Kinematics in Two Dimensions Chapter 3. Expectations After Chapter 3, students will: generalize the concepts of displacement, velocity, and acceleration.

Kinematics in Two Dimensions

Chapter 3

Expectations

After Chapter 3, students will: generalize the concepts of displacement, velocity,

and acceleration from one to two dimensions use the projectile-motion principle to analyze two-

dimensional constant-acceleration problems apply the kinematic equations to solve constant-

acceleration problems in two dimensions add relative velocities in two dimensions

Two-Dimensional Displacement

Sologdin the Turtle moves in the X-Y plane from the point (x0, y0) to the point (x1, y1).

R1 and R2 are vectors from the origin to his starting and finishing point, respectively.

y

x

R0

R1

(x0, y0)

(x1, y1)


Sologdin’s displacement vector D is the vector that extends from the starting of his motion to its endpoint.

In keeping with our Chapter 2 definition of displacement, we will define Sologdin’s displacement by

y

x

R1

(x0, y0)

(x1, y1)

R0

D

01 RRD


We can see the truth of this by solving for R1:

Our picture shows this addition being done, graphically.

y

x

R1

(x0, y0)

(x1, y1)

R0

D

DRRRRD

0101


Since both R0 and R1 start at the origin, it is very easy to write down the magnitudes of their X and Y components:

y

x

R1

(x0, y0)

(x1, y1)

R0

D

1111

0000

yRxR

yRxR

YX

YX


Now we perform the subtraction by subtracting the component magnitudes:

The magnitude of D:

y

x

R1

(x0, y0)

(x1, y1)

R0

D

0101

0101

01

yyRRD

xxRRD

RRD

YY

XX

Y

X

201

201

22

yyxxD

DDD YX


The direction of D is more clearly seen if we translate D so that its starting point is at the origin. (Remember that vectors are not changed by translating them.)

We can always write

01

01

01

01

arctan

tan

xx

yy

xx

yy

D

D

X

Y

- Y

+ X

D

x1 – x0

y1 – y0


We used the absolute value signs here because we want simply the lengths of the two legs of the right triangle. The sense of the angle is apparent from the drawing. We would describe the direction of D as “60° below the positive X axis,” or “60° clockwise from +X.”

01

01arctanxx

yy

- Y

+ X

D

x1 – x0

y1 – y0

Two-Dimensional Velocity

Let’s return to the travels of Sologdin … and this time, we’ll use our stopwatch.

Solgdin’s journey begins at time t0, and ends at a later time t1.

y

x

R1

(x0, y0)

(x1, y1)

R0

D

t = t0

t = t1


Sologdin’s average velocity is a vector. Its magnitude is the magnitude of his displacement, D, divided by the elapsed time:

Its direction is the same as the direction of the displacement vector.

01 tt

Dv

y

x

R1

R0

D

t = t0

t = t1

V


Sologdin’s average speed is not merely the magnitude of his average velocity. Let S be the curvilinear length of his meandering path. Then his average speed is

Note that his average speed exceeds his average velocity.

01

speed ave.tt

S

y

x

R1

R0

D

t = t0

t = t1

V


The critical difference between speed and velocity:

Consider the race car that wins the Indianapolis 500.

It completes 200 laps at 2.5 miles each in 2.75 hours. Its average speed is 500 mi / 2.75 hr = 182 mi/hr.

What is its average velocity?


What is its average velocity? Zero magnitude; pick a direction.

Its displacement is zero (the race both starts and ends at the start-finish line). And zero divided by any amount of time yields a magnitude of zero for the average velocity.

Two-Dimensional Acceleration

A truck travels east at 12 m/s. Following a bend in the road, the truck turns so that it is headed north, still at 12 m/s. The truck spends 8.0 s negotiating the curve. What is its average acceleration?

From the definition of acceleration:

Here, we have graphically added

-v0 to v1.

y

(east)

(north)

x

- V0

V1

V0

V1 – V0

t

vva 01


Having obtained v1 - v0, we need to multiply that vector by the scalar 1/t to obtain a:

2

2220

21

m/s 2.1

s 0.8

m/s 12m/s 12

a

t

vva y

(east)

(north)

x

- V0

V1

V0

V1 – V0a


The direction of a is the same as the direction of v1 - v0. Since the magnitudes of v1 and v0 are equal, we can write

So, the average acceleration is

2.1 m/s2, 45° west of north.

Note that the speed did not change – only the direction.

y

(east)

(north)

x

- V0

V1

V0

V1 – V0a

45 1arctan

1tan1

0

v

v

Projectile Motion

A projectile is something that is launched or thrown.

In the general case, it has a nonzero horizontal component of initial velocity.

v0

v0x

v0y

Projectile Motion

In most cases, the projectile experiences a negative vertical acceleration (g, due to gravity).

