06-1

Note 06 Motion in Two DimensionsSections Covered in the Text: Chapters 6 & 7, except 7.5 & 7.6

It is time to extend the definitions we developed inNote 03 to describe motion in 2D space. In doing sowe shall find that the vector nature of the definitions—displacement, velocity and acceleration—take onmore importance than before. Having developedthese tools we then apply them to describe animportant, special example of motion in 2D space,namely the motion of a projectile. As before, we shallmodel the object as a point particle.

Displacement, Velocity and AccelerationConsider an object modeled as a point particlemoving along an arbitrary path in the xy-plane(Figure 6-1). We assume that we are able to detect theparticle’s position at any point and to measure thecorres-ponding clock time. Two positions i and f inthe particle’s path are shown. Let the vectors thatlocate these positions be ri and r f , respectively. Thetime that elapses between the two positions is ∆t = tf –ti.

x

y

i

f path of objectΔr = r f – riri

rf

d

Figure 6-1. An object modelled as a point particle movesalong an arbitrary path in 2D space.

Distance TraveledThe distance d the particle travels between the twopositions is indicated approximately in the figure.This would be the number obtained if it were possibleto lay a flexible tape measure along the particle’s path.d is difficult to measure or calculate in all but thesimplest of paths. By its nature, d is always a positivenumber.

DisplacementThe displacement of the particle between the twopositions is different from the distance travelled; it isthe difference between the corresponding positionvectors:

Δr = r f – ri .

Obviously, in contrast to motion in 1D space, positionvectors in 2D space do not, in general, always point inthe same direction. Moreover, in general, the displace-ment vector points in a direction different from theposition vectors that define its limits.

VelocityTwo types of velocity are defined in 2D space as in 1Dspace: average velocity and instantaneous velocity.

Average VelocityThe average velocity of the particle is defined as thedisplacement divided by the corresponding elapsedtime ∆t:

v = ΔrΔt

. …[6-1]

Since average velocity is proportional to displacement(the proportionality factor being 1/∆t, a scalar), it is avector pointing in the same direction as the displace-ment vector. In any elapsed time the average velocitydepends on the end positions of the interval (as doesthe displacement), and is therefore independent of thepath taken by the particle.

Instantaneous VelocityThe instantaneous velocity of the particle is defined asthe limit of the average velocity as the elapsed time ∆tbecomes vanishingly small, i.e., as ∆t → 0:

v = limΔt→0

v = limΔ t→ 0

ΔrΔt

=drdt

. …[6-2]

Thus the instantaneous velocity vector equals the firstderivative of the position vector with respect to time(or the slope of the tangent to the r(t) function). Thedirection of the instantaneous velocity vector at anyposition in a particle’s path is along a tangent line tothe path at that position and in the direction ofmotion. The instantaneous speed is the magnitude ofthe instantaneous velocity.

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06-2

Figure 6-1 depicts the motion of the particle in termsof position or displacement vectors; but the motioncan also be described in terms of velocity vectors. Letthe instantaneous velocity vectors at positions i and fbe as drawn in Figure 6-2. In general, the change invelocity vector (small diagram in the figure) points ina direction different from the instantaneous velocityvectors. Since, in general, the instantaneous velocityvectors at positions i and f are unequal, the particle is,by definition, undergoing an acceleration.

x

y

i

fpath of object

vi

v f

Δv = v f – vi

vi

v f

Figure 6-2. The motion of a particle described in terms ofvelocity vectors. The velocity and position vectors (comparethis figure with Figure 6-1) are drawn with different headstyles to emphasize their different natures.

AccelerationTwo types of acceleration are defined in 2D space:average acceleration and instantaneous acceleration.

Average AccelerationThe average acceleration of a particle over an elapsedtime ∆t is defined as the change in velocity divided bythe elapsed time:

a =vf – vitf – ti

=ΔvΔt

. …[6-3]

The average acceleration is a vector divided by a scal-ar, and is therefore a vector. The average accelerationvector points in the same direction as the change invelocity vector (see Figure 6-3).

Instantaneous AccelerationThe instantaneous acceleration of a particle is defined asthe limit of the average acceleration as the elapsedtime becomes vanishingly small, i.e., as ∆t → 0:

x

y

i

fpath of object

vi

v f

Δv = v f – vi

vi

v f

a = ΔvΔt

Figure 6-3. The average acceleration vector of the particle hasthe direction of the change in velocity vector. Accelerationis a different vector type than is velocity or displacement.

a = limΔ t→ 0

a = limΔt→0

ΔvΔt

=dvdt

. …[6-4]

Thus the instantaneous acceleration vector is the firstderivative of the instantaneous velocity vector withrespect to time. The direction of the instantaneousacceleration vector at any position in a particle’s pathis along a tangent line to the velocity function at thatposition.

