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Motion in One Dimension 2.2
Ch 2.2
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Acceleration Avg acceleration = change in velocity
time needed for change a = v = vf – vi t tf – ti
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Acceleration variable unit Acceleration a m/s2 Velocity v m/s
(final or initial) Time t s
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Acceleration is a vector
Acceleration has direction and magnitude Velocity time graph has a slope of acceleration See figure 10 on page 50 Check out Table 3 on page 51 Conceptual Challenge #2 pg 50
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Acceleration A car is traveling at 4 m/s and accelerates to 20 m/s in 4 seconds. What is the acceleration? A = 20m/s – 4 m/s = + 4 m/s2 4 s
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Acceleration Time 0 s 1s 2s 3s 4s Accel +4 +4 +4 +4 Velocity 4 8 12 16
What would the graph look like?
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Deceleration A car is traveling at 10 m/s and accelerates to 0 m/s in 5 seconds. What is the acceleration? A = 0m/s – 10 m/s = - 2 m/s2 5 s Deceleration- negative acceleration (slowing down) Practice: page Practice: graphing sheet
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Acceleration Time 0 s 1s 2s 3s 4s 5s Accel -2 -2 -2 -2 -2
Velocity 10m/s What would the graph look like?
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Fundamental Equations
V (avg) = 1/2 (vf + Vi) (Used mainly for deriving) V = Dx = xf - xi Dt Dt a = vf – vi rearranged vf = vi + at t
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Derived Equations Derived equations - equations formed
from other equations Dx = ½ (vi + vf) t Dx = vi t + ½ at2 Vf2 = vi a Dx
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Deriving Dx = ½ (vi + vf) t
Put V = Dx and V = 1/2 (vf + Vi) Dt together Dx = V Dt use substitution Dx = 1/2 (vf + Vi) Dt
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Deriving Dx = vi t + ½ at2 Put Dx = 1/2 (vf + Vi) Dt and vf = vi + a Dt together Dx = 1/2 (vf + vi) Dt Use substitution Dx = 1/2 (vi + aDt + vi) Dt Dx = ½ (2 vi + a Dt ) Dt Dx = ½ Dt (2 vi) + ½ Dt (a Dt ) Dx = Vi Dt + ½ a Dt 2
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Deriving Vf2 = vi2 + 2a Dx Dx = v D t v = vf + vi and t = vf - vi 2 a
Dx = vf + vi multiplied vf - vi a Dx = vf2 – vi2 rearrange and get 2a Vf2 = vi a Dx
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Practice Problems Find the acceleration of an amusement park ride that falls from rest to a speed of 28 m /s in 3.0 s. a = vf – vi = m/s m/s tf – ti s a = m/s2
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Practice Problems A bicyclist accelerates from 5.0 m /s to 16 m /s in 8.0 s. Assuming uniform acceleration, what distance does the bicyclist travel during this time interval. Dx = ½ (vi + vf) t = ½ (5.0 m/s + 16 m/s) 8.0 s = 84 m
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Practice Problems An aircraft has a landing speed of 83.9 m /s. The landing area of an aircraft carrier is 195 m long. What is the minimum uniform acceleration required for a safe landing? a = Vf2 - vi2 = (0)2 - (83.9m/s) Dx (195 m) = m/s2
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Practice Problems An electron is accelerated uniformly from rest in an accelerator at 4.5 X 107 m/s2 over a distance of 95 km. Assuming constant acceleration, what is the final velocity of the electron? Vf2 = vi a Dx Vf2 = (0)2 + 2(4.5 X 107 m/s2) 95,000 m Take the square root of each side Vf = 2.9 X 106 m/s
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