# 1-Dimensional Kinematics

PhysicsUniform Acceleration Equations

# 1-Dimensional Kinematics

Created 3 years ago

Duration 0:09:10
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Slide Content
1. ### Slide 1 - 1-Dimensional Kinematics

• Honors Physics
2. ### Slide 2 - Changes in Position

• Displacement
• ∆ x : change in position
• Velocity: rate of change in position
• vav = ∆ x
• ∆ t
• “average” : over an interval
• Slope on “x vs. t” graph: velocity
3. ### Slide 3 - Velocity

• “instantaneous” : at a given moment
• v = ∆ x as ∆ t  zero
• ∆ t
• v = lim ∆ x a.k.a. dx
• ∆ t dt
• Slope of tangent to curve on “x vs. t” graph
• ∆ t  zero
4. ### Slide 4 - Changes in Velocity

• ∆v : change in velocity
• Acceleration: rate of change in velocity
• aav = ∆ v
• ∆ t
• “average” : over an interval
• Slope on “v vs. t” graph: acceleration
• Area under “v vs. t” graph: displacement
5. ### Slide 5 - Acceleration

• “instantaneous” : at a given moment
• a = ∆v as ∆t  zero
• ∆t
• a = lim ∆ v a.k.a. dv
• ∆ t dt
• Slope of tangent to curve on “v vs. t” graph
• ∆ t  zero
6. ### Slide 6 - Changes in Acceleration

• Jerk: rate of change in acceleration
• Slope on “a vs. t” graph: Jerk
• Area under “a vs. t” graph: change in velocity
7. ### Slide 7 - Uniformly Accelerated Motion

• Since vav = vi + vf
• 2
• and vav = ∆ x thus.. ∆ x = vav ∆ t
• ∆ t
• ∆ x = vi + vf ∆ t
• 2
8. ### Slide 8 - Uniformly Accelerated Motion

• aav = ∆v = vf – vi
• ∆t tf – ti
• a = vf – vi
• t – 0
• vf = vi + at
9. ### Slide 9 - Uniformly Accelerated Motion

• Since vav = ½ (vi + vf)
• and vf = vi + at
• vav = ½ (vi + vf) = ½ [vi + (vi + at)] = vi + ½ at
• Since ∆ x = vav ∆ t
• ∆ x = (vi + ½ a ∆ t) ∆ t
• ∆ x = vi ∆ t + ½ a (∆ t)2
10. ### Slide 10 - Uniformly Accelerated Motion

• Since ∆ x = vi t + ½ at2
• And a = vf – vi so: t = vf – vi
• t a
11. ### Slide 11 - Uniformly Accelerated Motion

• Since ∆ x = vi t + ½ at2
• And a = vf – vi so: t = vf – vi
• t a
• ∆ x = vi vf – vi + ½ a vf – vi 2
• a a
• vf 2 = vi 2 + 2a ∆ x
• (
• )
• )
• (
12. ### Slide 12 - Uniformly Accelerated Motion

• Full Equations
• ∆ x = [(vi + vf)/2]∆ t
• vf = vi + at
• ∆ x = vi ∆ t + ½ a (∆ t)2
• vf 2 = vi 2 + 2a ∆ x
• Zero Initial Velocity
• ∆ x = (vf/2)∆ t
• vf = at
• ∆ x = ½ a (∆ t)2
• vf 2 = 2a ∆ x