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A variety of mathematical operations can be performed with and upon

vectors. One such operation is the addition of vectors. Two vectors can be

added together to determine the result (or resultant). This process of adding

two or more vectors has already been discussed in an earlier unit. Recall in

our discussion of Newton's laws of motion that the net force e!perienced by

an ob"ect was determined by computing the vector sum of all the individual

forces acting upon that ob"ect. That is the net force was the result

(or resultant) of adding up all the force vectors. #uring that unit the rules for

summing vectors (such as force vectors) were $ept relatively simple.

Observe the following summations of two force vectors%

 

 These rules for summing vectors were applied to free&body diagrams in order

to determine the net force (i.e. the vector sum of all the individual forces).

ample applications are shown in the diagram below.

n this unit the tas$ of summing vectors will be e!tended to more

complicated cases in which the vectors are directed in directions other than

purely vertical and horiontal directions. *or e!ample a vector directed up

and to the right will be added to a vector directed up and to the left.

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 The vector sum will be determined for the more complicated cases shown in

the diagrams below.

 There are a variety of methods for determining the magnitude and directionof the result of adding two or more vectors. The two methods that will be

discussed in this lesson and used throughout the entire unit are%

• the +ythagorean theorem and trigonometric methods

• the head&to&tail method using a scaled vector diagram

 

The Pythagorean Theorem

 The +ythagorean theorem is a useful method for determining the result of

adding two (and only two) vectors that ma$e a right angle to each other. The

method is not applicable for adding more than two vectors or for adding

vectors that are not at ,-&degrees to each other. The +ythagorean theorem is

a mathematical euation that relates the length of the sides of a right

triangle to the length of the hypotenuse of a right triangle.

 To see how the method wor$s consider the following problem%

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/ric leaves the base camp and hi$es 00 $m north and then hi$es 00 $m

east. #etermine /ric's resulting displacement.

 This problem as$s to determine the result of adding two displacement

vectors that are at right angles to each other. The result (or resultant) of

wal$ing 00 $m north and 00 $m east is a vector directed northeast as shownin the diagram to the right. ince the northward displacement and the

eastward displacement are at right angles to each other the +ythagorean

theorem can be used to determine the resultant (i.e. the hypotenuse of the

right triangle).

 The result of adding 00 $m north plus 00 $m east is a vector with a

magnitude of 01.2 $m.3ater the method of determining the direction of the

vector will be discussed.

3et's test your understanding with the following two practice problems. n

each case use the +ythagorean theorem to determine the magnitude of

the vector sum. 4hen 5nished clic$ the button to view the answer.

 

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ee Answer to +ractice A

 

ee Answer to +ractice 6

 

Using Trigonometry to Determine a Vector's Direction

 The direction of a resultant  vector can often be determined by use of

trigonometric functions. 7ost students recall the meaning of the usefulmnemonic O8 9A8 TOA from their course in trigonometry. O8 9A8 TOA is

a mnemonic that helps one remember the meaning of the three common

trigonometric functions & sine cosine and tangent functions. These three

functions relate an acute angle in a right triangle to the ratio of the lengths

of two of the sides of the right triangle. The sine function relates the

measure of an acute angle to the ratio of the length of the side opposite the

angle to the length of the hypotenuse. The cosine function relates the

measure of an acute angle to the ratio of the length of the side ad"acent the

angle to the length of the hypotenuse. The tangent function relates the

measure of an angle to the ratio of the length of the side opposite the angleto the length of the side ad"acent to the angle. The three euations below

summarie these three functions in euation form.

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 These three trigonometric functions can be applied to the hi$er problem in

order to determine the direction of the hi$er's overall displacement. The

process begins by the selection of one of the two angles (other than the right

angle) of the triangle. Once the angle is selected any of the three functions

can be used to 5nd the measure of the angle. 4rite the function and proceedwith the proper algebraic steps to solve for the measure of the angle. The

wor$ is shown below.

Once the measure of the angle is determined the direction of the vector can

be found. n this case the vector ma$es an angle of :1 degrees with due

/ast. Thus the direction of this vector is written as :1 degrees. (Recallfrom earlier in this lesson that the direction of a vector is the

countercloc$wise angle of rotation that the vector ma$es with due /ast.)

