Motion in Two Dimensions

1.The Position, Velocity, and Acceleration Vectors

2.Two-Dimensional Motion with Constant Acceleration

3.Projectile Motion

The position of an object is described by its position vector, r

The displacementof the object is defined as the change in its position◦ Δr = rf - ri

The average velocity is the ratio of the displacement to the time interval for the displacement

◦ The direction of the average velocity is the direction of the displacement vector, Δr

t

r

v

The instantaneous velocity is the limit of the average velocity as Δt approaches zero◦ The direction of the instantaneous velocity is along

a line that is tangent to the path of the particle’s direction of motion

0limt

d

t dt

r rv

The average acceleration of a particle as it moves is defined as the change in the instantaneous velocity vector divided by the time interval during which that change occurs.

f i

f it t t

v v va

As a particle moves, Δv can be found in different ways

The average acceleration is a vector quantity directed along Δv

The instantaneous acceleration is the limit of the average acceleration as Δv/Δt approaches zero

0limt

d

t dt

v va

Various changes in a particle’s motion may produce an acceleration◦ The magnitude of the velocity vector may change

◦ The direction of the velocity vector may change

Even if the magnitude remains constant

◦ Both may change simultaneously

When the two-dimensional motion has a constant acceleration, a series of equations can be developed that describe the motion

These equations will be similar to those of one-dimensional kinematics

Position vector

Velocity

◦ Since acceleration is constant, we can also find an expression for the velocity as a function of time:

vf = vi + at

The velocity vector can be represented by its components

vf is generally not along the direction of either vi or at

vf = vi + at

The position vector can also be expressed as a function of time:◦ rf = ri + vit + ½ at2

◦ This indicates that the position vector is the sum of three other vectors: The initial position vector

The displacement resulting from vi t

The displacement resulting from ½ at2

• Equation (4.8) and (4.9) in component form (becausethey are vector expressions) :

tif avv tavv

tavv

yyiyf

xxixf

(4.8a)

2

21

iif tt avrr 2

y21

yiif

2

x21

xiif

tatvyy

tatvxx

(4.9a)

Example (4.1) : Motion in aPlane

A particle starts from the originat t = 0 with an initial velocityhaving an x component of 20m/s and a y component of –15m/s. The particle moves in thexy plane with an x componentof acceleration only, given byax=4.0 m/s2. (a) Determinethe components of the velocityvector at any time and thetotal velocity vector at anytime.

An object may move in both the x and ydirections simultaneously

The form of two-dimensional motion we will deal with is called projectile motion

0 svvoxox

Vertical Motion is the Same for Each BallVertical Motion is the Same for Each Ball

1 s

2 s

3 s

vvyy

vvxx

vvxx

vvxx

vvyy

vvyy

vvyy

vvyy

vvyy

The free-fall acceleration g is constant over the range of motion◦ And is directed downward

The effect of air friction is negligible With these assumptions, an object in

projectile motion will follow a parabolic path◦ This path is called the trajectory

Consider the motion as the superposition of the motions in the x- and y-directions

The x-direction has constant velocity◦ ax = 0

The y-direction is free fall◦ ay = -g

The actual position at any time is given by: rf = ri + vit + ½ gt2

rf = ri + vi t + ½ g t2

The final position is the vector sum of the initial position, the position resulting from the initial velocity and the position resulting from the acceleration

The y-component of the velocity is zero at the maximum height of the trajectory

The accleration stays the same throughout the trajectory

When analyzing projectile motion, two characteristics are of special interest

The range, R, is the horizontal distance of the projectile

The maximum height the projectile reaches is h

The maximum height of the projectile can be found in terms of the initial velocity vector:

This equation is valid only for symmetric motion

The range of a projectile can be expressed in terms of the initial velocity vector:

This is valid only for symmetric trajectory

2 2sin

2

i ivh

g

2 sin 2i ivR

g

The maximum range occurs at i = 45o

Complementary angles will produce the same range◦ The maximum height will be different for the two

angles

◦ The times of the flight will be different for the two angles

Select a coordinate system Resolve the initial velocity into x and y

components Analyze the horizontal motion using

constant velocity techniques Analyze the vertical motion using constant

acceleration techniques Remember that both directions share the

same time

E 4.2 The Long JumpAngle 20, at the speed of 11m/sa) X =?b) Hmax =?

E4.3 A Bull’s-Eye Every Time

Example (4.4) : That’s Quite an Arm!

A stone is thrown from the top of a building upward at an angle of 30.0o tothe horizontal and with an initial speed of 20.0 m/s, as shown in Figure(4.12). If the height of the building is 45.0 m, (a) how long is it before thestone hits the ground? (b) What is the speed of the stone just before itstrikes the ground?

Example (4.5) : The End of the Ski Jump

A ski jumper leaves the ski track moving in the horizontal direction with aspeed of 25.0 m/s, as shown in Figure (4.14). The landing incline belowhim falls off with a slope of 35.0o. Where does he land on the incline?

x = 50.0 mx = 50.0 m y = -19.6 my = -19.6 m

25 m/s

xy

Example :Example : A ball rolls off the top of a table A ball rolls off the top of a table 1.2 m1.2 m high and lands high and lands on the floor at a horizontal distance of on the floor at a horizontal distance of 2 m2 m. What was the velocity . What was the velocity as it left the table?as it left the table? What will be its speed when it strikes the floor?

1.2 m2 m

voy 28 m/s

vox30o

ymaxvy = 0

vox = 24.2 m/svoy = + 14 m/s