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Motion in One DimensionChapter 2Pg. 40-65

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2.1 Displacement and Velocity

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What do you think?

Is the book on your instructor’s desk in motion?Explain your answer.

What are some examples of motion?

How would you define motion?

+One-Dimensional Motion

It is the simplest form of motion

This is motion that happens in one direction

Example:Train on the tracts moving forward or

backward

Displacement (x)Change in position

Can be positive or negativeIt describes the direction of motion

You will see me use d instead of xfor displacement

+Displacement (x)

Displacement is not always equal to the distance traveled

A gecko runs up a tree from the 20 cm marker to the 80 cm marker, then he retreats to the 50 cm marker?

+Displacement (x)

In total it traveled 90 cm, however, the displacement is only 30 cm

x or d= 50cm- 20cm = 30 cm

If the gecko goes back to the startThe displacement is zero

Displacement What is the

displacement for the objects shown?

Answer: 70 cm

• Answer: -60 cm

+Displacement - Sign Conventions

Right (or east) ---> +

Left (or west) ---> –

Up (or north) ----> +

Down (or south) ---> –

+Average Velocity

The total displacement divided by the time interval during which the displacement occurred (Vavg)

Basically velocity is distance/time

• Average velocity is displacement divided by the time interval.

+Average Velocity

The units can be determined from the equation.SI Units: m/sOther Possible Units: mi/h, km/h,

cm/year

Velocity can be positive or negative but time must always be positive

+Average Velocity

You travel 370 km west to a friends house. You left at 10am and arrived at 3pm. What was your average velocity?

Vavg = d = -370 km

Δt 5.0h

Vavg= -74 km/h or 74 km/h west

+Classroom Practice Problems

Joey rides his bike for 15 min with an average velocity of 12.5 km/h, how far did he ride?

Vavg = d

Δt

d = (12.5 km/h) (0.25h)

d = 3.125 km

+Speed

Speed does not include direction while velocity does.

Speed uses distance rather than displacement.

In a round trip, the average velocity is zero but the average speed is not zero.

Graphing Motion How would you describe the motion

shown by this graph? Answer: Constant speed (straight

line)

What is the slope of this line? Answer: 1 m/s

What is the average velocity? Answer: 1 m/s

Graphing Motion

Describe the motion of each object.

Answers Object 1: constant

velocity to the right or upward

Object 2: constant velocity of zero (at rest)

Object 3: constant velocity to the left or downward

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2.2 Acceleration

Pg. 48 to 59

+What do you think?

Which of the following cars is accelerating?A car shortly after a stoplight turns greenA car approaching a red lightA car with the cruise control set at 80 km/hA car turning a curve at a constant speed

Based on your answers, what is your definition of acceleration?

Acceleration

What are the units? SI Units: m/s2

Other Units: (km/h)/s or (mi/h)/s

Acceleration = 0 implies a constant velocity (or rest)

+Classroom Practice Problem

A bus slows down with an average acceleration of -1.8m/s2. how long does it take the bus to slow from 9.0m/s to a complete stop?

Find the acceleration of an amusement park ride that falls from rest to a velocity of 28 m/s downward in 3.0 s.9.3 m/s2 downward

+Acceleration

Displacement with constant acceleration

d = ½ (vi + vf) Δt

Displacement= ½ (initial velocity + final velocity) (time)

+Example

A racecar reaches a speed of 42m/s. it uses its parachute and breaks to stop 5.5s later. Find the distance that the car travels during breaking.

Given: vi = 42 m/s vf= 0 m/s

t = 5.5 sec d = ??

+Example

d = ½ (vi + vf) Δt

d = ½ (42+0) (5.5)

d = 115.5 m

+Acceleration

Final velocityDepends on initial velocity,

acceleration and time

Velocity with constant acceleration

vf = vi + aΔt

Final velocity = initial velocity + (acceleration X time)

+Acceleration

Displacement with constant acceleration

d = viΔt + ½ a (Δt)2

Displacement = (initial velocity X time) + ½ acceleration X time2

+Example

A plane starting at rest at one end of a runway undergoes uniform acceleration of 4.8 m/s2 for 15s before takeoff. What is its speed at takeoff? How long must the runway before the plane to be able to take off?

