you need: binder for notes. describe motion in terms of frame of reference, displacement, time...
TRANSCRIPT
Motion with Constant Acceleration
McNutt - Physics
You need:
Binder For Notes
Objectives
Describe motion in terms of frame of reference, displacement, time interval and velocity.
Calculate displacement, average velocity, time interval and acceleration.
Draw and interpret position vs. time and velocity vs. time graphs.
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
))((21
if 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 Problems #1- 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
Remember to list the GIVENS & UNKOWNS when setting up your equations!
? ?
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 Problem Continued
tavv if
221 )( tatvx i
v
t
Practice Problems
#2 - 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 Problems 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 Problems
#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