 

The Calculated Angle is Not Always the Direction

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 The measure of an angle as determined through use of O8 9A8 TOA

is not always the direction of the vector. The following vector addition

diagram is an e!ample of such a situation. Observe that the angle within the

triangle is determined to be ;2.2 degrees using O8 9A8 TOA. This angle is

the southward angle of rotation that the vector R ma$es with respect to

4est. <et the direction of the vector as e!pressed with the 994

(countercloc$wise from /ast) convention is ;-2.2 degrees.

 

 Test your understanding of the use of O8 9A8 TOA to determine the vector

direction by trying the following two practice problems. n each case useO8 9A8 TOA to determine the direction of the resultant. 4hen 5nished

clic$ the button to view the answer.

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ee Answer to +ractice A

 

ee Answer to +ractice 6

 

n the above problems the magnitude and direction of the sum of two

vectors is determined using the +ythagorean theorem and trigonometric

methods (O8 9A8 TOA). The procedure is restricted to the addition of two

vectors that ma$e right angles to each other. 4hen the two vectors that areto be added do not ma$e right angles to one another or when there are

more than two vectors to add together we will employ a method $nown as

the head&to&tail vector addition method. This method is described below.

 

Use of Scaled Vector Diagrams to Determine a Resultant

 The magnitude and direction of the sum of two or more vectors can also be

determined by use of an accurately drawn scaled vector diagram. =sing a

scaled diagram the head-to-tail method is employed to determine the

vector sum or resultant. A common +hysics lab involves a vector walk . /ither

using centimeter&sied displacements upon a map or meter&sied

displacements in a large open area a student ma$es several consecutive

displacements beginning from a designated starting position. uppose that

you were given a map of your local area and a set of 0> directions to follow.

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tarting at home base these 0> displacement vectors could be added

together  in consecutive fashion to determine the result of adding the set of

0> directions. +erhaps the 5rst vector is measured 1 cm /ast. 4here this

measurement ended the ne!t measurement would begin. The process would

be repeated for all 0> directions. /ach time one measurement ended the

ne!t measurement would begin. n essence you would be using the head&to&

tail method of vector addition.

 

 The head&to&tail method involves drawing a vector to scale on a sheet of

paper beginning at a designated starting position. 4here the head of this

5rst vector ends the tail of the second vector begins (thus head-to-

tail method). The process is repeated for all vectors that are being added.

Once all the vectors have been added head&to&tail the resultant is then

drawn from the tail of the 5rst vector to the head of the last vector? i.e. from

start to 5nish. Once the resultant is drawn its length can be measured and

converted to real units using the given scale. Thedirection of the resultant

can be determined by using a protractor and measuring its countercloc$wise

angle of rotation from due /ast.

A step&by&step method for applying the head&to&tail method to determine the

sum of two or more vectors is given below.

0. 9hoose a scale and indicate it on a sheet of paper. The best choice ofscale is one that will result in a diagram that is as large as possible yet

5ts on the sheet of paper.;. +ic$ a starting location and draw the 5rst vector to scale in the

indicated direction. 3abel the magnitude and direction of the scale on thediagram (e.g. 9A3/% 0 cm @ ;- m).

. tarting from where the head of the 5rst vector ends draw the secondvector to scale in the indicated direction. 3abel the magnitude anddirection of this vector on the diagram.

:. Repeat steps ; and for all vectors that are to be added

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1. #raw the resultant from the tail of the 5rst vector to the head of thelast vector. 3abel this vector as Resultant or simply R.

2. =sing a ruler measure the length of the resultant and determine itsmagnitude by converting to real units using the scale (:.: cm ! ;- mB0cm @ >> m).

C. 7easure the direction of the resultant using the countercloc$wiseconvention discussedearlier in this lesson.

 

An e!ample of the use of the head&to&tail method is illustrated below. The

problem involves the addition of three vectors%

 

;- m :1 deg. D ;1 m -- deg. D 01 m ;0- deg.

 

9A3/% 0 cm @ 1 m

 The head&to&tail method is employed as described above and the resultant is

determined (drawn in red). ts magnitude and direction is labeled on the

diagram.

9A3/% 0 cm @ 1 m

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nterestingly enough the order in which three vectors are added has no

eEect upon either the magnitude or the direction of the resultant. Theresultant will still have the same magnitude and direction. *or e!ample

consider the addition of the same three vectors in a diEerent order.

 

15 m, 210 deg. + 25 m, 300 deg. + 20 m, 45 deg.

 

SCAL! 1 cm " 5 m

 

4hen added together in this diEerent order these same three vectors still

produce a resultant with the same magnitude and direction as before (;-. m

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0; degrees). The order in which vectors are added using the head&to&tail

method is insigni5cant.

