CHAPTER 3

MOTION IN TWO AND THREE DIMENSIONS

• General properties of vectors • displacement vector • position and velocity vectors • acceleration vector • equations of motion in 2- and 3-dimensions

• Projectile motion • path of a projectile • time of flight • range

Examples of motion in 2- and 3-dimensions:

• any non-linear motion (circular motion, orbits of

planets, etc.)

• flight of projectiles (artillery shells, golf balls,

footballs, etc.)

Motion in 2- and 3-dimensions requires the use of

vectors. Basically, there are two main types of physical

quantities we deal with:

Scalars have size (magnitude) only and they can be

added, subtracted, multiplied and divided according to

the rules of simple math, i.e., Examples

include time, ordinary numbers, mass, distance (not

displacement) and speed (not velocity).

+, −, ×, ÷ .

Examples of vector quantities:

• Displacement. (The result of going is not

the same as going ).

• Force. (The result of pushing an object up is not the

same as pushing it down.)

• Velocity.

• Acceleration.

You cannot use the rules of simple math with vectors. As

an example take the displacement vector …

E → W

W → E

Military Trail

Glades Road

Spanish River Blvd

Palmetto Park Road

US 1 A1A

A B

C D

E

F

BBQ at Spanish River Park! Follow the map.

Military Trail

Glades Road

Spanish River Blvd

Palmetto Park Road

US 1 A1A

A B

C D

E

F

You have several choices:

A B

C

6.0 km

4.2 km

Displacement (resultant)

7.3 km

Note that:

but . . . . vector equation.

The displacement vector is 7.3 km,

AB+ BC = 10.2 km ≠ AC

35!

35! N of E.

AB! "!!

+ BC! "!!

= AC! "!!

Military Trail

Glades Road

Spanish River Blvd

Palmetto Park Road

US 1 A1A

A B

C D

E

F

55!

A

D C

A E

F C

In all three cases the displacement vectors are the same

and so are the distances.

55!( E of N) AD! "!!

+ DC! "!!

= AC! "!!

AE! "!!

+ EF! "!

+ FC! "!

= AC! "!!

All parallel vectors with the same lengths are equivalent.

These two vectors are not equivalent.

ADDITION

SUBTRACTION

!A +!B =!C

!A + (−

!B) =

!D

i.e., !A −!B =!D

Multiplication by a scalar.

A vector multiplied by a scalar n is another vector

parallel to but with magnitude (size) n times as large:

i.e.,

and

!C = n

!A

!A

!C

!A

!C = n

!A .

When working with vectors we will either

• draw them to scale, or

• use trigonometry and components:

!A

Ax

Ay

θ

y

x

In component form: and

The components are: and

where

!A = (Ax ,Ay)

!A = Ax

2 + Ay2 .

Ax =!A cosθ

Ay =

!A sinθ,

tanθ =

Ay

Ax.

Ax Bx

By

Ay

!A

!B

!C

y

xθ

Adding two vectors in component form:

In component form: and

The resultant is

where

Also and

and

!A = (Ax ,Ay)

!B = (Bx ,By).

!C = (Cx ,Cy),

!C = (Ax + Bx )2 + (Ay + By)2 .

Cx = (Ax + Bx ) Cy = (Ay + By),

tanθ =

Cy

Cx =

Ay + By

Ax + Bx

⎛

⎝⎜

⎞

⎠⎟ .

!A +!B =!C

Question 3.1: A yacht sails from a port (A) for 7.07 km in

a north-easterly direction to point B. Then, it sails in a

south-easterly direction for a further 4.24 km to point C.

(a) What is the displacement of the ship relative to the

port (A) when it reaches point C?

(b) If the yacht travels at 10.0 km/h, what is its average

velocity for the journey from A to C?

!A

!B

!C

Ax Bx

By

Ay

θ

(a)  We have to determine The components of

and are:

But

and N of E.

!C =!A +!B.

Ax =!A cos45" = (7.07 km)(0.707) = 5.0 km.

Ay =

!A sin 45" = (7.07 km)(0.707) = 5.0 km.

