PHY061 Chapter 2. 2D Motion

CHAPTER 2
Motion in Two Dimensions
Prepared by HM
Chapter 2
2.1 Displacement, Velocity, and

Acceleration in Two Dimensions
2.2 Motion in Two Dimensions
Projectile Motion
Learning Objectives
At the end of this presentation, you should be able
to:
Define instantaneous velocity and instantaneous
acceleration;
Describe the path and motion of a projectile
motion;
Write expressions for horizontal and vertical
components of velocity and acceleration; and
Use equations to solve problems involving
projectile motion.
2.1 Displacement, Velocity, and
Acceleration in Two Dimensions
➢Suppose an object travelled
the path shown in the figure
(dashed green line).
➢In the xy-plane, the initial
position at t1 has a position
vector 𝒓𝟏 , while the final
position at t2 has the position
vector 𝒓𝟐 .
➢The vector from t1 to t2, which is
denoted as ∆𝒓, is the
displacement of the object.
Based on the vector diagram, we can write the
relationship of the position vectors and the
displacement as,
𝒓2 = 𝒓1 + ∆𝒓
Rearranging the equation above, we

get the definition of displacement in
two dimensions, that is,
∆𝒓 = 𝒓𝟐 − 𝒓𝟏
Using this description of displacement, we can now
generalize the definition of velocity and
acceleration in two dimensions:
𝒓𝟐 − 𝒓𝟏 ∆𝒓
Average velocity: 𝒗𝒂𝒗 = ⟹ 𝒗𝒂𝒗 =
𝒕𝟐 − 𝒕𝟏 ∆𝒕
∆𝒓
Instantaneous velocity: 𝒗= lim
∆𝑡→0 ∆𝒕
𝒗𝟐 −𝒗𝟏 ∆𝒗
Average acceleration: 𝒂𝒂𝒗 = ⟹ 𝒂𝒂𝒗 =
𝒕𝟐 −𝒕𝟏 ∆𝒕
∆𝒗
Instantaneous acceleration: 𝒂 = lim
∆𝑡→0 ∆𝒕
Keep in mind that the SI units of velocity is meter
per second (m/s) while that of acceleration is
meter per square second (m/s2).
From the definition of acceleration, we can

say that acceleration occurs when an
object:
(1)changes speed (either speeds up or slows
down);
(2)changes direction; or
(3)changes both speed and direction.
SELF-ASSESSMENT QUESTION 3.2
Which of the following objects is NOT accelerating?
(a)A car that is traveling with a constant speed
(b)A car that is moving with a constant velocity
(c)A car that is moving along a curved path
Answer: b
2.2 Motion in Two Dimensions
Recall the following kinematics Equations

along the x-axis: along the y-axis:
𝑣𝑥 = 𝑣0𝑥 + 𝑎𝑥 𝑡 𝑣𝑦 = 𝑣0𝑦 − 𝑔𝑡
1 1 2
𝑥 − 𝑥0 = 𝑣0𝑥 𝑡 + 𝑎𝑥 𝑡 2 𝑦 − 𝑦0 = 𝑣0𝑦 𝑡 − 𝑔𝑡
2 2
𝑣𝑥 2 = 𝑣0𝑥 2 + 2𝑎𝑥 (𝑥 − 𝑥0 ) 𝑣𝑦 2 = 𝑣0𝑦 2 − 2𝑔(𝑦 − 𝑦0 )
Projectile Motion
• A special case of motion in two
dimensions is the so-called projectile
motion.
• Similar to free fall, we ignore air resistance
and the rotation of the Earth, and only
consider the effect of gravity on an
object’s motion.
• The figure illustrates the parabolic
trajectory of an object undergoing
projectile motion with an initial velocity 𝒗𝟎 .
• Observe that the velocity at any point
can be expressed through its
components.
Projectile Motion
Equations to be used:
Two characteristics of an
object in projectile motion In the x-axis:
that you need to remember 𝑣𝑥 = 𝑣0𝑥 = 𝑣0 cos 𝜃0 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Eq. 3.1
are:
𝑥 − 𝑥0 = 𝑣0𝑥 𝑡 = 𝑣0 cos 𝜃0 𝑡 Eq. 3.2
✓the x-component of its
velocity is constant; thus, its In the y-axis:
acceleration in the x-axis is
zero (ax = 0); and 𝑣𝑦 = 𝑣0 sin 𝜃0 − 𝑔𝑡 Eq. 3.3
✓its acceleration in the y-axis 1 2

is constant and is equal to 𝑦 − 𝑦0 = (𝑣0 sin 𝜃0 )𝑡 − 𝑔𝑡 Eq. 3.4
the acceleration due to 2
gravity (ay = –g = –9.8 m/s2). 𝑣𝑦 2 = (𝑣0 sin 𝜃0 )2 −2𝑔(𝑦 − 𝑦0 ) Eq. 3.5
Example 2.2.1
A ball is kicked with an initial velocity of 15 m/s and

leaves the foot at ground level with an angle θ0=60º as
illustrated in the figure. Calculate (a) the maximum
height the ball reaches, (b) the time it takes the ball to
hit the ground, (c) the distance the ball travels, (d) the
velocity vector at the maximum height, and (e) the
acceleration vector at the maximum height. Ignore air
resistance.
Diagram of the Problem:
Solution Let us write first what are given in the problem:
Initial velocity: vo = 15 m/s Initial angle: θo = 60º
(a) To solve for the maximum height (ymax),

