829d ago

Two Dimensional Kinematics  

Vectors

  • Quantities with both magnitude and direction

  • Magnitude is a number value (with units)

  • Directions can be compass points, left, right, up, down, etc.

  • You can assume that every vector makes a right angle. You can use trigonometry to find the resultant vector (R).

  • This is the sum of both vectors.

  • To subtract a vector you add the negative of that vector.

  • The negative of that vector has an opposite direction (180)

Projectile Motion

  • Projectile motion describes objects travelling through space under the influence of gravity

  • This applies from when the object is launched to right before it hits the ground

  • ==The only force is the weight (force of gravity)==

  • Ay = -9.80 m/s

  • Ax = 0 m/s

  • In the X-Axis we can only use the equation vx = x/t

  • In the Y-Axis we can use 4 kinematic equations:

  • The easiest way to solve for projectile motion is to use a table!

  • Example: If a projectile was launched at 22 m/s at an angle of 35 what is the displacment after 2.50s?


X


Y

Vx:

22cos35 = 18.0

Voy:

22sin35 =12.6

x:

18.0/2.50 = 45.0 m

Vy:


t:

2.50s

t:

2.50s



a:

-9.80 m/s



y:

(12.6)(2.5) + 1/2(-9.8)(2.5)^2 = .88m

  • The displacement can now be solved for using vector trigonometry

  • Take the square root of 45^2 and .88^2 = 45m

Uniform Circular Motion

  • Uniform circular motion describes objects moving in a circular path with a constant speed. The magnitude of the velocity is a constant.

  • At any moment the direction of the velocity is tangent to the path that it is following

  • This is called the tangential velocity

  • The magnitude can be found using: v = d/t → 2πr/T

  • ==Acceleration can cause a change in direction but does not change the magnitude of the velocity==

  • It is always directed towards the center of the path

  • Centripetal means directed to the center

  • Centripetal acceleration can be found using v^2/r

  • Example: A car traveling at 25.0 m/s around a curve with radius 15 meters

  • If we were to find the centripetal acceleration we use Ac = v^2/r

  • Ac = 25^2/15 = 42 m/s


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Two Dimensional Kinematics  

Vectors

  • Quantities with both magnitude and direction
  • Magnitude is a number value (with units)
  • Directions can be compass points, left, right, up, down, etc.
  • You can assume that every vector makes a right angle. You can use trigonometry to find the resultant vector (R).
  • This is the sum of both vectors.
  • To subtract a vector you add the negative of that vector.
  • The negative of that vector has an opposite direction (180)

Projectile Motion

  • Projectile motion describes objects travelling through space under the influence of gravity

  • This applies from when the object is launched to right before it hits the ground

  • ==The only force is the weight (force of gravity)==

  • Ay = -9.80 m/s

  • Ax = 0 m/s

  • In the X-Axis we can only use the equation vx = x/t

  • In the Y-Axis we can use 4 kinematic equations:

  • The easiest way to solve for projectile motion is to use a table!

  • Example: If a projectile was launched at 22 m/s at an angle of 35 what is the displacment after 2.50s?

XY
Vx:22cos35 = 18.0Voy:22sin35 =12.6
x:18.0/2.50 = 45.0 mVy:
t:2.50st:2.50s
a:-9.80 m/s
y:(12.6)(2.5) + 1/2(-9.8)(2.5)^2 = .88m
  • The displacement can now be solved for using vector trigonometry
  • Take the square root of 45^2 and .88^2 = 45m

Uniform Circular Motion

  • Uniform circular motion describes objects moving in a circular path with a constant speed. The magnitude of the velocity is a constant.

  • At any moment the direction of the velocity is tangent to the path that it is following

  • This is called the tangential velocity

  • The magnitude can be found using: v = d/t → 2πr/T

  • ==Acceleration can cause a change in direction but does not change the magnitude of the velocity==

  • It is always directed towards the center of the path

  • Centripetal means directed to the center

  • Centripetal acceleration can be found using v^2/r

  • Example: A car traveling at 25.0 m/s around a curve with radius 15 meters

  • If we were to find the centripetal acceleration we use Ac = v^2/r

  • Ac = 25^2/15 = 42 m/s