Bx =!B cos(−45") = (4.24 km)(0.707) = 3.0 km.

By =

!B sin(−45") = (4.24 km)( − 0.707) = −3.0 km.

!B

!A

!C = (Cx ,Cy) = (Ax + Bx ,Ay + By)

= (8.0 km,2.0 km).

∴!C = (8.0 km)2 + (2.0 km)2 = 8.24 km

θ = tan−1 Cy

Cx

⎛

⎝⎜

⎞

⎠⎟ = tan−1 2.0 km

8.0 km⎛⎝⎜

⎞⎠⎟= 14.0!

(b) Total distance traveled by yacht was

and the speed was So, the time taken was

The average velocity is defined as

Velocity is a vector (in the same direction as the

displacement) so to be precise the average velocity of the

yacht is in a direction N of E.

7.07 km + 4.24 km = 11.31 km,

10.0 km/h.

11.31 km10.0 km/h

= 1.13 h.

displacement

time= 8.24 km

1.13 h= 7.29 km/h.

7.29 km/h 14.0!

In vector addition

if and

then and

So:

!A = (Ax ,Ay)

!B = (Bx ,By)

!A = Axi + Ay j

!B = Bxi + By j.

!A ±!B = (Ax ± Bx )i + (Ay ± By) j+ (Az ± Bz )k.

Often, when we use vectors in component form, it is

more convenient to use “unit vectors”. Unit vectors

indicate direction only (and have magnitude ).

i k

j

i = j = k = 1.

unit vectors

= 1

Question 3.2: If and

what are:

(a) and

(b)

!L = 6i + 3j− k,

!M = 4i − 5j+ 8k

!N = i − 4 j− 4k,

!L + 2

!M −

!N,

!L + 2

!M −

!N ?

(a)

(b)

!L + 2

!M −

!N = (6i + 3j− k) + 2(4i − 5j+ 8k)

−(i − 4 j− 4k)

= 6i + 3j− k + 8i −10 j+16k − i + 4 j+ 4k

= 13i − 3j+19k.

!L + 2

!M −

!N = (13)2 + (−3)2 + (19)2 = 539

= 23.2.

Question 3.3: If and

what is the angle between !A = 3.8i +1.5 j

!B = 1.4i + 3.5 j,

!A and

!B?

From earlier, tanθA =

AyAx

⎛ ⎝

⎞ ⎠ , where θA is the angle

between ! A and the i direction (x-axis).

∴θA = tan−1 Ay

Ax⎛ ⎝

⎞ ⎠ = tan−1 1.5

3.8( ) = 21.5".

Similarly, from the i direction (x-axis),

θB = tan−1 By

Bx⎛ ⎝

⎞ ⎠ = tan−1 3.5

1.4( ) = 68.2".

So, the angle between ! A and

! B is

θB − θA = 68.2" − 21.5" = 46.7".

I will show you another way to solve this problem later.

! A

! B

21.5"

68.2"

i

j

0 1 2 3 4

1

2

3

4

Position and velocity vectors

The position vector of a point

with coordinates is (x, y,z)

!r = xi + yj+ zk.

Consider an object with position vector

that moves from to Then, if

and

The average velocity vector is

(parallel to ).

The instantaneous velocity vector at any point is

i.e.,

!r = xi + yj+ zk

P1(t1) P2(t2 ).

Δ!r = !r2 −

!r1 Δt = t2 − t1

!vav = Δ!r

Δt Δ!r

!r

!v = LimitΔt→0

Δ!rΔt

⎛⎝⎜

⎞⎠⎟ !r

= d!rdt

= dxdt

i + dydt

j+ dzdt

k,

!v = vx i + vy j+ vzk.

The velocity vector at the point is the tangent to the

position-time graph at and is parallel to the

instantaneous direction of motion.

The instantaneous speed is the magnitude of the

instantaneous velocity vector.

!r

!r

Acceleration vector

Consider an object with the velocity vector

that moves from to Then, if

and

the average acceleration vector is

(parallel to ).

The instantaneous acceleration vector is

i.e.,

v = vx i + vy j+ vzk

P1(v1, t1) P2(v2, t2 ).

