Note 06
Motion in Two Dimensions
Sections Covered in the Text: Chapters 6 & 7, except 7.5 & 7.6
It is time to extend the definitions we developed in
Note 03 to describe motion in 2D space. In doing so
we shall find that the vector nature of the definitions
—displacement, velocity and acceleration—take on
more importance than before. Having developed
these tools we then apply them to describe an
important, special example of motion in 2D space,
namely the motion of a projectile. As before, we shall
model the object as a point particle.
Displacement, Velocity and Acceleration
Consider an object modeled as a point particle
moving along an arbitrary path in the xy-plane
(Figure 6-1). We assume that we are able to detect the
particle’s position at any point and to measure the
corres-ponding clock time. Two positions i and f in
the particle’s path are shown. Let the vectors that
locate these positions be r i and r f , respectively. The
time that elapses between the two positions is ∆t = tf –
ti.
Displacement
The displacement of the particle between the two
positions is different from the distance travelled; it is
the difference between the corresponding position
vectors:
Δr = r f – ri .
Obviously, in contrast to motion in 1D space, position
vectors in 2D space do not, in general, always point in
the same direction. Moreover, in general, the displacement vector points in a direction different from the
position vectors that define its limits.
Velocity
Two types of velocity are defined in 2D space as in 1D
space: average velocity and instantaneous velocity.
Average Velocity
The average velocity of the particle is defined as the
displacement divided by the corresponding elapsed
time ∆t:
v=
y
d
Δr = r f – ri
f
rf
…[6-1]
Since average velocity is proportional to displacement
(the proportionality factor being 1/∆t, a scalar), it is a
vector pointing in the same direction as the displacement vector. In any elapsed time the average velocity
depends on the end positions of the interval (as does
the displacement), and is therefore independent of the
path taken by the particle.
i
ri
Δr
.
Δt
path of object
x
Figure 6-1. An object modelled as a point particle moves
along an arbitrary path in 2D space.
Instantaneous Velocity
The instantaneous velocity of the particle is defined as
the limit of the average velocity as the elapsed time ∆t
becomes vanishingly small, i.e., as ∆t → 0:
v = lim v = lim
Δt →0
Distance Traveled
The distance d the particle travels between the two
positions is indicated approximately in the figure.
This would be the number obtained if it were possible
to lay a flexible tape measure along the particle’s path.
d is difficult to measure or calculate in all but the
simplest of paths. By its nature, d is always a positive
number.
Δ t→ 0
Δr dr
=
.
Δt dt
…[6-2]
Thus the instantaneous velocity vector equals the first
derivative of the position vector with respect to time
(or the slope of the tangent to the r(t) function). The
direction of the instantaneous velocity vector at any
position in a particle’s path is along a tangent line to
the path at that position and in the direction of
motion. The instantaneous speed is the magnitude of
the instantaneous velocity.
06-1
Note 06
Figure 6-1 depicts the motion of the particle in terms
of position or displacement vectors; but the motion
can also be described in terms of velocity vectors. Let
the instantaneous velocity vectors at positions i and f
be as drawn in Figure 6-2. In general, the change in
velocity vector (small diagram in the figure) points in
a direction different from the instantaneous velocity
vectors. Since, in general, the instantaneous velocity
vectors at positions i and f are unequal, the particle is,
by definition, undergoing an acceleration.
y
vi
i
vi
Δv = v f – vi
vf
f
a=
Δv
Δt
path of object
vf
x
y
i
vi
vi
Δv = v f – vi
vf
f
path of object
Figure 6-3. The average acceleration vector of the particle has
the direction of the change in velocity vector. Acceleration
is a different vector type than is velocity or displacement.
Δ t→ 0
vf
x
Figure 6-2. The motion of a particle described in terms of
velocity vectors. The velocity and position vectors (compare
this figure with Figure 6-1) are drawn with different head
styles to emphasize their different natures.
Acceleration
Average Acceleration
The average acceleration of a particle over an elapsed
time ∆t is defined as the change in velocity divided by
the elapsed time:
v f – vi Δv
=
.
t f – ti
Δt
Thus the instantaneous acceleration vector is the first
derivative of the instantaneous velocity vector with
respect to time. The direction of the instantaneous
acceleration vector at any position in a particle’s path
is along a tangent line to the velocity function at that
position.
