Motion in One Motion in One Dimension 2.2Dimension 2.2

Ch 2.2Ch 2.2

AccelerationAcceleration

Avg acceleration = Avg acceleration = change in velocitychange in velocity

time needed for time needed for changechange

a = a = v v = = vvff – v – vii

t tt tff – t – tii

AccelerationAcceleration

variablevariable unitunit

Acceleration Acceleration aa m/sm/s22

Velocity Velocity vv m/sm/s

(final or initial)(final or initial)

Time Time tt ss

Acceleration is a vectorAcceleration is a vector

Acceleration has direction and magnitudeAcceleration has direction and magnitude

Velocity time graph has a slope of accelerationVelocity time graph has a slope of acceleration

See figure 10 on page 50See figure 10 on page 50

Check out Table 3 on page 51Check out Table 3 on page 51

Conceptual Challenge #2 pg 50Conceptual Challenge #2 pg 50

AccelerationAcceleration

A car is traveling at 4 m/s and accelerates A car is traveling at 4 m/s and accelerates to 20 m/s in 4 seconds. What is the to 20 m/s in 4 seconds. What is the acceleration?acceleration?

A = A = 20m/s – 4 m/s20m/s – 4 m/s = + 4 m/s = + 4 m/s22

4 s4 s

AccelerationAcceleration

TimeTime 0 s0 s 1s 1s 2s 2s 3s3s 4s 4s

AccelAccel +4+4 +4 +4 +4 +4 +4 +4

VelocityVelocity 44 8 8 12 12 16 16

What would the graph look like?What would the graph look like?

DecelerationDeceleration

A car is traveling at 10 m/s and accelerates to 0 A car is traveling at 10 m/s and accelerates to 0 m/s in 5 seconds. What is the acceleration?m/s in 5 seconds. What is the acceleration?

A = A = 0m/s – 10 m/s0m/s – 10 m/s = - 2 m/s = - 2 m/s22

5 s5 s

Deceleration- negative acceleration Deceleration- negative acceleration (slowing down)(slowing down)

Practice: page 49 1-5Practice: page 49 1-5Practice: graphing sheetPractice: graphing sheet

AccelerationAcceleration

TimeTime 0 s0 s 1s 1s 2s 2s 3s 3s 4s 5s 4s 5s

AccelAccel -2 -2 -2 -2 -2 -2 -2 -2 -2 -2

Velocity 10m/s 8 6 Velocity 10m/s 8 6 4 2 0 4 2 0

What would the graph look like?What would the graph look like?

Velocity VS Time Graph

0

2

4

6

8

10

12

0 1 2 3 4 5 6

Time (s)

Velo

city

(m/s

)

Acceleration

Fundamental EquationsFundamental Equations

V V (avg)(avg) = 1/2 (vf + Vi) ( = 1/2 (vf + Vi) (Used mainly for Used mainly for

deriving)deriving)

V = V = x x = x = xf - xif - xi

t t tt

a = a = vf – vivf – vi rearrangedrearranged vf = vi + at vf = vi + at

tt

Derived EquationsDerived Equations

Derived equationsDerived equations - equations formed - equations formed

from other equationsfrom other equations

x = ½ (vi + vf) tx = ½ (vi + vf) t

x = vi t + ½ atx = vi t + ½ at22

VfVf22 = vi = vi22 + 2a + 2a xx

DerivingDerivingx = ½ (vi + vf) tx = ½ (vi + vf) t

Put Put V = V = xx and and V = 1/2 (vf + Vi)V = 1/2 (vf + Vi)

tt togethertogether

x = x = VV t t useuse substitutionsubstitution

x = x = 1/2 (vf + Vi)1/2 (vf + Vi) tt

DerivingDerivingx = vi t + ½ atx = vi t + ½ at22

Put Put x = 1/2 (vf + Vi) x = 1/2 (vf + Vi) tt and and vf = vi + a vf = vi + a tt

togethertogether

x = 1/2 (x = 1/2 (vf vf + vi) + vi) t t Use substitutionUse substitution

x = 1/2 (x = 1/2 (vi + avi + att + vi) + vi) t t x = ½ x = ½ ((2 vi + a 2 vi + a tt ) ) t t x = ½x = ½ t t ((2 vi2 vi) ) + ½+ ½ t t ((a a tt ))

x = Vi x = Vi t + ½ a t + ½ a t t 22

DerivingDerivingVfVf22 = vi = vi22 + 2a + 2a xx

v = v = vf + vivf + vi and t = and t = vf - vivf - vi 22 a a

x = x = vv t tx = x = vf + vivf + vi multiplied multiplied vf - vivf - vi

22 a a

x = x = vfvf22 – vi – vi22 rearrange and get rearrange and get 2a2a

VfVf22 = vi = vi22 + 2a + 2a xx

Practice ProblemsPractice Problems

Find the acceleration of an amusement Find the acceleration of an amusement park ride that falls from rest to a speed of park ride that falls from rest to a speed of 28 m /s in 3.0 s.28 m /s in 3.0 s.

a = a = vvff – v – vi i = = 28m/s - 0 m/s 28m/s - 0 m/s

ttff – t – tii 3.0 s3.0 s

a = 9.3 m/sa = 9.3 m/s22

Practice ProblemsPractice Problems

A bicyclist accelerates from 5.0 m /s to 16 A bicyclist accelerates from 5.0 m /s to 16 m /s in 8.0 s. Assuming uniform m /s in 8.0 s. Assuming uniform acceleration, what distance does the acceleration, what distance does the bicyclist travel during this time interval.bicyclist travel during this time interval.

x = ½ (vi + vf) t x = ½ (vi + vf) t

= ½ (5.0 m/s + 16 m/s) 8.0 s= ½ (5.0 m/s + 16 m/s) 8.0 s

= 84 m= 84 m

Practice ProblemsPractice Problems

An aircraft has a landing speed of 83.9 m /s. An aircraft has a landing speed of 83.9 m /s. The landing area of an aircraft carrier is 195 m The landing area of an aircraft carrier is 195 m long. What is the minimum uniform acceleration long. What is the minimum uniform acceleration required for a safe landing?required for a safe landing?

a = a = VfVf22 - vi - vi22 = (= (0)0)22 - (83.9m/s) - (83.9m/s)22 2 2 x 2(195 m)x 2(195 m)

= - 18.0 m/s= - 18.0 m/s22

Practice ProblemsPractice Problems

An electron is accelerated uniformly from rest in An electron is accelerated uniformly from rest in an accelerator at 4.5 X 10an accelerator at 4.5 X 1077 m/s m/s22 over a distance over a distance of 95 km. Assuming constant acceleration, what of 95 km. Assuming constant acceleration, what is the final velocity of the electron?is the final velocity of the electron?

VfVf22 = vi = vi22 + 2a + 2a x x VfVf22 = (0) = (0)2 2 + 2(4.5 X 10 + 2(4.5 X 1077 m/s m/s22) 95,000 m ) 95,000 m

Take the square root of each sideTake the square root of each sideVf = 2.9 X 10Vf = 2.9 X 1066 m/s m/s