9A3/% 0 cm @ 1 m

 

Additional e!amples of vector addition using the head&to&tail method are

given on a separate web page.

On occasion ob"ects move within a medium that is moving with respect to an

observer. *or e!ample an airplane usually encounters a wind & air that is

moving with respect to an observer on the ground below. As another

e!ample a motorboat in a river is moving amidst a river current & water that

is moving with respect to an observer on dry land. n such instances as this

the magnitude of the velocity of the moving ob"ect (whether it be a plane or

a motorboat) with respect to the observer on land will not be the same as

the speedometer reading of the vehicle. That is to say the speedometer on

the motorboat might read ;- miBhr? yet the motorboat might be movingrelative to the observer on shore at a speed of ;1 miBhr. 7otion is relative to

the observer. The observer on land often named (or misnamed) the

Fstationary observerF would measure the speed to be diEerent than that of

the person on the boat. The observed speed of the boat must always be

described relative to who the observer is.

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Tailwinds, eadwinds, and Side !inds

 To illustrate this principle consider a plane Gying amidst a tail#ind. A

tailwind is merely a wind that approaches the plane from behind thus

increasing its resulting velocity. f the plane is traveling at a velocity of 0--

$mBhr with respect to the air and if the wind velocity is ;1 $mBhr then what

is the velocity of the plane relative to an observer on the ground belowH The

resultant velocity of the plane (that is the result of the wind velocity

contributing to the velocity due to the plane's motor) is the vector sum of the

velocity of the plane and the velocity of the wind. This resultant velocity is

uite easily determined if the wind approaches the plane directly from

behind. As shown in the diagram below the plane travels with a resultingvelocity of 0;1 $mBhr relative to the ground.

 

f the plane encounters a headwind the resulting velocity will be less than

0-- $mBhr. ince a headwind is a wind that approaches the plane from the

front such a wind would decrease the plane's resulting velocity. uppose a

plane traveling with a velocity of 0-- $mBhr with respect to the air meets a

headwind with a velocity of ;1 $mBhr. n this case the resultant velocity

would be C1 $mBhr? this is the velocity of the plane relative to an observer on

the ground. This is depicted in the diagram below.

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Now consider a plane traveling with a velocity of 0-- $mBhr outh that

encounters a side #ind of ;1 $mBhr 4est. Now what would the resulting

velocity of the plane beH This uestion can be answered in the same manner

as the previous uestions. The resulting velocity of the plane is the vector

sum of the two individual velocities. To determine the resultant velocity the

plane velocity (relative to the air) must be added to the wind velocity. This is

the same procedure that was used above for the headwind and the tailwind

situations? only now the resultant is not as easily computed. ince the two

vectors to be added & the southward plane velocity and the westward wind

velocity & are at right angles to each other the +ythagorean theorem can be

used. This is illustrated in the diagram below.

 

n this situation of a side wind the southward vector can be added to the

westward vector using the usual methods of vector addition. The magnitudeof the resultant velocity is determined using +ythagorean theorem. The

algebraic steps are as follows%

(0-- $mBhr); D (;1 $mBhr); @ R;

0- --- $m;Bhr; D 2;1 $m;Bhr; @ R;

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0- 2;1 $m;Bhr; @ R;

IRT(0- 2;1 $m;Bhr;) @ R

103.1 $m%h& " R

 

 The direction of the resulting velocity can be determined using

a trigonometric function. ince the plane velocity and the wind velocity form

a right triangle when added together in head&to&tail fashion the angle

between the resultant vector and the southward vector can be determined

using the sine cosine or tangent functions. The tangent function can be

used? this is shown below%

 

tan (theta) @ (oppositeBad"acent)

tan (theta) @ (;1B0--)

theta @ invtan (;1B0--)

theta " 14.0 deg&ees

f the resultant velocity of the plane ma$es a 0:.- degree angle with thesouthward direction (theta in the above diagram) then the direction of the

resultant is ;12 degrees. 3i$e any vector the resultant's direction is

measured as a countercloc$wise angle of rotation from due /ast.

 

Analysis of a Ri"er#oat's $otion

 The eEect of the wind upon the plane is similar to the eEect of the river

current upon the motorboat. f a motorboat were to head straight across a

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river (that is if the boat were to point its bow straight towards the other

side) it would not reach the shore directly across from its starting point. The

river current inGuences the motion of the boat and carries it downstream.