Given: vi = 0 m/s a= 4.8m/ss

t= 15s vf = ?? d= ??

+Example

What do we need to figure out?Speed at takeoffHow long the runway needs to be

vf = vi + aΔt

vf = 0m/s + (4.8 m/s2)(15 s)

vf = 72m/s

+Example

d = viΔt + ½ a (Δt)2

d = (0m/s)(15s) + ½ (4.8m/s2)(15s)2

d = 540 m

+Acceleration

Final velocity after any displacement

vf2 = vi

2 +2ad

Final velocity2 = initial velocity2 + 2(acceleration)(displacement)

+Example

An aircraft has a landing speed of 83.9 m/s. The landing area of an aircraft carrier is 195 m long. What is the minimum uniform acceleration required for safe landing?Answer: -18.0 m/s2

+Useful Equations

1.

2.

3.

4.

5.

d = 1 (vi + vf) t 2

avg

dv

t

avg

va

t

f iv v a t

2i f

avg

v vv

21

2id v t a t

2 2 2f iv v ad

+Classroom Practice Problems

A bicyclist accelerates from 5.0 m/s to 16 m/s in 8.0 s. Assuming uniform acceleration, what distance does the bicyclist travel during this time interval?Answer: 84 m

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2.3 Falling Objects

+Free Fall

The motion of a body when only the force due to gravity is acting

Acceleration is constant for the entire fall

Acceleration due to gravity (ag or g)

Has a value of -9.81 m/s2 (we will use 10m/s2)Negative for downward

Roughly equivalent to -22 (mi/h)/s

+Free Fall

Acceleration is constant during upward and downward motion

Objects thrown into the air have a downward acceleration as soon as they are released.

+Free Fall

The instant the velocity of the ball is equal to 0 m/s is the instant the ball reaches the peak of its upward motion and is about to begin moving downward.

REMEMBER!!! Although the velocity is 0m/s the

acceleration is still equal to 10 m/s2

+Example

Jason hits a volleyball so that it moves with an initial velocity of 6m/s straight upward. If the volleyball starts from 2.0m above the floor how long will it be in the air before it strikes the floor?

We want to use our velocity and acceleration equations

+Example

Given: vi = 6m/s a= 10m/s2

d= -2.0m t = ?? vf = ??

vf2= vi

2 + 2ad

vf= at + vi

+Free Fall

For a ball tossed upward, make predictions for the sign of the velocity and acceleration to complete the chart.

Velocity

(+, -, or zero)

Acceleration

(+, -, or zero)

When halfway up

When at the peak

When halfway down

+ -

zero -

- -

Graphing Free Fall

Based on your present understanding of free fall, sketch a velocity-time graph for a ball that is tossed upward (assuming no air resistance). Is it a straight line? If so, what is the slope?

Compare your predictions to the graph to the right.

Classroom Practice Problem

A ball is thrown straight up into the air at an initial velocity of 25.0 m/s upward. Create a table showing the ball’s position, velocity and acceleration each second for the first 5 s.

20.1 +15.2 -9.81

30.4 +5.4 -9.81

30.9 -4.4 -9.81

21.6 -14.2 -9.81

2.50 -24.0 -9.81t (s) y (m) v (m/s) a (m/s2)

1.00

2.00

3.00

4.00

5.00

+Direction of Acceleration

Describe the motion of an object with

vi and a as shown to the left.

Moving right as it speeds up

Moving right as it slows down

Moving left as it speeds up

Moving left as it slows down

Vi a

+ +

+ -

- -

- +

Graphing Velocity The slope (rise/run) of

a velocity/time graph is the acceleration. Rise is change in v Run is change in t

This graph shows a constant acceleration.

Average speed is the midpoint.

Graph of v vs. t for a train

Describe the motion at points A, B, and C.

Answers A: accelerating (increasing

velocity/slope) to the right B: constant velocity to the

right C: negative acceleration

(decreasing velocity/slope) and still moving to the right