (Answer)
6. Solution: From the definition of acceleration for a straight line motion
dv
a dt =,
and the given condition a =-
dv dt
-=
. Apply chain rule to d v /d t , the equation can be rewritten as
d v d x d v
v d x d t d x
-=
= Separating the variables gives
v k d
x =- Take definite integration for both sides of the equation with initial conditions,
we have
x
v d v k d
x =-⎰
⎰, or 3/2
023x v k = (Answer)
Problem 2. Answers:
1. 13.0 m/s 2, 5.7m/s, 7.5 m/s 2,
2.
1)R .
Solution: At initial moment
when the ball is just kicked out:
2i
r v
a g ρ
=
=, 2i
v g
ρ=
. In order the ball not to hit the rock,
2
i v R g ρ=≥,
For the vertical motion:
2
12
gt R =
, t i v t R -
=1)R (Answer)
3. d.
4. d.
5. Solution: at any moment the speed of the projectile is
v =
Tangential acceleration:2t dv a dt ==,
Radial acceleration:r a ==
,
From 2
r v a ρ
=
,
we have the radius of curvature is given by
2223/2
20
()r v g t v a gv ρ+=
= (Answer)
6. 86.7θ=
Solution: sd sg gd v v v =+ sg dg v v =-
3
510/3600t a n
17.36
0.8
dg sg
v v θ⨯=== 86.78642
θ︒'==
Problem 3. Answers : 1. Solution: 2
12cos cos v T T m R
θθ+=
12sin sin T T mg θθ=+
(a)21108N 2cos 2sin mv mg
T R θθ
=
+=, (b) 2156N sin mg
T T θ
=-
= 2. (a) Mg ,
(b)
3. d.
4. e
5. Solution: from the given condition ct i v v e -= and with 10.0m/s i v =, at 20 s,
5.0m/s v =, we have (a)10.0347s -,
With 40.0 s, we can get (b) 2.50 m/s,
Solution: (c) From ct i v v e -=, we have
ct i dv
cv e dt
-=- Therefore a cv =-, (Answer)
which means the acceleration of the boat is proportional to the speed at any time.
6.
Solution: From Newton ’s second law, we have
2
k m v
m a -= By the definition of acceleration, this equation can be rewritten as
2dv
kmv m dt
-=
Separating the variables obtains
2dv
kdt v
-=
Take the definition integral with the initial conditions, we hve
020v t v dv
kdt v -=⎰⎰
11
kt v v -= Then 00/(1)v v ktv =+.
(Answer)
Fig. 3-1