Δ!v = !v2 −

!v1 Δt = t2 − t1

!aav = Δ!v

Δt Δ!v

!a = Limit

Δt→0

Δ!vΔt

⎛⎝⎜

⎞⎠⎟= d!v

dt=

dvxdt

i +dvy

dtj+

dvzdt

k,

!a = ax i + ay j+ azk.

Since ! a is parallel to Δ

! v , the direction of the acceleration is towards the “inside” of a curve. Note: on a curve or turn, there is always an acceleration even though the magnitude of the velocity (i.e., speed) may be constant.

! v

! a

! v

! a

Special case of circular motion

Shown here are the position and velocity vector diagrams of an object in uniform circular motion, i.e., motion at constant speed (v) and constant radius (r). The average acceleration from P1 to P2 is:

aav =

Δ! v Δt

=2v sin θ

2( )Δt

,

where Δt is the time to travel from P1 to P2. Now,

Δt =

arc length P1P2v

=rθv

.

∴aav =

2v2 sin θ2( )

rθ.

θ •

•

P1 t1( )

P2 t2( ) ! v 2

! v 1

! v 1 = ! v 2 = v

! r 1

! r 2

! r 1 = ! r 2 = r

−! v 1

! v 2

Δ! v

θ

Δ! v = 2v sin θ

2( ) Δ! v = ! v 2 − ! v 1

The instantaneous acceleration at the point P1 occurs when

Δt → 0, i.e., as θ → 0. Then the radial acceleration is

a r = Limitθ→0

2v2 sin θ2( )

rθ

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ =

2v2

rLimitθ→0

sin θ2( )

θ

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ .

Put φ = θ2, then

a r = Limitφ→0

sin φ2φ

⎡

⎣ ⎢

⎤

⎦ ⎥

= Limitφ→0

12sin φφ

⎡

⎣ ⎢

⎤

⎦ ⎥ =

12

.

∴ar =

2v2

r12

=v2

r.

Note that the direction of the acceleration is parallel to Δ! v ,

i.e., it is radial. It is called the centripetal (i.e., center-directed) acceleration.

We’ll return to this topic in chapter 5.

The instantaneous acceleration at the point P1 occurs when

Δt → 0, i.e., as θ → 0. Then the radial acceleration is

a r = Limitθ→0

2v2 sin θ2( )

rθ

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ =

2v2

rLimitθ→0

sin θ2( )

θ

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ .

Put φ = θ2, then

a r = Limitφ→0

sin φ2φ

⎡

⎣ ⎢

⎤

⎦ ⎥

= Limitφ→0

12sin φφ

⎡

⎣ ⎢

⎤

⎦ ⎥ =

12

.

∴ar =

2v2

r12

=v2

r.

Note that the direction of the acceleration is parallel to Δ! v ,

i.e., it is radial. It is called the centripetal (i.e., center-directed) acceleration.

We’ll return to this topic in chapter 5.

Put , then

But

φ = θ

2 ar =

2v2

rLimitφ→0

sinφ2φ

⎡

⎣⎢

⎤

⎦⎥.

Limitφ→0sinφ2φ

⎡

⎣⎢

⎤

⎦⎥ = Limitφ→0

12

sinφφ

⎡

⎣⎢

⎤

⎦⎥

= 12

Limitφ→0sinφφ

⎡

⎣⎢

⎤

⎦⎥ =

12

.

Question 3.4: An object experiences a constant

acceleration of At time , it is at

rest at Find

(a) the velocity and position vectors at any later time

t, and

(b) the equation describing the path of the object in

the xy-plane.

!a = (6i + 4 j) m/s2. t = 0

!r" = (10,0) m.

When :

(a) From chapter 2:

Also:

(b) But

Hence, so

i.e.,

t = 0 !r" = (10,0) m and !v" = (0,0) m/s,

!a = (ax ,ay) = (6i + 4 j) m/s2.

v = v! + at ⇒ "v = "v! +"at.

∴!v = (6i + 4 j)t = (6ti + 4tj) m/s.