The kinematic equations of motion we developed in
Note 03 (eqs[3-7], [3-11] and [3-12])) can be generalized by writing them in full vector notation. We shall
simply state the results here. For convenience we
assume the particle is moving with constant acceleration in 2D space. The instantaneous position vector of
a particle at a point (x,y) is
)
)
r = xi + yj .
…[6-5]
…[6-3]
The average acceleration is a vector divided by a scalar, and is therefore a vector. The average acceleration
vector points in the same direction as the change in
velocity vector (see Figure 6-3).
Instantaneous Acceleration
The instantaneous acceleration of a particle is defined as
the limit of the average acceleration as the elapsed
time becomes vanishingly small, i.e., as ∆t → 0:
06-2
…[6-4]
Two-Dimensional Motion with Constant
Acceleration
Two types of acceleration are defined in 2D space:
average acceleration and instantaneous acceleration.
a=
Δv dv
=
.
Δt →0 Δt
dt
a = lim a = lim
If, over an elapsed time t, the velocity of the particle
changes from v i to v f then
v f = v i + at .
…[6-6]
If, over the elapsed time t the position of the particle
changes from r i to r f then
1
2
r f = ri + v it + at 2 .
…[6-7]
Note 06
In 2D space eq[6-6] gives rise to an equation for each
of the x- and y-components
v xf = vxi + a xt
…[6-8]
and
v yf = vyi + a yt .
Eq[6-7] gives rise to the equations
tion is to derive this kind of expression for projectile
motion from kinematics.
To begin our description, we model the problem in
xy-space with the y-direction positive upward (Figure
6-4). We assume that the particle is projected from the
origin (0,0) with an initial velocity v i at an angle θi
with respect to the horizontal. The position of the
particle at any subsequent clock time t is given by the
position vector r .
1
2
x f = x i + v xit + axt 2
and
1
2
y
…[6-9]
Projectile Motion
One of the early successes of Natural Science in the
renaissance was the correct description of the motion
of a projectile. This problem was pursued by the best
minds of the day, including Da Vinci, Galileo and
others, as it had important usage in the art of war. The
burning question was if a cannonball is fired with a
certain velocity at a certain angle above the horizon,
how high will the ball go and how far will it travel? In
other words, what will be the cannonball’s maximum
height and range? This was important in putting a
cannonball where you wanted it to go.
We have all seen a similar motion in baseball. When
a player hits or throws the ball, the ball travels in a
curved path of a certain maximum height and range.
One can demonstrate photographically that if the
resistance of the air is negligible then the path of the
projectile is a parabola.1 In other words the functional
form of the path is
y(x) = ax + bx 2 ,
ri
v yi
vi
v yf
θi
x
v xi
Figure 6-4. The path of a projectile in 2D space. The position
and velocity vectors are drawn with different head styles to
emphasize their different natures.
We make the following assumptions:
1 the acceleration of the particle is a constant of
value ay = –g directed downwards
2 the effect of air resistance is negligible.2
This means that the acceleration of the particle has a
non-zero component in the y-direction only; the xcomponent of the acceleration is zero. The initial
velocity vector can be resolved into its x- and y-components
v xi = vi cos θi
and
v yi = vi sin θi .
Substituting these expressions into eqs[6-8] and [6-9]
and taking ax = 0 and ay = –g yields
…[6-10]
where a and b are constants. One objective of this secAs the astronomers are quick to point out the correct path is an
ellipse. However, if the particle moves close to the Earth so that its
acceleration remains constant then to a good approximation the
path can be described by a parabola.
v xi
v xi
These equations emphasize that motion with constant
acceleration in 2D space is equivalent to two
independent motions in the x- and y-directions having constant accelerations ax and ay, respectively. The
component of motion in the x-direction does not affect
the component of motion in the y-direction and viceversa. With these tools we are ready to describe the
motion of a projectile.
1
(R/2,h)
y f = yi + vyi t + ayt 2 .
v xf = vxi = vi cosθ i = constant
…[6-11]
v yf = vyi – gt = vi sin θi – gt
…[6-12]
2
If we do not make this assumption the problem is more
complicated. Anyone who has watched a baseball game has seen
baseballs veer downwards, to the left or right in response to air
friction or the wind.
