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