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Vectors  (Page 7/10)

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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.

Resultant vectors

  1. Harold walks to school by walking 600 m Northeast and then 500 m N 40 ∘ W. Determine his resultant displacement by using accurate scale drawings.
  2. A dove flies from her nest, looking for food for her chick. She flies at a velocity of 2 m · s - 1 on a bearing of 135 ∘ and then at a velocity of 1,2 m · s - 1 on a bearing of 230 ∘ . Calculate her resultant velocity by using accurate scale drawings.
  3. A squash ball is dropped to the floor with an initial velocity of 2,5 m · s - 1 . It rebounds (comes back up) with a velocity of 0,5 m · s - 1 .
    1. What is the change in velocity of the squash ball?
    2. What is the resultant velocity of the squash ball?

Remember that the technique of addition and subtraction just discussed can only be applied to vectors acting along a straight line. When vectors are not in a straight line, i.e. at an angle to each other, the following method can be used:

A more general algebraic technique

Simple geometric and trigonometric techniques can be used to find resultant vectors.

A man walks 40 m East, then 30 m North. Calculate the man's resultant displacement.

  1. As before, the rough sketch looks as follows:

  2. Note that the triangle formed by his separate displacement vectors and his resultant displacement vector is a right-angle triangle. We can thus use the Theorem of Pythagoras to determine the length of the resultant. Let x R represent the length of the resultant vector. Then:

    x R 2 = ( 40 m ) 2 + ( 30 m ) 2 x R 2 = 2 500 m 2 x R = 50 m

  3. Now we have the length of the resultant displacement vector but not yet its direction. To determine its direction we calculate the angle α between the resultant displacement vector and East, by using simple trigonometry:

    tan α = opposite side adjacent side tan α = 30 40 α = tan - 1 ( 0 , 75 ) α = 36 , 9 ∘

  4. The resultant displacement is then 50 m at 36,9 ∘ North of East.

    This is exactly the same answer we arrived at after drawing a scale diagram!
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Source:  OpenStax, Physics - grade 10 [caps 2011]. OpenStax CNX. Jun 14, 2011 Download for free at http://cnx.org/content/col11298/1.3
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