 The motorboat may be moving with a velocity of : mBs directly across the

river yet the resultant velocity of the boat will be greater than : mBs and at

an angle in the downstream direction. 4hile the speedometer of the boat

may read : mBs its speed with respect to an observer on the shore will be

greater than : mBs.

 The resultant velocity of the motorboat can be determined in the same

manner as was done for the plane. The resultant velocity of the boat is the

vector sum of the boat velocity and the river velocity. ince the boat heads

straight across the river and since the current is always directed straight

downstream the two vectors are at right angles to each other. Thus

the +ythagorean theorem can be used to determine the resultant velocity.

uppose that the river was moving with a velocity of mBs North and the

motorboat was moving with a velocity of : mBs /ast. 4hat would be the

resultant velocity of the motorboat (i.e. the velocity relative to an observer

on the shore)H The magnitude of the resultant can be found as follows%

(:.- mBs); D (.- mBs); @ R;

02 m;Bs; D , m;Bs; @ R;

;1 m;Bs; @ R;

IRT (;1 m;Bs;) @ R

5.0 m%s " R

 

 The direction of the resultant is the countercloc$wise angle of rotation that

the resultant vector ma$es with due /ast. This angle can be determined

using a trigonometric function as shown below.

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tan (theta) @ (oppositeBad"acent)

tan (theta) @ (B:)

theta @ invtan (B:)

theta " 3'.( deg&ees

 

Jiven a boat velocity of : mBs /ast and a river velocity of mBs North the

resultant velocity of the boat will be 1 mBs at 2., degrees.

 

7otorboat problems such as these are typically accompanied by threeseparate uestions%

a. 4hat is the resultant velocity (both magnitude and direction) of the boatH

b. f the width of the river is X  meters wide then how much time does it ta$e

the boat to travel shore to shoreH

c. 4hat distance downstream does the boat reach the opposite shoreH

 The 5rst of these three uestions was answered above? the resultant velocity

of the boat can be determined using the +ythagorean theorem (magnitude)

and a trigonometric function (direction). The second and third of these

uestions can be answered using the average speed euation (and a lot of

logic).

a)e. s*eed " distance%time

9onsider the following e!ample.

%&amle (

A motorboat traveling : mBs /ast encounters a current traveling .- mBs

North.

a. 4hat is the resultant velocity of the motorboatH

b. f the width of the river is >- meters wide then how much time does it ta$e

the boat to travel shore to shoreH

c. 4hat distance downstream does the boat reach the opposite shoreH

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 The solution to the 5rst uestion has already been shown in the above

discussion. The resultant velocity of the boat is 1 mBs at 2., degrees. 4e

will start in on the second uestion.

 The river is >-&meters wide. That is the distance from shore to shore as

measured straight across the river is >- meters. The time to cross this >-&

meter wide river can be determined by rearranging and substituting into the

average speed euation.

time " distance %a)e. s*eed

 The distance of >- m can be substituted into the numerator. 6ut what about

the denominatorH 4hat value should be used for average speedH hould

mBs (the current velocity) : mBs (the boat velocity) or 1 mBs (the resultant

velocity) be used as the average speed value for covering the >- metersH

4ith what average speed is the boat traversing the >- meter wide riverH

7ost students want to use the resultant velocity in the euation since that is

the actual velocity of the boat with respect to the shore. <et the value of 1

mBs is the speed at which the boat covers the diagonal dimension of the

river. And the diagonal distance across the river is not $nown in this case. f

one $new the distance C in the diagram below then the a)e&age s*eed

C could be used to calculate the time to reach the opposite shore. imilarly

if one $new the distance in the diagram below then the a)e&age s*eed

 could be used to calculate the time to reach the opposite shore. And

5nally if one $new the distance A in the diagram below then thea)e&ages*eed A could be used to calculate the time to reach the opposite shore.

n our problem the >- m corresponds to the distance A and so the average

speed of : mBs (average speed in the direction straight across the river)

should be substituted into the euation to determine the time.

time " 0 m%4 m%s " 20 s

t reuires ;- s for the boat to travel across the river. #uring this ;- s of

crossing the river the boat also drifts downstream. +art c of the

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problem as$s F4hat distance downstream does the boat reach the opposite

shoreHF The same euation must be used to calculate this downstream

distance. And once more the uestion arises which one of the three

average speed values must be used in the euation to calculate the distance

downstreamH The distance downstream corresponds to /istance  on the

above diagram. The speed at which the boat covers this distance

corresponds to A)e&age S*eed  on the diagram above (i.e. the speed at

which the current moves & mBs). And so the average speed of mBs

(average speed in the downstream direction) should be substituted into the

euation to determine the distance.

distance " a)e. s*eed time " 3 m%s 20 s

distance " '0 m

 The boat is carried 2- meters downstream during the ;- seconds it ta$es to

cross the river.