Two-Dimensional Motion with ConstantAcceleration

The kinematic equations of motion we developed inNote 03 (eqs[3-7], [3-11] and [3-12])) can be general-ized by writing them in full vector notation. We shallsimply state the results here. For convenience weassume the particle is moving with constant accelera-tion in 2D space. The instantaneous position vector ofa particle at a point (x,y) is

r = x) i + y

) j . …[6-5]

If, over an elapsed time t, the velocity of the particlechanges from vi to v f then

v f = vi + at . …[6-6]

If, over the elapsed time t the position of the particlechanges from r i to r f then

r f = ri + vit + 12at2 . …[6-7]

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06-3

In 2D space eq[6-6] gives rise to an equation for eachof the x- and y-components

vxf = vxi + axt…[6-8]

and vyf = vyi + ayt .

Eq[6-7] gives rise to the equations

x f = xi + vxit +12axt

2

…[6-9]and yf = yi + vyit +

12ayt

2 .

These equations emphasize that motion with constantacceleration in 2D space is equivalent to twoindependent motions in the x- and y-directions hav-ing constant accelerations ax and ay, respectively. Thecomponent of motion in the x-direction does not affectthe component of motion in the y-direction and vice-versa. With these tools we are ready to describe themotion of a projectile.

Projectile MotionOne of the early successes of Natural Science in therenaissance was the correct description of the motionof a projectile. This problem was pursued by the bestminds of the day, including Da Vinci, Galileo andothers, as it had important usage in the art of war. Theburning question was if a cannonball is fired with acertain velocity at a certain angle above the horizon,how high will the ball go and how far will it travel? Inother words, what will be the cannonball’s maximumheight and range? This was important in putting acannonball where you wanted it to go.

We have all seen a similar motion in baseball. Whena player hits or throws the ball, the ball travels in acurved path of a certain maximum height and range.One can demonstrate photographically that if theresistance of the air is negligible then the path of theprojectile is a parabola.1 In other words the functionalform of the path is

y(x) = ax + bx2 , …[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 anellipse. However, if the particle moves close to the Earth so that itsacceleration remains constant then to a good approximation thepath can be described by a parabola.

tion is to derive this kind of expression for projectilemotion from kinematics.

To begin our description, we model the problem inxy-space with the y-direction positive upward (Figure6-4). We assume that the particle is projected from theorigin (0,0) with an initial velocity vi at an angle θiwith respect to the horizontal. The position of theparticle at any subsequent clock time t is given by theposition vector r .

y

x

vxi

vxi

vyf

(R/2,h)

θi

vxi

vyivi

ri

Figure 6-4. The path of a projectile in 2D space. The positionand velocity vectors are drawn with different head styles toemphasize their different natures.

We make the following assumptions:1 the acceleration of the particle is a constant of

value ay = –g directed downwards2 the effect of air resistance is negligible.2

This means that the acceleration of the particle has anon-zero component in the y-direction only; the x-component of the acceleration is zero. The initialvelocity vector can be resolved into its x- and y-com-ponents

vxi = vi cosθi and vyi = vi sinθi .

Substituting these expressions into eqs[6-8] and [6-9]and taking ax = 0 and ay = –g yields

vxf = vxi = vi cosθ i = constant …[6-11]

vyf = vyi – gt = vi sinθi – gt …[6-12]

2 If we do not make this assumption the problem is morecomplicated. Anyone who has watched a baseball game has seenbaseballs veer downwards, to the left or right in response to airfriction or the wind.

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06-4

x f = xi + vxit = vi cosθi( )t , …[6-13]

yf = yi + vyit –12gt2 = vi sinθ i( )t – 1

2gt2 …[6-14]

If we solve for t in eq[6-13] and substitute it into eq[6-14] we get

yf = tanθ x( )x f –g

2vi2 cos2θi

x f

2 …[6-15]

This expression has the form of eq[6-10] and provesthat the path of the projectile is, indeed, a parabolaaccording to the assumptions we have made.

Horizontal Range and Maximum HeightWe are now in a position to derive expressions for themaximum height h and horizontal range R. Theseoccur, 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 theparticle takes to reach this position:

th =vi sinθ ig

.

Substituting this expression for th into eq[6-14] andreplacing yf with h, we have

h = vi sinθ i( ) vi sinθ ig

– 12g vi sinθ i

g

2

,

which reduces to h = vi2 sin2θ i2g

. …[6-16]

It can be seen that, other things being equal, maxim-um h occurs for θi= 90˚. h can be increased by increas-ing v i or by operating in an environment (on a planetor small astronomical body) where g is very small.

Now the range R is the horizontal distance traveledby the particle in twice the time it takes to reach thepeak, that is, in a time 2th. Using th from above andnoting that xh = R at t = 2th, we have

R = vi cosθi( )2th = vi cosθ i( ) 2vi sinθig

,

=2vi2 sinθi cosθ i

g ,

which reduces to R = vi2 sin 2θ ig

, …[6-17]

since sin2θ = 2sinθcosθ. It can be seen that, otherthings being equal, a maximum R occurs for θi = 45˚.As before, R can be increased by increasing vi or byoperating in an environment where g is small.

We now have the equations we need to describe anyaspect of the motion of a projectile. Let us consider anexample.

