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.

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  Moving while changing velocity  Can be changing the rate or the direction  If you close your eyes, you feel as though you are not moving at all  You feel like you are being pushed or pulled

Changing Velocity

 Consider the following motion (particle model) diagram

Not 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) 0 1 2 3 4 5 Velocity (m/s) 0 5 10 15 20 25 Slope???

Area???

Velocity-Time Graphs

Velocity vs. Tim e

30  Analyze the units 25  Slope = rise over run    20 15 m = ∆y / ∆x 10 5 Slope = m/s/ s = m/s^2 0 m/s^2 is the unit for acceleration 0  Area = ½ b h  b = s * m/s = m  m is the unit for displacement 2

Tim e (s)

 The slope of a velocity-time graph is the

ACCELERATION

and the area is

DISPLACEMENT

. 4 6

30 25 20 15 10 5 0 0

Velocity vs. Tim e Slope = Acceleration

2

Tim e (s) Area = Displacement

4 6

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 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

A

•

Constant velocity

•

Zero Acceleration

•

Positive displacement B

•

Constant Acceleration

•

Starts from Rest

•

Positive displacement

 Describe the motion of each sprinter.

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

Sample Question

On the basis of the velocity-time graph of a car moving up a hill, as shown on the right, determine the average acceleration of the car?

A.

0.5 m/s 2

B.

-0.5 m/s 2

C.

2 m/s 2

D.

-2 m/s 2

Acceleration

 The rate at which an object’s velocity changes  Variable: a  Units: m/s^2  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: Δ

v

=

v

f -

v

i =

v

f + (-

v

i ).

Then divide by the time interval, Δ

t

. The time interval, Δ

t

, is 1 s. This vector, (

v

f -

v

i )/1 s, shown in violet, is the average acceleration during that time interval.

Average Acceleration Vectors

v

i = velocity at the beginning of a chosen time interval

v

f = 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 vs. Instantaneous Acceleration

 Average Acceleration  Change in velocity during some measurable time interval divided by the time interval  Found by plugging into the equation  a = ∆v / t  Instantaneous Acceleration  Change in velocity at an instant of time  Found by calculating the slope of a velocity time graph at that instant

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

Average vs. Instantaneous Acceleration

 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



v i



= 0 m/s; v f = 11 m/s; t = 4s a = (11m/s – 0 m/s)/ 4s



a = 2.75 m/s 2

Average vs. Instantaneous Acceleration

 What is the acceleration at 5s?

 Refers to instantaneous acceleration because it is looking for acceleration at an instant  Need to find the slope of the line to solve for acceleration  Slope is zero; thus instantaneous acceleration is zero at the instant of 5s.

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 direction  Displacement is positive  Thus, velocity is positive  Object is getting faster  Acceleration is positive

 Object is moving in the positive direction  Displacement is positive  Thus, velocity is positive  Object is getting slower  Acceleration is negative

 Object is moving in the negative direction  Displacement is negative  Thus, velocity is negative  Object is getting faster  Acceleration is negative

 Object is moving in the negative direction  Displacement is negative  Thus, velocity is negative  Object is getting slower  Acceleration 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 always have the same sign Displacement Velocity Acceleration

+

Speeding UP Or Slowing Down

UP + Down UP

Up = same Down = Different

Sample Question

How can the instantaneous acceleration of an object with varying acceleration be calculated?

A.

B.

C.

D.

by calculating the slope of the tangent on a distance versus time graph by calculating the area under the graph on a distance versus time graph by calculating the area under the graph on a velocity versus time graph by calculating the slope of the tangent on a velocity versus time graph

Practice v-t graph

B C A D E

Segment t (s) A B C D E v i (m/s) v f (m/s) ∆v (m/s) avg. a (m/s 2 ) ins. A (m/s 2 ) X i (m) 0 X f (m) ∆X (m)

**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.  Results in a curve that is parabolic.

Position vs. Time

350.0

300.0

250.0

200.0

150.0

100.0

50.0

0.0

0 2 4 6

Time (s)

8 10 12

Constant acceleration: x-t Graphs

x (m) x (m) Concave UP = +a Concave UP = + a x (m) t (s) x (m) t (s) Concave Down = -a t (s) t (s) Concave Down = 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), velocity), a v i (initial velocity), v f (acceleration), and t (time). (final  Must know any three the other two. in order to solve for

First Kinematics Equation

 Remember that the slope of a v-t graph is the average acceleration.

Replace t f with t – t i

 Rearranging the equation, gives us the first kinematics equation.

v f = v i + at

Second Kinematics Equation

30 25 20 15 10 5 0 0 2

Velocity vs. Time

4 6 8 10

We remember that area of a v-t graph equals displacement Time (s)

   Break into two known shapes (rectangle & triangle). Area = Area of rectangle + area of triangle Δx = v i t + ½ (v f –v i )t

v f – v i = at (substitute)

Δx = v

i

t + ½ at

2

Third Kinematics Equation

   First equation substituted into the second to cancel out the time variable. v f = v i + at Δx = v i t + ½ at 2 t = (v f – v i ) / a

Simplify

    Δx = v i ((v f Δx = v i v f – v – v i 2 i )/a) + ½ a ((v + ½ a (v f 2 f – v – 2 v i v f i )/a) + v i 2 2 )/a 2 2a Δx = 2 v i v f - 2 v i 2 + v f 2 – 2 v i v f 2a Δx = - v i 2 + v f 2 (rearrange) + v i 2

Multiply by 2a to get rid of fraction Combine like terms v f 2 = v i 2 + 2a Δx