we can use and transform equation 3.5. Take note that at the maximum
(As you can see, it is not convenient to height, the velocity vector is
use equations 3.3 and 3.4 because time t parallel to the x-axis. Hence,
is not given in the problem.)
the y-component of velocity is
transpose zero (vy = 0). Also, we let y2 =
ymax and y1 = 0.
𝑣𝑦 2 = (𝑣0 sin 𝜃0 )2 −2𝑔(𝑦2 − 𝑦1 )
0 0 Simplifying our equation,
2𝑔(𝑦2 − 𝑦1 ) = (𝑣0 sin 𝜃0 )2 −𝑣𝑦 2 2𝑔𝑦𝑚𝑎𝑥 = (𝑣0 sin 𝜃0 )2
Solution
Dividing both sides of
2𝑔𝑦𝑚𝑎𝑥 = (𝑣0 sin 𝜃0 )2

Substituting the given values,
by 2𝑔, we get
15 𝑚Τ𝑠 (sin 60)2
2𝑔𝑦𝑚𝑎𝑥 (𝑣0 sin 𝜃0 )2 𝑦𝑚𝑎𝑥 =
= 2 9.8𝑚/𝑠 2
2𝑔 2𝑔
𝑦𝑚𝑎𝑥 = 8.6 𝑚
(𝑣0 sin 𝜃0 )2
𝑦𝑚𝑎𝑥 =
2𝑔
Solution
1
(b) The total travel time can be 𝑔𝑡 = (𝑣0 sin 𝜃0 )
2
calculated using equation 3.4,
0 0 2
1
𝑦2 − 𝑦1 = (𝑣0 sin 𝜃0 )𝑡 − 𝑔𝑡 2 𝑡 = (𝑣0 sin 𝜃0 )
2 𝑔
Simplifying, we get 2
𝑡= 2
(15𝑚/𝑠 sin 60)
9.8 𝑚/𝑠
1
0 = (𝑣0 sin 𝜃0 ) − 𝑔𝑡
2 𝑡 = 2.65 𝑠.
Solution
(c) The distance travelled by

the ball is equal to the ball’s Use the answer in part (b), which is
displacement (∆𝑥)in the x- 𝑡 = 2.65 𝑠, to solve for x2.
axis. So, we can use Equation
3.2, ∆𝑥 = 𝟏𝟓 𝒎Τ𝒔 𝐜𝐨𝐬 𝟔𝟎 (2.65s)
∆𝑥 = 𝑥2 − 𝑥1 = (𝑣0 cos 𝜃0 )𝑡 ∆𝑥 = 𝟏𝟗. 𝟗 𝒎
∆𝑥 = (𝑣0 cos 𝜃0 )𝑡
Solution
Meanwhile, the y-component of
(d) To find the velocity vector the velocity at the maximum height
at the highest point, we need is zero, 𝑣𝑦𝑚𝑎𝑥 = 0.
to calculate first its Thus, the magnitude of the velocity
components. For the x- at the highest point is:
component, we have
𝑣= (𝑣𝑥 )2 +(𝑣𝑦 )2
𝑣𝑥 = 𝑣0 cos 𝜃0
𝑣𝑥 = 15𝑚/𝑠 (cos 60) = 7.5 𝑚/𝑠 𝑣= (7.5𝑚/𝑠)2 +(0)2

𝑣 = 7.5 𝑚/𝑠
Solution
(d) SO, the final velocity (e) Recall that, for projectile
vector is motion, the acceleration in the x-
axis is zero while in the y-axis, it is
𝑚
𝒗 = 7.5 , 𝑒𝑎𝑠𝑡 equal to –g. Hence, the
𝑠 acceleration vector at the highest
(𝑜𝑟 𝑡𝑜𝑤𝑎𝑟𝑑𝑠 𝑡ℎ𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑥
− 𝑎𝑥𝑖𝑠) point is
𝒂 = 9.8 𝑚/𝑠 2 , 𝑠𝑜𝑢𝑡ℎ

References:
Giancoli, D. (2009). Physics for Scientists and Engineers with

Modern Physics, 4th Edition. USA: Pearson Education, Inc.
Serway, R. A. & Vuille, C. (2018). College Physics, 11th Edition.
USA: Cengage Learning.
Young, H. D. & Freedman, R. A. (2016). Sears & Zemansky’s
University Physics with Modern Physics, 14th Edition. USA:
Pearson Education, Inc.

PHY061 Chapter 2. 2D Motion

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CHAPTER 2

Motion in Two Dimensions

2.1 Displacement, Velocity, and

Rearranging the equation above, we

From the definition of acceleration, we can

Recall the following kinematics Equations

✓its acceleration in the y-axis 1 2

A ball is kicked with an initial velocity of 15 m/s and

(a) To solve for the maximum height (ymax),

2𝑔𝑦𝑚𝑎𝑥 = (𝑣0 sin 𝜃0 )2

(c) The distance travelled by

∆𝑥 = 𝑥2 − 𝑥1 = (𝑣0 cos 𝜃0 )𝑡 ∆𝑥 = 𝟏𝟗. 𝟗 𝒎

𝑣𝑥 = 15𝑚/𝑠 (cos 60) = 7.5 𝑚/𝑠 𝑣= (7.5𝑚/𝑠)2 +(0)2

𝒂 = 9.8 𝑚/𝑠 2 , 𝑠𝑜𝑢𝑡ℎ

Giancoli, D. (2009). Physics for Scientists and Engineers with

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