(!r − !r") =

!v"t +12!at2 ⇒ !r = !r" +

!v"t +12!at2.

∴!r = 10i + 1

2(6i + 4 j)t2 = (10 + 3t2)i + 2t2 j m.

!r = (x, y) = xi + yj.

∴x = (10 + 3t2) and y = 2t2.

t2 = y

2 x = 10 + 3

y2

⎛⎝⎜

⎞⎠⎟

,

y = 2

3x − 20

3.

Question 3.5: A ship departs from point A heading to a

port 52 km due north at point B. The speed of the boat

relative to the water is 10 km/h. If there is a steady

current flowing at 2 km/h in a north-westerly direction,

(a) what is the proper heading to make the trip from

A to B?

(b) How long does the journey take?

(c) HOME PROBLEM: If the captain of the ship had

failed to take into account of the wind, how far due west

of B would the ship had been?

(a) Draw the velocity vector diagram. The resultant path

is the result of the ship setting

out along but being re-directed

by the current along .

So,

Since the resultant heading is due

north, the resultant velocity

component in the E-W direction is zero, i.e.,

east of north.

(b) The resultant speed of the ship is

Since the journey time is

θ

45!

A

B

C N

D

!vAB = !vAC + !vCB.

∴(10 km/h)sinθ = (2 km/h)cos45!,

i.e., θ = sin−1 (2 km/h)cos45!

10 km/h

⎛

⎝⎜

⎞

⎠⎟ = 8.1!

(10 km/h)sinθ − (2 km/h)sin45! = 0

= (10 km/h)cos8.1! + (2 km/h)sin45! = 11.3 km/h.

AB = 52 km,

52 km

11.3 km/h= 4.60 h ⇒ 4 h 36 min.

AB! "!!

AC! "!!

CB! "!!

!vAB = !vAD + !vDB = !vAC cosθ + !vCB sin45"

Projectile motion

• thrown objects (baseball, arrows, shells, etc.)

• bouncing balls, etc.

• Investigated first by Galileo.

• Historically important for the military.

vy

vy

vx

vx !v

!v

!v

!v"

v!x

v!y

(x!, y!) x

y

θ!

Important concept: we can treat the x and y components

separately. The initial velocity components are:

If we ignore air resistance and other drag forces, the only

force acting on the projectile is the gravitational force, so

the acceleration components of the projectile are:

v!x = "v! cosθ! and v!y = "v! sinθ!.

ax = 0 and ay = −g.

Analysis of projectile motion Since!!!!!!!!!!!!!!!!!!!!

•

Since

•

The components of the displacements are:

• and

•

These latter two equations represent a parabola.

The four equations listed above are the equations of

motion for a projectile (in the absence of air resistance).

Although the equations for the horizontal and vertical

components are treated separately, they are connected by

time (t).

ax = 0,

vx (t) ⇒ constant = v!x .

ay = −g,

vy(t) = v!y − gt.

x(t) = x! + v!t

y(t) = y! + v!yt − 1

2gt2.

If a rock is dropped from the same height as a rifle at the

same time it is fired, no matter how powerful the rifle, in

the absence of air resistance and assuming the ground is

horizontal and level, the bullet and the rock will hit the

ground at the same time, since all objects fall the same

distance in the same time, i.e., a distance of

in t seconds. 12

gt2

At every point on the path: tanθ =

vyvx

.

At maximum height: vy = v!y − gt = 0.

∴v!y = v! sinθ! = gt .

So, the time to reach maximum height is

t =

v! sin θ!g

.

If the take-off and landing are in the same horizontal plane, the total time of flight is:

T = 2

v! sin θ!g

.

vxˆ i

vyˆ j

Range

θ

" v

If the take-off and landing are in the same horizontal plane the horizontal range is:

R = v!xT

where T is the total flight time.

∴R = (v! cosθ!) × 2

v! sin θ!g

=

v!22sin θ! cosθ!g

.

i.e., R =

v!2

gsin 2θ!.

Maximum range ⇒ when θ! = 45!.

