1. Motion in One Dimension 2.2
Ch 2.2

2. Acceleration Avg acceleration = change in velocity
time needed for change
a = v = vf – vi
t tf – ti

3. Acceleration variable unit Acceleration a m/s2 Velocity v m/s
(final or initial)
Time t s

4. 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

5. 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

6. Acceleration Time 0 s 1s 2s 3s 4s Accel +4 +4 +4 +4 Velocity 4 8 12 16
What would the graph look like?

7. 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

8. Acceleration Time 0 s 1s 2s 3s 4s 5s Accel -2 -2 -2 -2 -2
Velocity 10m/s
What would the graph look like?

10. 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

11. Derived Equations Derived equations - equations formed
from other equations
Dx = ½ (vi + vf) t
Dx = vi t + ½ at2
Vf2 = vi a Dx

12. 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

13. 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

14. 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

15. 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

16. 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

17. 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

18. 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