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Accelerated Motion
Chapter 3
Chapter Objectives
Describe accelerated motion
Use graphs and equations to solve problems involving moving objects
Describe the motion of objects in free fall. (separate presentation - not on this test)
Section 3.1 Acceleration
Define acceleration
Relate velocity and acceleration to the motion of an object
Create velocity-time graphs
Uniform Motion Nonuniform Motion
Moving at a constant velocity
If you close your eyes, you feel as though you are not moving at all
Moving while changing velocity
Can be changing the rate or the direction
You feel like you are being pushed or pulled
Changing Velocity Consider the following motion (particle
model) diagramNot moving Constant Velocity
Increasing Velocity Decreasing Velocity
Changing velocity You can indicate change in velocity by
the motion diagram spacing the magnitude (length) of the velocity vectors.
If the object speeds up, each subsequent velocity vector is longer.
If the object slows down, each vector is shorter than the previous one.
Velocity-Time Graphs
Distance being covered is longer, thus the runner is speeding up.
Velocity-Time Graphs
Time (s) Velocity (m/s)
0 0
1 5
2 10
3 15
4 20
5 25
0 1 2 3 4 5 60
5
10
15
20
25
30
Velocity vs. Time
Time (s)
Ve
loc
ity
(m
/s)
Slope???Area???
Velocity vs. Time
0
5
10
15
20
25
30
0 2 4 6
Time (s)
Vel
oci
ty (
m/s
)
Velocity-Time Graphs Analyze the units
Slope = rise over run m = ∆y / ∆x Slope = m/s/ s = m/s^2 m/s^2 is the unit for acceleration
Area = ½ b h b = s * m/s = m m is the unit for displacement
The slope of a velocity-time graph is the ACCELERATION and the area is DISPLACEMENT.
Change in Position → can’t
get position
Velocity vs. Time
0
5
10
15
20
25
30
0 2 4 6
Time (s)
Velo
cit
y (
m/s
)
Slope =
Acceleration
Area =
Displacement
Velocity – Time Graphs How Fast something is moving at a given time? Average Acceleration
Use the information on the x & y axis to plug into the equation
a = ∆v / t Slope
Instantaneous Acceleration Find the slope of the line (straight line) Find the slope of the tangent (curve)
Displacement Find the area under the curve You do not know the initial or final position of the
runner, just the displacement.
Velocity-Time Graphs
Describe the motion of each sprinter.
A• Constant velocity
• Zero Acceleration• Positive displacement
B• Constant Acceleration• Starts from Rest• Positive displacement
Velocity-Time Graphs D• Constant Acceleration• Positive Acceleration• Comes to a Stop • Zero displacement
C• Constant Acceleration• Negative Acceleration• Comes to a Stop • Positive displacement
E• Constant Velocity• Zero Acceleration• Negative displacement
What is the average acceleration of the car shown on the graph below?
A. 0.5 m/s2
B. -0.5 m/s2
C. 2 m/s2
D. -2 m/s2
Sample Question
Acceleration The rate at which an object’s velocity changes
Variable: a Units: m/s2
It is the change in velocity which measures the change in position. Thus, it is measuring a change of a change, hence why the square time unit.
When the velocity of an object changes at a constant rate, it has constant acceleration
Motion Diagrams & Acceleration In order for a motion diagram to display a full
picture of an object’s movement, it should contain information about acceleration by including average acceleration vectors. The vectors are average acceleration vectors
because motion diagrams display the object at equal time INTERVALS (intervals always mean average)
Average acceleration vectors are found by subtracting two consecutive velocity vectors.
Average Acceleration Vectors
You will have:
Δv = vf - vi = vf + (-vi).
Then divide by the time interval,
Δt. The time interval, Δt, is 1 s.
This vector, (vf - vi)/1 s, shown in
violet, is the average acceleration
during that time interval.
Average Acceleration Vectors vi = velocity at the beginning of a chosen time interval
vf = velocity at the end of a chosen time interval.
∆v = change in velocity
* Acceleration is equal to the change in velocity over the time interval
** Since the time interval is 1s, the acceleration is equal to the change in velocity
***Anything divided by 1 is equal to itself…
Average AccelerationChange in velocity during some measurable
time interval divided by the time interval Found by plugging into the equation
Instantaneous Acceleration Change in velocity at an instant of timeFound by calculating the slope of a velocity-
time graph at that instant
Average vs. Instantaneous Acceleration
𝐚=∆𝒗𝒕
Velocity & Acceleration
How would you describe the sprinter’s velocity and acceleration as shown on the graph?
Velocity & Acceleration
Sprinter’s velocity starts at zero Velocity increases rapidly for the first four
seconds until reaching about 10 m/s Velocity remains almost constant
What is the acceleration for the first four seconds? Refers to average acceleration because there is a
time interval Solve using the equation a = ∆v /t vi = 0 m/s; vf = 11 m/s; t = 4s
a = (11m/s – 0 m/s)/ 4s
a = 2.75 m/s2
Average vs. Instantaneous Acceleration
What is the acceleration at 5s?Refers to instantaneous acceleration because
it is looking for acceleration at an instantNeed to find the slope of the line to solve for
acceleration Slope is zero; thus instantaneous acceleration is
zero at the instant of 5s.
Average vs. Instantaneous Acceleration
Instantaneous Acceleration
Solve for the acceleration at 1.0 s Draw a tangent to the curve at t = 1s
The slope of the tangent is equal to the instantaneous acceleration at 1s.
a = rise / run
Instantaneous Acceleration The slope of the line at 1.0 s is equal to
the acceleration at that instant .
