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Vector Addition and Magnitude - Prof. Rose A. Finn, Exams of Physics

This chapter discusses the addition of vectors and the relationship between vector magnitude and its components. It covers the cases where vectors are in the same direction, opposite directions, and other configurations. The text also explains how a particle's speed and velocity relate to acceleration.

Typology: Exams

Pre 2010

Uploaded on 08/09/2009

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Ch 3: Solutions, Part b
Questions
6. If the two vectors are in the same direction, the magnitude o f their sum wi ll be a maximum, and wi ll
be 7.5 km. If the two vectors are in the opposite direction, the magnitude of their sum will be a
minimum, and will be 0.5 km. If the two vec tors are oriented in any other configuration, the
magnitude of their su m will be betwe en 0.5 km and 7.5 km.
8. (a) The magnitude of a v ector can equ al the length of one of its components if the other
components of the vector are all 0; i.e., if the vector lies along one of the coordinate axe s.
(b) The magnitude of a vector can never be less th an one of its components, because each
component contributes a positive amount to the over all magnitude, through the Pythagorean
relationship. The square root of a sum of squares is never less than the absolute value of any
individual term.
9. A particle with constant speed can be accelerating, if its direction is changing. Driving on a curved
roadway at constant speed would be an example. However, a particle with constant velocity cannot
be accelerating – its acceleration mus t be zero. It has bo th constant speed and constant direction.
11. Assume that the bul let was fired from behind and below the airplane. As the bullet rose in the air, its
vertical speed would be slowed by both gravity and air resistance, and its horizontal speed would be
slowed by air resistance. If the altitude of the a irplane was slightly below the maximum height of the
bullet, then at the altitude of the airplane, the bullet would be moving qu ite slowly in th e vertical
direction. If th e bullet’s horizontal speed had also s lowed enough to approximately match the speed
of the a irplane, then the bullet’s velocity relative to the airplane would be small. With the bullet
moving slowly, it could safely be caught by hand.
Problems:
17. Choose downward to be the positive y direction. The origin will be at the point wh ere the tiger leaps
from the rock. In the horizontal direction,
sm 5.3
0=
x
v
and
.0=
x
a
In the vertical direction,
,0 ,sm 80.9 ,0 0
2
0=== yav yy
and the f inal location
m. 5.6=y
The time for the tiger to rea ch the
pf2

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Ch 3: Solutions, Part b Questions

  1. If the two vectors are in the same direction, the magnitude of their sum will be a maximum, and will be 7.5 km. If the two vectors are in the opposite direction, the magnitude of their sum will be a minimum, and will be 0.5 km. If the two vectors are oriented in any other configuration, the magnitude of their sum will be between 0.5 km and 7.5 km.
  2. ( a ) The magnitude of a vector can equal the length of one of its components if the other components of the vector are all 0; i.e., if the vector lies along one of the coordinate axes. ( b ) The magnitude of a vector can never be less than one of its components, because each component contributes a positive amount to the overall magnitude, through the Pythagorean relationship. The square root of a sum of squares is never less than the absolute value of any individual term.
  3. A particle with constant speed can be accelerating, if its direction is changing. Driving on a curved roadway at constant speed would be an example. However, a particle with constant velocity cannot be accelerating – its acceleration must be zero. It has both constant speed and constant direction.
  4. Assume that the bullet was fired from behind and below the airplane. As the bullet rose in the air, its vertical speed would be slowed by both gravity and air resistance, and its horizontal speed would be slowed by air resistance. If the altitude of the airplane was slightly below the maximum height of the bullet, then at the altitude of the airplane, the bullet would be moving quite slowly in the vertical direction. If the bullet’s horizontal speed had also slowed enough to approximately match the speed of the airplane, then the bullet’s velocity relative to the airplane would be small. With the bullet moving slowly, it could safely be caught by hand.

Problems:

  1. Choose downward to be the positive y direction. The origin will be at the point where the tiger leaps from the rock. In the horizontal direction, vx 0 = 3. 5 ms and ax = 0. In the vertical direction, v (^) y 0 = 0 , ay = 9. 80 ms^2 , y 0 = 0 , and the final location y = 6. 5 m.The time for the tiger to reach the

ground is found from applying Eq. 2-11b to the vertical motion.

y = yo + voyt + 12 ayt^2! 6.5m = 0 + 0 + 12 (9.8 m s^2 ) t^2! t = 2 9.8^ (^ 6.5 m^ m s 2 ) = 1.15 sec

The horizontal displacement is calculated from the constant horizontal velocity.

! x = vxt =( 3. 5 ms)(1.15 sec) = 4. 0 m