06-3
Note 06
x f = x i + v xit = (vi cos θi )t ,
y f = yi + vyi t –
1
2
…[6-13]
=
1
2
2vi2 sin θi cosθ i
,
g
gt 2 = (vi sin θ i )t – gt 2 …[6-14]
which reduces to
If we solve for t in eq[6-13] and substitute it into eq[614] we get

 2
g
y f = (tanθ x ) x f –  2
 x …[6-15]
2
 2vi cos θi  f
This expression has the form of eq[6-10] and proves
that the path of the projectile is, indeed, a parabola
according to the assumptions we have made.
Horizontal Range and Maximum Height
We are now in a position to derive expressions for the
maximum height h and horizontal range R. These
occur, respectively, at the special positions (R/2, h)
and (R, 0) in the particle’s path.
At the highest point of the particle’s path, vyf = 0.
Thus from eq[6-12] we can calculate the time the
particle takes to reach this position:
th =
R=
vi2 sin 2θ i
,
g
…[6-17]
since sin2θ = 2sinθcosθ. It can be seen that, other
things being equal, a maximum R occurs for θi = 45˚.
As before, R can be increased by increasing vi or by
operating in an environment where g is small.
We now have the equations we need to describe any
aspect of the motion of a projectile. Let us consider an
example.
Example Problem 6-1
An Example of Projectile Motion
A ball is thrown from the top of one building toward
a tall building 50.0 m away (Figure 6-5). The initial
velocity of the ball is 20.0 m.s–1, 40.0˚ above the horizontal. How far above or below the original level does
the ball strike the opposite wall?
vi sin θ i
.
g
Substituting this expression for th into eq[6-14] and
replacing yf with h, we have
40.0˚
2
h = (vi sin θ i )
which reduces to
vi sin θ i 1  vi sin θ i 
– g
 ,
2 
g
g 
h=
vi2 sin2 θ i
.
2g
…[6-16]
It can be seen that, other things being equal, maximum h occurs for θi= 90˚. h can be increased by increasing v i or by operating in an environment (on a planet
or small astronomical body) where g is very small.
Now the range R is the horizontal distance traveled
by the particle in twice the time it takes to reach the
peak, that is, in a time 2th. Using th from above and
noting that xh = R at t = 2th, we have
R = ( vi cos θi )2th = (vi cosθ i )
2vi sin θi
,
g
50.0 m
Figure 6-5. A ball is projected at 40.0˚ above the horizontal.
Solution:
We have the following components of the initial
velocity:
vix = (20.0m.s –1 )cos40.0 o = 15.3m.s –1
viy = (20.0m.s –1 )sin 40.0o = 12.9m.s –1
The horizontal component of the initial velocity
remains unchanged at
06-4
Note 06
vix = v fx = vx = 15.3m.s
–1
Since the x-component of acceleration is zero, we can
use the expression x = v x t to solve for t where t is the
time that elapses between launch and impact on the
opposite wall. Thus t is
t=
same event with the vector r po' .
O’ and his/her reference frame is moving with
velocity v o'o relative to O. The two position vectors
are related by
r PO = r PO' + v O' Ot .
50m
= 3.27s .
15.3m.s–1
For the vertical motion taking “up” as positive and
“down” as negative we have
1
2
y = v0 t + ay t 2
Figure 6-6. Two drivers A and B move past an observer O.
1
2
= (12.9m.s–1 )(3.27s) + (–9.80m.s –2 )(3.27s)2
y'
y
P
= –10.2m .
y is the vertical distance above the original level. Since
y is negative here, the ball hits the opposite wall at a
point 10.2 m below the original level.
Another, special acceleration is defined for objects
moving with uniform speed in a circle. Though this
topic is presented at this point in the textbook, we
postpone it until a later note.
Relative Velocity
Thus far we have taken the origin of our coordinate
system as a fixed point (the point we have taken as 0
in a 1D frame or (0,0) in a 2D frame). There will be
times when we need to relate certain measurements
made in different systems, in particular when one
system is moving with respect to the another.
Take for example the motion illustrated in Figure 66. Two cars with drivers A and B move past a third
observer O who is standing beside the road. O
measures the speeds of A and B at 60 km/hr. But
according to observer A what is the speed of B? As we
all know from experience A would say that B’s speed
is 0 km/hr. In other words the speed of B relative to O
is 60 km/hr, but relative to A is 0 km/hr.
