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.
The displacement of the particle between the two positions is different from the distance travelled; it is the difference between the corresponding position vectors:
Δ r = r f
– r i
.
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 ri and r f , respectively. The time that elapses between the two positions is ∆ t = t f
– t i
.
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.
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 =
Δ r
.
Δ t
…[6-1] y r i i
Δ r = r f
– r i r f d f path of object x
Figure 6-1.
An object modelled as a point particle moves along an arbitrary path in 2D space.
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.
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
Δ t → 0 v = lim
Δ t → 0
Δ r
Δ t
= dr
.
dt
…[6-2]
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.
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 a = i
Δ
Δ v t v i v i v f
Δ v = v f f path of object
– v i v f x
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.
y i v i v i v f
Δ v = v f f path of object
– v i v f 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.
Two types of acceleration are defined in 2D space: average acceleration and instantaneous 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: a = v f t f
– v i
– t i
=
Δ v
.
Δ t
…[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: a = lim
Δ t → 0 a = lim
Δ t → 0
Δ v
=
Δ t dv
.
dt
…[6-4]
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 = x
) i + y
) j .
…[6-5]
If, over an elapsed time t , the velocity of the particle changes from vi 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 r f
= r i
+ v i t +
1
2 at
2
.
…[6-7]
06-2

In 2D space eq[6-6] gives rise to an equation for each of the x- and y-components v xf
= v xi
+ a x t
…[6-8] and v yf
= v yi
+ a y t .
Eq[6-7] gives rise to the equations x f
= x i
+ v xi t +
1
2 a x t
2 and y f
= y i
+ v yi t +
1
2 a y t
2
.
…[6-9]
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 a x
and a y
, 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.
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
, …[6-10] where a and b are constants. One objective of this sec-
1
As 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.
Note 06 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 vi at an angle θ i with respect to the horizontal. The position of the particle at any subsequent clock time t is given by the position vector r .
y
(R/2,h) v xi v xi v yi
i v i v xi r i v yf x
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 a y
= –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
= v i cos θ i and v yi
= v i sin θ i
.
Substituting these expressions into eqs[6-8] and [6-9] and taking a x
= 0 and a y
= –g yields v xf
= v xi
= v i cos θ i
= constant …[6-11] v yf
= v yi
gt = v i
θ i
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 xi t = ( v i cos θ i
) t , …[6-13] y f
= y i
+ v yi t –
1
2 gt
2
= ( v i sin θ i
) t –
1
2 gt
2
…[6-14]
If we solve for t in eq[6-13] and substitute it into eq[6-
14] we get y f
= (
θ x
) x f


v i
2 g
2
θ i

 x
2 f
…[6-15]
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, v yf
= 0.
Thus from eq[6-12] we can calculate the time the particle takes to reach this position: t h
= v i sin θ i g
.
Substituting this expression for t h replacing y f
with h , we have
into eq[6-14] and h = ( v i sin θ i
) v i sin θ i g
–
1
2 g
 v i
 sin θ i g


2
, which reduces to h = v i
2
2
θ i
g
.
…[6-16]
It can be seen that, other things being equal, maximum h occurs for θ i ing v
= 90˚. h can be increased by increasi
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 2 t h noting that x h
= R at t = 2 t h
. Using t h
from above and
, we have
R = ( v i cos θ i
) 2 t h
= ( v i cos θ i
)
2 v i sin θ i g
,
=
2 v i
2 sin θ i g cos θ i
, which reduces to R = v i
2
θ i g
, …[6-17] since sin2 θ = 2sin θ cos θ . It can be seen that, other things being equal, a maximum R occurs for θ i
= 45˚.
As before, R can be increased by increasing v i operating in an environment where g is small.
or by
We now have the equations we need to describe any aspect of the motion of a projectile. Let us consider an example.
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?
40.0˚
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: v ix
= (20.0
m .
s
–1
)cos40.0
o
= 15.3
m .
s
–1 v iy
= (20.0
m .
s
–1
)sin 40.0
o
= 12.9
m .
s
–1
The horizontal component of the initial velocity remains unchanged at
06-4

v ix
= v fx
= v x
= 15.3
m .
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 =
50 m
15.3
m .
s
–1
= 3.27
s .
For the vertical motion taking “up” as positive and
“down” as negative we have y = v
0 t +
1
2 a y t
2
= (12.9
m .
s
–1
)(3.27
s ) +
1
2
(–9.80
m .
s
–2
)(3.27
s )
2
= –10.2
m .
y is the vertical distance above the original level. Since y is negative here, the ball hits the opposite wall at a
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.
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 6-
6. 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
Note 06 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 ' O t .
Figure 6-6.
Two drivers A and B move past an observer O.
y y'
P r
PO r
PO '
O
O'
S' v
O ' O x
S v
O ' O t
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 dt r
( )
= d dt
( r
PO '
+ v
O ' O t
) 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
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
θ = sin
–1

 v v
WO
BW


= sin
–1

0.300

0.500


= 37.0
o
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.
d =
60.0 m v v
BW
WO
= 0.300
m .
s
= 0.500
m .
s
–1
–1
θ v
BO
O downriver
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
)(0.500
m .
s
–1
) = 0.400
m .
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: d t = v
BO
=
60.0
m
0.400
m .
s
–1
= 150 s .
The boat crosses the river in 150 seconds.
• 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
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
C
B
100.0 m
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 + cx
2 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