<Chap.

3 Motion in Two and Three Dimensions: Suggested Problems and Solutions>

3.11, 3.25, 3.34, 3.41, 3.64, 3.81
11.
(a) Magnitude of the displacement vector
Setting east as x-axis and north as y-axis, the displacement vector is given by
𝑥𝑥⃗ = (1.57𝚥𝚥̂ + 0.846𝚤𝚤̂) km
The magnitude can be calculated from the Pythagorean theorem.

|𝑥𝑥⃗| = �(1.57)2 + (0.846)2 km = 1.78 km

(b) Angle relative to the north
From the drawing below and using trigonometric function,
0.846
𝜃𝜃 = tan−1 = 28.3°
1.57
25.
Since there’s no acceleration in x-direction, the velocity in x-direction is constant.
𝑣𝑣⃗𝑥𝑥 = 1.1 m/s
In y-direction, the initial velocity is zero and the acceleration is constant.
𝑣𝑣⃗𝑦𝑦 = 5.2 s × 0.52 m⁄s 2 = 2.7 m/s

Therefore, the velocity vector after 5.2 seconds is

𝑣𝑣⃗ = (1.1𝚤𝚤̂ + 2.7𝚥𝚥̂) m⁄s
34.
Following quantities are given.
 Initial velocity, 𝑣𝑣⃗ = (35𝚤𝚤̂) m⁄s
 Initial position, 𝑟𝑟⃗(𝑡𝑡 = 0) = ℎ𝚥𝚥̂ where ℎ is the initial height.
 Distance traveled in x-direction, 𝑑𝑑 = 23 m
 The gravitational acceleration 𝑔𝑔 = 9.8 m⁄s 2
As a vector equation, position as function of time is given by the following.
1
𝑟𝑟⃗(𝑡𝑡) = 𝑟𝑟⃗(𝑡𝑡 = 0) + 𝑣𝑣⃗ 𝑡𝑡 − 𝑔𝑔𝑡𝑡 2 𝚥𝚥̂
2
i) Motion in x-direction
In x-direction, it is a constant-speed motion that moves 23 m.
𝑣𝑣𝑥𝑥 𝑡𝑡 = 𝑑𝑑
𝑑𝑑
𝑡𝑡 =
𝑣𝑣𝑥𝑥
23 m
=
35 m⁄s
= 0.66 s
ii) Motion in y-direction
In y-direction, it is a constant-acceleration motion that ends at 0 height.
1
ℎ − 𝑔𝑔𝑡𝑡 2 = 0
2
1 2
ℎ = 𝑔𝑔𝑡𝑡
2
1
= × (9.8 m⁄s 2 ) × (0.66 s)2
2
= 0.17 m
41.
Given following quantities,
 The radius of orbit, 𝑟𝑟 = 384.4 Mm = 384.4 × 106 m
 The period of orbital motion, 𝑇𝑇 = 27.3 days = 2.36 × 106 s
the acceleration is given by
4𝜋𝜋 2 𝑟𝑟 4𝜋𝜋 2 (384.4 × 106 m)
𝑎𝑎 = 2 =
𝑇𝑇 (2.36 × 106 s)2
= 2.72 × 10 m⁄s 2
−3
64.
The velocity and position as functions of time are given as follows. (In SI units)
𝑣𝑣⃗ (𝑡𝑡) = 𝑣𝑣⃗0 + 𝑎𝑎�𝑡𝑡
= (11 − 1.4𝑡𝑡)𝚤𝚤̂ + (18 + 0.27𝑡𝑡)𝚥𝚥̂
1
𝑟𝑟⃗(𝑡𝑡) = 𝑣𝑣⃗0 𝑡𝑡 + 𝑎𝑎�𝑡𝑡 2
2
0.27 2
= (11𝑡𝑡 − 0.70𝑡𝑡 2 )𝚤𝚤̂ + �18 + 𝑡𝑡 � 𝚥𝚥̂
2
(a) Setting 𝑥𝑥 = 0 and 𝑡𝑡 ≠ 0 gives
11 − 0.70𝑡𝑡 = 0
11
𝑡𝑡 =
0.70
= 16 (s)
(b) Plugging 𝑡𝑡 obtained in (a) into the y-position,
11 0.27 11 2
𝑦𝑦 = 18 × + ×� �
0.7 2 0.7
= 3.2 × 102 (𝑚𝑚)
Note that the scientific notation is used to correctly express significant figures.
(c) Plugging 𝑡𝑡 obtained in (a) into the velocity expression,
11 11 11
𝑣𝑣⃗ �𝑡𝑡 = � = �11 − 1.4 × � 𝚤𝚤̂ + �18 + 0.27 × � 𝚥𝚥̂
0.7 0.7 0.7
= −11𝚤𝚤̂ + 22𝚥𝚥̂ (m⁄s)
The speed is the magnitude of 𝑣𝑣⃗ , which is
|𝑣𝑣⃗ | = �(11)2 + (22)2 = 25 (m⁄s)
and the angle measured from positive x-direction is given by
22
𝜃𝜃 = tan−1
11
which is about 116 degrees.
81.
Two quantities are given.
 The (maximum) acceleration, 𝑎𝑎 = 0.795𝑔𝑔
 The speed of your car, 𝑣𝑣 = 80.0 km⁄h
To avoid truck with circular motion, you need distance more than 𝑟𝑟, which is a radius of the
circular motion.
𝑣𝑣 2
𝑎𝑎 =
𝑟𝑟
𝑣𝑣 2
𝑟𝑟 =
𝑎𝑎
(80.0 km⁄h)2
=
0.795𝑔𝑔
(80.0 ÷ 3.6 m⁄s)2
=
0.795 × 9.81 m⁄s 2
= 63.3 m