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The parallelogram method is restricted to the addition of just two vectors. However, it is arguably the most intuitive way of adding twoforces acting on a point.

Algebraic addition and subtraction of vectors

Vectors in a straight line

Whenever you are faced with adding vectors acting in a straight line (i.e. some directed left and some right, or some acting up and others down) you can use a very simple algebraic technique:

Method: Addition/Subtraction of Vectors in a Straight Line

  1. Choose a positive direction. As an example, for situations involving displacements in the directions west and east, youmight choose west as your positive direction. In that case, displacements east are negative.
  2. Next simply add (or subtract) themagnitude of the vectors using the appropriate signs.
  3. As a final step the direction of the resultant should be included in words (positive answers are in the positive direction, while negativeresultants are in the negative direction).

Let us consider a few examples.

A tennis ball is rolled towards a wall which is 10 m away from the ball. If after striking the wall the ball rolls a further 2,5 m along the ground away from the wall, calculate algebraically the ball's resultant displacement.

  1. We know that the resultant displacement of the ball ( x R ) is equal to the sum of the ball's separate displacements ( x 1 and x 2 ):

    x R = x 1 + x 2

    Since the motion of the ball is in a straight line (i.e. the ball moves towards and away from the wall), we can use the method of algebraic additionjust explained.

  2. Let's choose the positive direction to be towards the wall. This means that the negative direction is away from the wall.

  3. With right positive:

    x 1 = + 10 , 0 m · s - 1 x 2 = - 2 , 5 m · s - 1
  4. Next we simply add the two displacements to give the resultant:

    x R = ( + 10 m · s - 1 ) + ( - 2 , 5 m · s - 1 ) = ( + 7 , 5 ) m · s - 1
  5. Finally, in this case towards the wall is the positive direction , so: x R = 7,5 m towards the wall.

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Suppose that a tennis ball is thrown horizontally towards a wall at an initial velocity of 3 m · s - 1 to the right. After striking the wall, the ball returns to the thrower at 2 m · s - 1 . Determine the change in velocity of the ball.

  1. A quick sketch will help us understand the problem.

  2. Remember that velocity is a vector. The change in the velocity of the ball is equal to the difference between the ball's initial and finalvelocities:

    Δ v = v f - v i

    Since the ball moves along a straight line (i.e. left and right), we can use the algebraic technique of vector subtraction just discussed.

  3. Choose the positive direction to be towards the wall. This means that the negative direction is away from the wall.

  4. v i = + 3 m · s - 1 v f = - 2 m · s - 1
  5. Thus, the change in velocity of the ball is:

    Δ v = ( - 2 m · s - 1 ) - ( + 3 m · s - 1 ) = ( - 5 ) m · s - 1
  6. Remember that in this case towards the wall means a positive velocity , so away from the wall means a negative velocity : Δ v = 5 m · s - 1 away from the wall.

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Questions & Answers

What is a vector
Mercii Reply
vector is anything that has both a direction and a magnitude .they are usually drawn as pointed arrows ,the length of which represents a vector's magnitude
Tetteh
what is electronics?
Edward
how to calculate the reading on voltmeter or ammeter
mosima Reply
why is HCl considered a strong acid
Sphiwe Reply
it dissociate almost completely
Tetteh
what is metal displacement
Andile Reply
what is an electric field?
Noluthando Reply
is the charge of an electron always 1,6 ×10^-19? and the mass is always 9,1×10^-13?
Neo Reply
how to calculate a distance between charges
Celucolo Reply
juss apply the formula of the Electrostatic force
Noluthando
please what's the simplest way to derived Schrodinger's wavelength equation
Maimo
how to show polarity
Sandiso Reply
what can ii do to pass physics
Slindile Reply
Use previous years question papers to understand how questions are answered and asked
Joseph
but it hard ii am a slow learner
Slindile
Then the best thing to do is that immediately you are done reading through a certain topic, and you think you understood everything in that topic that's when you can use previous question papers and answer questions related to the topic. I think that's not difficult
Joseph
mm ii will try
Slindile
Good
Joseph
Here's a tip in reading textbooks, don't read it like a novel. First, flip through the pages—scan the chapter that you wanted to read. Second, go to the end of the chapter. Usually, there's a quiz at the end, so if will give you the important information that you need to know.
Sarah
Third, go to the beginning of the chapter and read through the words that were printed in bold: Titles, subtitles, headings, important words —because it helps to break down information.
Sarah
Fourth, read the first sentence of the chapter—if it is written by a good author; therefore, it will also have a good introduction. Check also the last sentence of the chapter to sum it up. Finally, read the whole chapter. You won't read it twice anymore.
Sarah
It looks hard, cause there are so much to do but read it thoroughly, it's easy and it will help you to save time and comprehend better. If you don't really have interest on reading—there are various of videos in youtube😊
Sarah
Oh, so em dashes turn into question marks :' (( nvm. Goodluck to all of us!
Sarah
How many ways can we calculate the empirical formula
Joseph Reply
How is current divided between resistors in parallel
Joseph
1/R=1/R+1/R
Slindile
tanx
Joseph
ii know 2 ways
Slindile
what's the other way
Joseph
What is the meaning of Coulomb's law
Nozipho Reply
Electronic magnetic field
Pride Reply
what is an emf ?
Nobuhle Reply
electromotive force
Tetteh
what are moments of a force
Mercy Reply
the moment of a force" is a measure of a tendency to cause a body to rotate about a specific point or axis....a moment is due to force not having equal and opposite force directly along its line of action.
Bhuboy
how to calculate the magnitude of the force of repulsion
Vicky Reply

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Source:  OpenStax, Siyavula textbooks: grade 11 physical science. OpenStax CNX. Jul 29, 2011 Download for free at http://cnx.org/content/col11241/1.2
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