In most cases (assuming we can ignore resistance due to the air), the projectile experiences zero horizontal acceleration. Its horizontal velocity is constant.

The general approach to all problems of this kind: apply the kinematic equations separately to the vertical and horizontal motion.

Projectile Motion: Example

Illustrative example: the golfer in the picture has launched his ball with an initial velocity of 45 m/s, 31° above the horizontal. We will analyze the flight of the ball, in order to be able to say:

How far will the ball “carry” over level ground? How high above the ground will the ball reach at its

highest point? What is the total time the ball will be in the air?


First step: resolve the ball’s initial velocity into horizontal and vertical components.

v0

v0y

30

v0x

y

x

sin

cos

00

00

vv

vv

y

x


Keep firmly in mind that v0y is not going to remain constant … but that v0x will.

v0

v0y

30

v0x

y

x

sin

cos

00

00

vv

vv

y

x


During the time the ball takes to rise vertically, stop vertically, and fall vertically back to the ground, the ball moves horizontally with constant velocity.

We must determine just how long that time of flight is.

v0

v0y

30

v0x

y

x

sin

cos

00

00

vv

vv

y

x


Vertically, the ball obeys the first kinematic equation:

In our case, the initial vertical velocity is:

The acceleration is downward: g, the acceleration due to gravity. If we define our coordinate system with +Y pointing up, we rewrite the first kinematic equation:

atvv 0

sin00 vv y

gtvgtvv yy sin00


In terms of its vertical motion, the ball stops momentarily at the top of its trajectory. We also showed, in chapter 2, that its vertical travel is symmetric: it rises for half the time of flight, and falls for half the time. So, if t is the full time of flight, we can rewrite our kinematic equation for the upward travel by setting the final (y) velocity to zero:

02

1sin

2

100

tgvtgvv yy


Solving for t, the total time of flight:

s 4.730

m/s 8.9

31sinm/s 452

sin2

sin2

02

1sin

20

00

g

vt

vtg

tgv


The ball’s horizontal motion is governed by another kinematic equation:

In the horizontal, however, the acceleration is zero, and the initial velocity is

so we can rewrite our equation:

20 2

1attvx

cos00 vv x

m 182s 730.431cosm/s 45 cos0 tvx


How high does the ball go at the top of its trajectory?

Recall that we already calculated the total time of flight:

t = 4.730 s. Recall also that the ball spent half that time

falling (with zero initial vertical velocity) from its highest

point. How far did it fall?


We can calculate this maximum height directly from the third kinematic equation:

Notice that the initial y velocity (for the falling part of the trajectory) is zero, reducing the equation to:

The time we used was half the time-of-flight, t.

20 2

1gttvy y

m 27s 2

730.4m/s 8.9

2

1

2

12

22

gty

Parametric Representation of a Trajectory

So far, we have written both x and y as independent functions of time, t. We can combine these descriptions to express y as a function of x: the equation of the projectile’s trajectory.

Suppose our object is launched from the origin: x = y = 0.

Then:

(2) 2

1sin

(1) cos

20

0

gttvy

tvx

Parametric Representation of a Trajectory

Solve eq’n (1) for t:

Substitute this result for t in eq’n. 2:

cos

cos0

0 v

xttvx

220

2

2

00

0

cos2tan

cos2

1

cos

sin

v

gxxy

v

xg

v

xvy

Parametric Representation of a Trajectorytrajectory

0

5

10

15

20

25

30

0 20 40 60 80 100 120 140 160 180 200

x, m

y, m

Projectile Motion: Things to Remember Basic principle: the vertical motion and the horizontal

motion are treated separately. In the vertical motion, there is acceleration (usually, g). In the horizontal motion, acceleration is zero and velocity

is constant. Resolve the initial velocity into horizontal and vertical

components. Apply the kinematic equations to each motion, separately. If “g” appears in an equation with “x”, slap your own hand.

Relative Motion

Consider an airplane flying through air, when the air is in motion relative to the ground (wind).

The wind is blowing eastward. The pilot points his airplane straight north. In what direction does the airplane move over the ground? How fast does it go?

(In other words, what is the airplane’s velocity over the ground … magnitude and direction?)

Relative Motion

Principle of relative velocity:

Suppose the airplane’s speed through the air is 55 m/s, and the air moves over the ground with a wind speed of 8.9 m/s.

ground -air air - planeground - plane vvv

x (east)

y (north)

va-g

vp-gvp-a

Relative Motion

Velocity of plane over groundMagnitude (speed):

Direction:

(9.2° east of north)

m/s 56 m/s 8.9m/s 55 22gp

2ga

2apgp

v

vvv

x (east)

y (north)

va-g

vp-g

vp-a

2.9m/s 55

m/s 9.8arctanarctan

ap

ga

v

v

Kinematics in Two Dimensions Chapter 3. Expectations After Chapter 3, students will: generalize the concepts of displacement, velocity, and acceleration.

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