SECTION 3'2 MOTION WITH CONSTANT ACCELERATION

1
SECTION 3.2 MOTION WITH CONSTANT ACCELERATION

•  Objectives
• Interpret position-time graphs for motion with
  constant acceleration.
• Determine mathematical relationships among
  position, velocity, acceleration, and time.
• Apply graphical and mathematical relationships to
  solve problems related to constant acceleration.

2
INTRO/VELOCITY WITH AVERAGE ACCELERATION

• The definition of average acceleration can be
  manipulated similar to how average velocity can
  be manipulated to show the new velocity after a
  period of time given the initial velocity and the
  average acceleration.
• ?a vf vi ?v Average Acceleration
  Equation
• tf ti ?t
•  
• So ?v a?t and thus vf vi a?t
•  
• Final Velocity is equal to the initial velocity
  plus the product of the acceleration and time.
  It can be found using the equation
• vf vi a?t or vf vi at
•  
• Do Practice Problems p. 65 18-21

3
POSITION WITH CONSTANT ACCELERATION

• You have learned that an object experiencing
  constant acceleration changes its velocity at a
  constant rate.
•  
• The area under the curve of a Velocity Time graph
  is the Displacement.
• Example 3 p. 67
• v ?d / ?t gt 75 ?d / 1 gt 75 m ?d
• v ?d / ?t gt 75 ?d / 2 gt 150 m ?d
• Do Practice Problems p. 67 22-25

4
POSITION WITH CONSTANT ACCELERATION

• ?d ?drectangle ?dtriangle vi?t ½ a?t2
•  
• df di vi?t ½ a?t2 or df di vi?t
  ½ a?t2
•  
• Position with Average Acceleration an objects
  position at a time after the initial time is
  equal to the sum of its initial position, the
  product of the initial velocity and the time, and
  half the product of the acceleration and the
  square of the time.
• df di vi?t ½ a?t2
•  
• Or from the old book the equation is
• d vit ½ at2

5
SOME OTHER FORMULAS

• v d / t or d vt
• ?v ½ (vf vi)
•  d ½ (vf vi)t

6
AN ALTERNATIVE EXPRESSION

• t ?v / a and d vit ½ at2
• And if we combine them we get
• d vi(vf vi) / a ½ a(vf vi) / a 2
• and this will rearrange to the Velocity with
  Constant Acceleration
•  
• Velocity with Constant Acceleration the square
  of the final velocity equals the sum of the
  square of the initial velocity and twice the
  product of the acceleration and the displacement
  since the initial time.
• vf2 vi2 2a(df di) or vf2 vi2
  2ad

7
AN ALTERNATIVE EXPRESSION

• Table 3.3 on p. 68 shows the 3 equations also the
  3 from old book.
• Do Example 4 p. 69
• vf2 vi2 2ad gt 252 02 2(3.5)d gt 625
  7d gt 89.286 m d
•  
• Do Practice Problems p. 69 26-29

8
AN ALTERNATIVE EXPRESSION

• Do Example 5 p. 70
• v d / t And vf2 vi2 2ad So total
  distance
• 25 d / .45 02 252 2(-8.5)d
  11.25 36.765
• 11.25 m d -625 -17d 48.015 m
• 36.765 m d
• Do Practice Problems p. 71 30-33
• Do 3.2 Section Review p. 71 34-41