 

 The mathematics of the above problem is no more diKcult than dividing or

multiplying two numerical uantities by each other. The mathematics is

easyL The diKculty of the problem is conceptual in nature? the diKculty lies

in deciding which numbers to use in the euations. That decision emerges

from one's conceptual understanding (or unfortunately one's

misunderstanding) of the comple! motion that is occurring. The motion of

the riverboat can be divided into two simultaneous parts & a motion in the

direction straight across the river and a motion in the downstream direction.

 These two parts (or components) of the motion occur simultaneously for the

same time duration (which was ;- seconds in the above problem). The

decision as to which velocity value or distance value to use in the euation

must be consistent with the diagram above. The boat's motor is what carries

the boat across the river the/istance A? and so any calculation involving

the /istance A must involve the speed value labeled as S*eed A (the boat

speed relative to the water). imilarly it is the current of the river thatcarries the boat downstream for the /istance ? and so any calculation

involving the/istance  must involve the speed value labeled as S*eed

 (the river speed). Together these two parts (or components) add up to

give the resulting motion of the boat. That is the across&the&river component

of displacement adds to the downstream displacement to eual the resulting

displacement. And li$ewise the boat velocity (across the river) adds to the

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river velocity (down the river) to eual the resulting velocity. And so any

calculation of the #istance 9 or the Average peed 9 (FResultant MelocityF)

can be performed using the +ythagorean theorem.

 

Now to illustrate an important point let's try a second e!ample problem that

is similar to the5rst e!ample problem. 7a$e an attempt to answer the three

uestions and then clic$ the button to chec$ your answer.

am*le 2

A motorboat traveling : mBs /ast encounters a current traveling C.- mBs

North.

a. 4hat is the resultant velocity of the motorboatH

b. f the width of the river is >- meters wide then how much time does it ta$e

the boat to travel shore to shoreH

c. 4hat distance downstream does the boat reach the opposite shoreH

ee Answer

 

An important concept emerges from the analysis of the two e!ample

problems above. n /!ample 0 the time to cross the >-&meter wide river

(when moving : mBs) was ;- seconds. This was in the presence of a mBs

current velocity. n /!ample ; the current velocity was much greater & C mBs

& yet the time to cross the river remained unchanged. n fact the current

velocity itself has no eEect upon the time reuired for a boat to cross the

river. The river moves downstream parallel to the ban$s of the river. As such

there is no way that the current is capable of assisting a boat in crossing a

river. 4hile the increased current may aEect the resultant velocity & ma$ingthe boat travel with a greater speed with respect to an observer on the

ground & it does not increase the speed in the direction across the river. The

component of the resultant velocity that is increased is the component that

is in a direction pointing down the river. t is often said that Fperpendicular

components of motion are independent of each other.F As applied to

riverboat problems this would mean that an across&the&river variable would

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be independent of (i.e. not be aEected by) a downstream variable. The time

to cross the river is dependent upon the velocity at which the boat crosses

the river. t is only the component of motion directed across the river (i.e.

the boat velocity) that aEects the time to travel the distance directly across

the river (>- m in this case). The component of motion perpendicular to this

direction & the current velocity & only aEects the distance that the boat

travels down the river. This concept of perpendicular components of motion

will be investigated in more detail in the ne!t part of 3esson 0.

 

Riverboat Problems with Examples

Ri"er#oat Pro#lems

Example Boat has velocity 5m/s with respect to the river. It aims to reach point A, however,

because of the river speed it reaches another point B. If the speed of current is 2m/s to the east

calculate the time of trip and the distance between A and B.

Boat follows the path shown in dashed line which is the direction of the resultant vector.

 

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his example can be examined under two part vertical and

hori!ontal motion as in the case of pro"ectile motion.

 

Example #elocity of the boat with respect to river is $% m/s. It passes the river and reaches

opposite shore at point &. If the velocity of the river is 'm/s, find the time of the trip and distance

between B and &.

 

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http://slidepdf.com/reader/full/a-variety-of-mathematical-operations-can-be-performed-with-and-upon-vectors 22/22

 

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