But sin 2θ = 2sin θcosθ and sin θ = cos(90! −θ)

∴sin 2θ = 2cos(90! − θ)cosθ

i.e., R(90! − θ) = R(θ).

So ... R(30! ) = R(60!), etc.

75!

15!

60!

30! 45!

Question 3.6: The sketch above shows the trajectories of

two golf balls, A and B. If the balls reach the same

height, which one had the greater initial velocity? Ignore

air resistance.

The balls reach the same height, so the vertical components of the initial velocities are equal,

i.e., vA sin θA = vB sin θB,

where vA and vB are the initial velocities of A and B,

respectively. Clearly, the take-off angle for A θA( ) is greater than that for B θB( ), then

sin θA > sin θB, so vB > vA,

i.e., the initial velocity of B is greater than that of A.

•

•

A

B

Question 3.7: A cannonball is fired from the top of a

40 m high tower with an initial speed of 42.2 m/s at an

angle of above the horizontal.

(a) How long is the cannonball in the air before it

hits the ground?

(b) When it hits the ground, how far is it from the

base of the tower?

(c) What is its speed when it hits the ground?

30!

∴4.91t2 − 21.1t − 40 = 0,

t =

21.1± (−21.1)2 − 4 × 4.91× (−40)2 × 4.91

. i.e.,

The appropriate solution is

(b) Range

∴ t = +5.72 s and t = −1.42 s.

t = +5.72 s.

R = v!xt = (36.5 m/s)(5.72 s) = 209 m.

(c) 

The final speed is

But

and

Hence,

Vx

Vy

!V

!v = vx

2 + v2y .

vx = v!x = 36.5 m/s,

v2

y = v!y2 + 2(−g)(y − y!)

= (21.1 m/s)2 + 2(−9.81 m/s2)(−40 m)

= (1230 m/s)2.

∴vy = 35.1 m/s.

v = (36.5 m/s)2 + (35.1 m/s)2 = 50.6 m/s.

DISCUSSION QUESTION:

Question 3.8: During the Roman invasion of Britain, a

group of Roman soldiers (at A) locate a British camp on

the top of a hill (at B). Using a trebuchet they want to lob

a rock from A onto the British camp, a horizontal distance

of 500 m away. If the maximum angle of elevation of the

trebuchet is and the height of the hill is 16 m,

(a) what initial speed of the rock is required in order

to hit the camp?

(b) What is the time of flight of the rock?

(c) With what speed does the rock strike the camp?

(d) What is the maximum height (h) achieved by the

rock?

60!,

v!x = v! cos60! = 0.50v!,

v!y = v! sin60! = 0.866v!,

(x!, y!) = (0,0); (x,y) = (500 m, 16 m).

(a)  The horizontal range is i.e.,

Also,

i.e.,

(b) From above,

v!xt, t = 500 m

0.50v!= 1000 m

v!.

(y − y!) = v!yt − 12

gt2.

∴16 m = 0.866v!(1000 m)

v!− 1

2(9.81 m/s2) (1000 m)

v!

⎛

⎝⎜⎞

⎠⎟

2

= (866 m) − (4.91 m/s2)(106 m2)

v!2

.

∴v!2 = (4.91 m/s2)(106 m2)

850 m= 5776.5 (m/s)2,

v! = 76.0 m/s.

t = 1000 m

v!= 1000 m

76.0 m/s= 13.16 s.

(c) The rock strikes the camp with speed

But

(d) The maximum height is h, where

v = vx

2 + vy2 .

vx = v!x = 0.50v! = 38.0 m/s,and

vy2 = v!y

2 − 2g(y − y!)

= (0.866 × 76.0 m)2 − 2(9.81 m/s2)(16 m − 0)

= 4017.8 (m/s)2, i.e., vy = −63.4 m/s.

∴v = (38.0 m/s)2 + (−63.4 m/s)2

= 73.9 m/s.

vy = 0.

∴0 = v!y2 − 2g(h − y!), i.e., v!y

2 = 2gh.

∴h =v!y

2

2g = (0.866 × 76.0 m/s)2

2(9.81 m/s2)= 220.8 m.