We can quantify the physics of these observations
with the help of the position vectors in Figure 6-7.
Observers O and O’ in their own reference frames (S
and S’) observe the same event P. O locates the event
with the position vector r po , while O’ locates the
rPO
rPO'
O'
O
S
S'
vO' Ot
x
vO' O
Figure 6-7. Observers in stationary and moving reference
frames describe the same event P with different position and
velocity vectors.
If we differentiate this expression with respect to time
we obtain an expression for the corresponding
velocity vectors:
d
d
r PO =
r PO' + v O'Ot
dt
dt
( )
gives
(
)
v po = v po' + v o' o .
The subscripts po, po’ and O’O mean “p wrt O”, “p
wrt O’” and “O’ wrt O”. Thus the velocity of P wrt O
equals the velocity of P wrt O’ plus the velocity of O’
wrt O. This result is made clearer with the help of an
example.
06-5
Note 06
Example Problem 6-2
A Boat Crossing a River
vWO = 0.300m.s–1
A boat propelled with a speed of 0.500 m.s–1 in still
water moves directly across a river that is 60.0 m
wide. The river flows with a speed of 0.300 m.s–1
relative to the riverbank. (a) At what angle, relative to
the straight-across direction, must the boat be
pointed? (b) How long does it take the boat to cross
the river?
Solution:
Let the velocity of the water wrt an observer on the
riverbank (magnitude 0.300 m.s–1) be denoted v WO .
Let the velocity of the boat relative to the water (magnitude 0.500 m.s–1) be denoted v BW . Let the velocity
of the boat relative to the riverbank be denoted v BO .
The vectors are arranged as shown in Figure 6-8.
(a) The boat must therefore be steered upstream by
the angle
v 
0.300 
θ = sin –1  WO  = sin –1 
 v BW 
0.500 
d=
60.0 m
v BW = 0.500m.s –1
v BO
θ
O
Figure 6-8. Velocity diagram for a boat crossing a river.
(b) The speed of the boat with respect to the riverbank
is the magnitude of the vector v BO . This is
v BO = (cos37 o )(0.500m.s–1 ) = 0.400m.s –1 .
This is the resultant speed of the boat wrt the riverbank. The time required for the boat to go directly
across the river is the distance travelled divided by
this speed:
t=
= 37.0o upstream
with respect to the water. It is the subsequent drag of
the water flowing downstream that results in the boat
moving directly across the river.
downriver
d
60.0m
=
= 150s .
vBO 0.400m.s–1
The boat crosses the river in 150 seconds.
To Be Mastered
•
•
•
•
Definitions: distance travelled, displacement, average velocity, instantaneous velocity, instantaneous speed, average
acceleration, instantaneous acceleration
kinematic equations (committed to memory)
projectile motion problem
relative velocity problems
06-6
Note 06
Typical Quiz/Test/Exam Questions
1.
A child throws a ball straight up into the air and catches the ball on its return. Let the maximum height
reached by the ball be h and the total elapsed time be ∆t. Assume the ball is at the same, negligible height
above the Earth when thrown and caught and neglect the effect of air friction.
(a) What is the distance travelled by the ball?
(b) What is the speed of the ball at its highest point?
(c) What is the average velocity of the ball?
(d) What is the acceleration of the ball at its highest point?
2.
A pilot in an airplane is moving at the speed of 15.0 m.s –1 parallel to flat ground 100.0 m below as shown in
the figure. He releases a sack of flour intending to hit a target. Assume air friction is negligible and answer the
following questions:
sack
B
100.0 m
C
A
target
(a) Which of A, B or C most closely approximates the path of the sack of flour? Explain.
(b) How large must the distance x from plane to target be in order that the sack hits the target?
(c) How long is the sack in the air?
3.
A particle is projected horizontally from a position 10.0 m above ground (see the figure on the next page). It
hits the ground a horizontal distance of 100.0 m away.
It is proposed that the path of the particle in xy-space can be described by the function
y(x) = a + bx + cx2
where x and y are in m and a, b and c are constants. Based on the facts given find values for a, b and c.
06-7
Note 06
particle
10.0 m
100.0 m
06-8