Motion with Constant AccelerationMcNutt Physics – 09/16/2013

The Story so far….• The average velocity for any motion is

– Where Δx is the displacement and Δt is the time interval.

• The instantaneous velocity v is the velocity the object has at a particular time.– It is the average velocity over a very short time

interval.

t

xvAV

Position vs. Time for Constant Velocity Motion

0

2

4

6

8

10

12

0 2 4 6

x (m

)

t (s)

Position vs. Time• If the velocity is

constant, the instantaneous velocity is the average velocity.

• v = vAV

• The graph is a straight line.

• The position is given by the equation

iAV xtvx

Position vs. Time for Accelerated Motion

• Here the average velocity is not constant.

• For the instantaneous velocity, take the average velocity over a very short time interval.

• Graphically, this is the slope of the tangent line of the graph.

Acceleration• When velocity changes, we have an acceleration.

• Velocity can change in magnitude or direction.

• Average acceleration is given by the formula:

if

ifAV tt

vv

Δt

Δva

Accelerations can be positive or negative in 1-d motion.• v• a

• v• a

• v• a

• v • a

• v • a = 0

• v = 0 • a or

Constant Acceleration Model• Accelerations can vary with time.

• Many situations in physics can be modeled by a constant acceleration.– Constant acceleration means the object

changes velocity at a constant rate.• When dealing with a constant acceleration

situation, we will drop the subscript “AV”.

aaAV

0

2

4

6

8

10

12

0 2 4 6

Velocity vs. time for constant acceleration

• aAV is the slope of the velocity vs. time graph.

• If the velocity vs. time graph is a straight line, the acceleration is constant.

• In this case, the formula for velocity is

• t (s)

v (m/s)

ivatv

Displacement on a Velocity vs. Time graph

• Since and v is the height of the area under the velocity versus time graph, and t is the base of the velocity versus time graph, the area under a velocity versus time graph shows the displacement.

vtx

Δx

Displacement for constant acceleration

• The displacement from time 0 to time t is the area under the velocity graph from 0 to t.

• Area = ½ b h

t (s)

v (m/s)

fv

))((21

fvtx

Displacement for constant acceleration

0

2

4

6

8

10

12

14

0 2 4 6

• If the initial velocity is not zero, we have to include a rectangular piece.

• Triangle Area = ½ b h• Rectangle = l x w

t (s)

v (m/s)

fv

)())((21

iif vtvvtx 0v

))(())((

)())(())((

21

21

21

21

if

iif

vtvtx

vtvtvtx

))((21

if vvtx

Displacement for constant acceleration

0

2

4

6

8

10

12

14

0 2 4 6

• If we don’t know vf, we can calculate it from a.

• Area =l w + ½ b h

t (s)

v (m/s)

at

221

21 ))((

attvx

atttvx

i

i

atvv

vatv

if

if

Equations of Motion for Constant Acceleration• Now we have derived three equations that

apply to the motion with constant acceleration model

221 attvx i

atvv if ))((2

1if vvtx

Formulas for other time intervals

• If the motion begins at some other time other than t = 0, then we simply replace t with the time interval Δt.

221 )( tatvx i

tavv if

tvvx fi )(21

Practice 2D, p. 55• #2- An automobile with an initial speed of

4.3 m/s accelerates uniformly at the rate of 3.0 m/s2. Find the final speed and the displacement after 5.0 s.

s 0.5

m/s 0.3

m/s 30.42

t

a

viv

t

?

?

x

v fConstant Acceleration

? ?

s 0.5

m/s 0.3

m/s 30.42

xv

t

a

v

f

i tavv if s) )(5.0m/s 0.3( m/s 30.4 2fv

m/s 3.19fv2

21 tatvx i

2221 )s 5.0)(m/s 0.3( s) (5.0 m/s) 3.4( x

m 59 m 5.37 m 5.21 x

Practice 2D, # 2, p. 55

tavv if

221 )( tatvx i

v

t

Practice 2D, p. 55• #3- A car starts from rest and travels for 5.0

s with a uniform acceleration of -1.5 m/s2. What is the final velocity of the car? How far does the car travel in this time interval?

One Other Equation for Constant Acceleration• All of the equations we have so far for this

model involve time.• Sometimes, we are not told the time over

which the motion occurs.• We can use two of these equations to

eliminate time.

xavv if 222

Equations for the Constant Acceleration Model

221 tatvx i

xavv if 222

tavv if

tvvx fi )(21

Practice 2C, p. 53

• A jet plane lands with a speed of 100 m/s and can accelerate uniformly at a maximum rate of -5.0 m/s2 as it comes to rest. Can this airplane land at an airport where the runway is 0.80 km long?

Practice 2C, p. 53• #3

m 1000 10

10000

)m/s 0.5(2

m/s) 100(m/s) 0(

2

2

2

2

2

sm

sm

2

22

22

22

x

x

xa

vv

xavv

if

if

m/s 0.5 m/s 0 m/s 100 2 avv fi

?) km .80 (Is ? xx

xavv if 222

Constant Acceleration