Example Problem 6-1An Example of Projectile Motion

A ball is thrown from the top of one building towarda tall building 50.0 m away (Figure 6-5). The initialvelocity of the ball is 20.0 m.s–1, 40.0˚ above the horiz-ontal. How far above or below the original level doesthe ball strike the opposite wall?

50.0 m

40.0˚

Figure 6-5. A ball is projected at 40.0˚ above the horizontal.

Solution:We have the following components of the initialvelocity:

vix = (20.0m.s–1 )cos40.0o = 15.3m.s –1

viy = (20.0m.s–1)sin 40.0o = 12.9m.s –1

The horizontal component of the initial velocityremains unchanged at

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06-5

vix = v fx = vx = 15.3m.s–1

Since the x-component of acceleration is zero, we canuse the expression x = vxt to solve for t where t is thetime that elapses between launch and impact on theopposite wall. Thus t is

t = 50m15.3m.s–1

= 3.27s .

For the vertical motion taking “up” as positive and“down” as negative we have

y = v0t +12ayt

2

= (12.9m.s–1)(3.27s) + 12(–9.80m.s–2 )(3.27s)2

= –10.2m .

y is the vertical distance above the original level. Sincey is negative here, the ball hits the opposite wall at apoint 10.2 m below the original level.

Another, special acceleration is defined for objectsmoving with uniform speed in a circle. Though thistopic is presented at this point in the textbook, wepostpone it until a later note.

Relative VelocityThus far we have taken the origin of our coordinatesystem as a fixed point (the point we have taken as 0in a 1D frame or (0,0) in a 2D frame). There will betimes when we need to relate certain measurementsmade in different systems, in particular when onesystem 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 thirdobserver O who is standing beside the road. Omeasures the speeds of A and B at 60 km/hr. Butaccording to observer A what is the speed of B? As weall know from experience A would say that B’s speedis 0 km/hr. In other words the speed of B relative to Ois 60 km/hr, but relative to A is 0 km/hr.

We can quantify the physics of these observationswith the help of the position vectors in Figure 6-7.Observers O and O’ in their own reference frames (Sand S’) observe the same event P. O locates the eventwith the position vector r po , while O’ locates the

same event with the vector r po' .O’ and his/her reference frame is moving with

velocity vo'o relative to O. The two position vectorsare related by

rPO = rPO' + vO' Ot .

Figure 6-6. Two drivers A and B move past an observer O.

x

y

O O'S S'

y' P

vO' O

rPOrPO'

vO' OtFigure 6-7. Observers in stationary and moving referenceframes describe the same event P with different position andvelocity vectors.

If we differentiate this expression with respect to timewe obtain an expression for the correspondingvelocity vectors:

ddt

rPO( ) = ddt

rPO' + vO'Ot( )

gives v po = vpo' + vo' o .

The subscripts po, po’ and O’O mean “p wrt O”, “pwrt O’” and “O’ wrt O”. Thus the velocity of P wrt Oequals the velocity of P wrt O’ plus the velocity of O’wrt O. This result is made clearer with the help of anexample.

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06-6

Example Problem 6-2A Boat Crossing a River

A boat propelled with a speed of 0.500 m.s–1 in stillwater moves directly across a river that is 60.0 mwide. The river flows with a speed of 0.300 m.s–1

relative to the riverbank. (a) At what angle, relative tothe straight-across direction, must the boat bepointed? (b) How long does it take the boat to crossthe river?

Solution:Let the velocity of the water wrt an observer on theriverbank (magnitude 0.300 m.s–1) be denoted vWO .Let the velocity of the boat relative to the water (mag-nitude 0.500 m.s–1) be denoted vBW . Let the velocityof the boat relative to the riverbank be denoted vBO .The vectors are arranged as shown in Figure 6-8.

(a) The boat must therefore be steered upstream bythe angle

θ = sin–1 vWOvBW

= sin–1

0.3000.500

= 37.0o upstream

with respect to the water. It is the subsequent drag ofthe water flowing downstream that results in the boatmoving directly across the river.

d =60.0 m

downriver

O

vWO = 0.300m.s–1

vBW = 0.500m.s –1θ vBO

Figure 6-8. Velocity diagram for a boat crossing a river.

(b) The speed of the boat with respect to the riverbankis the magnitude of the vector vBO . This is

vBO = (cos37o)(0.500m.s–1) = 0.400m.s–1 .

This is the resultant speed of the boat wrt the river-bank. The time required for the boat to go directlyacross the river is the distance travelled divided bythis speed:

t = dvBO

=60.0m

0.400m.s–1= 150s .

The boat crosses the river in 150 seconds.

To Be Mastered

• Definitions: distance travelled, displacement, average velocity, instantaneous velocity, instantaneous speed, averageacceleration, instantaneous acceleration

• kinematic equations (committed to memory)• projectile motion problem• relative velocity problems

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06-7

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 heightreached by the ball be h and the total elapsed time be ∆t. Assume the ball is at the same, negligible heightabove 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 inthe figure. He releases a sack of flour intending to hit a target. Assume air friction is negligible and answer thefollowing questions:

100.0 mA

BC

target

sack

(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). Ithits 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.

Note 06

06-8

10.0 m

particle

100.0 m