Positive & Negative Acceleration
These four motion diagrams represent the four different possible ways to move along a straight line with constant acceleration.
Object is moving in the positive directionDisplacement is positiveThus, velocity is positive
Object is getting fasterAcceleration is positive
Object is moving in the positive directionDisplacement is positiveThus, velocity is positive
Object is getting slowerAcceleration is negative
Object is moving in the negative directionDisplacement is negativeThus, velocity is negative
Object is getting fasterAcceleration is negative
Object is moving in the negative directionDisplacement is negativeThus, velocity is negative
Object is getting slowerAcceleration is positive
Positive & Negative Acceleration
When the velocity vector and acceleration vector point in the SAME direction, the object is INCREASING SPEED
When the velocity vector and acceleration vector point in the OPPOSITE direction, the object is DECREASING SPEED
Displacement Velocity Acceleration Speeding UP Or
Slowing Down
+ UP
+ -
- Down
- UP
Displacement & Velocity always have the same sign
Up = sameDown = Different
How can the instantaneous acceleration of an object with varying acceleration be calculated?
Sample Question
A. by calculating the slope of the tangent on a distance versus time graph
B. by calculating the area under the graph on a distance versus time graph
C. by calculating the area under the graph on a velocity versus time graph
D. by calculating the slope of the tangent on a velocity versus time graph
A
B
C
D
E
Practice v-t graph
Segment t (s) vi (m/s)
vf
(m/s)∆v
(m/s)avg. a(m/s2)
ins. A(m/s2)
Xi
(m)Xf
(m)∆X(m)
A 0
B
C
D
E
**Can not assume position on graph. Velocity time graphs can only be used to figure out displacement. You must be given an initial position.
3.2 Motion with Constant Acceleration
Interpret position-time graphs for motion with constant acceleration
Determine mathematical relationships among position, velocity, acceleration, and time
Apply graphical and mathematical relationships to solve problems related to constant acceleration.
Constant acceleration: x-t Graphs Velocity is constantly increasing, which means more
displacement. Slope must be getting steeper.
Results in a curve that is parabolic.
Position vs. Time
0.0
50.0
100.0
150.0
200.0
250.0
300.0
350.0
0 2 4 6 8 10 12
Time (s)
Po
siti
on
(m
)
Constant acceleration: x-t Graphs
t (s)
Concave UP = + a
x (m)
Concave Down = -a
x (m)
t (s)
x (m)
t (s)
Concave Down = -a
x (m)
t (s)
Concave UP = +a
Kinematics Equations Three equations that relate position,
velocity, acceleration, and time. First two are derived from a v-t graph and
the third is a substitution. Total of five different variables.
Δx (displacement), vi (initial velocity), vf (final velocity), a (acceleration), and t (time).
Must know any three in order to solve for the other two.
First Kinematics Equation
Remember that the slope of a v-t graph is the average acceleration.
Rearranging the equation, gives us the first kinematics equation.
vf = vi + at
Replace tf – ti with t
Second Kinematics Equation
Break into two known shapes (rectangle & triangle). Total Area = Area of rectangle + area of triangle Δx = vit + ½ (vf –vi)t
Velocity vs. Time
0
5
10
15
20
25
30
0 2 4 6 8 10
Time (s)
Vel
ocity
(m/s
)
We remember that area of a v-t graph
equals displacement
Δx = vit + ½ at2vf – vi = at
(substitute)
Third Kinematics Equation First equation substituted into the second to
cancel out the time variable. vf = vi + at t = (vf – vi) / a Δx = vit + ½ at2
Δx = vi((vf – vi)/a) + ½ a ((vf – vi)/a) 2
Δx = vivf – vi2 + ½ a (vf
2 – 2 vivf + vi2 )/a2
2a Δx = 2 vivf - 2 vi2 + vf
2 – 2 vivf + vi2
2a Δx = - vi2 + vf
2 (rearrange)Combine like
terms
Multiply by 2a to get rid of
fraction
Simplify
vf2 = vi
2 + 2a Δx
Equation Use when:NO ∆x
No displacement (position)
No vf
No final velocity
No tNo time
Kinematics
vf2 = vi
2 + 2a Δx
Δx = vit + ½ at2
vf = vi + at
• Must have 3 of the 5 variables to start. (∆x, vi, vf, a, t) • Must always know acceleration.
• If acceleration is not given, then must solve for it first.
Sample Problem An automobile starts at rest and speeds up at 3.5 m/s2 after the traffic light turns green. How far will it have gone when it is traveling at 25 m/s?
vi = 0 m/s
vf = 25 m/s
a = 3.5 m/s2
∆x = ? m
vf2 = vi
2 + 2a ∆x
(25m/s)2 = (0 m/s)2 + 2 (3.5 m/s2)(∆x )
∆x = 89.3 m
Given Unknown Formula Substitution Answer
Sample Problem You are driving a car, traveling at constant velocity of 25 m/s, when you see a child suddenly run onto the road. It takes you 0.45 s to react and apply the brakes. As a result, the car slows with a steady acceleration of 8.5 m/s2 and comes to a stop. What is the total distance that the car moves before it stops?
Reaction Stopping
t =0.45 sv = 25 m/s
vi = 25 m/svf = 0 m/sa = -8.5 m/s2
∆xreaction = ? m ∆xstopping = ? m
(0 m/s)2 = (25 m/s)2 + 2(-8.5m/s2)(∆x)
∆x = 11.25 m ∆xstopping = 36.76 m
vf2 = vi
2 + 2a Δx
xtotal = 48.01 m