Lecture 2:One Dimensional Motion

Position, Distance, and DisplacementBefore describing motion, you must set up a coordinate system – define an origin and a positive direction.

The distance is the total length of travel; if you drive from your house to the grocery store and back, you have covered a distance of 8.6 mi. Your net displacement is zero.

Position, Distance, and Displacement

If you drive from your house to the grocery store and then to your friend’s house, your displacement is -2.1 mi and the distance you have traveled is 10.7 mi.

You and your dog go for a walk to the

park. On the way, your dog takes many

side trips to chase squirrels or examine

fire hydrants. When you arrive at the

park, do you and your dog have the

same displacement?

a) yes

b) no

Walking the Dog

You and your dog go for a walk to the

park. On the way, your dog takes many

side trips to chase squirrels or examine

fire hydrants. When you arrive at the

park, do you and your dog have the

same displacement?

a) yes

b) no

Yes, you have the same displacement. Because you and your dog

had

the same initial position and the same final position, then you have

(by definition) the same displacement.

Walking the Dog

Follow-up: have you and your dog traveled the same distance?

Average SpeedThe average speed is defined as the distance traveled divided by the time the trip took:

Average speed = distance / elapsed time

If you drive from your house to the grocery store and then to your friend’s house, your displacement is -2.1 mi and the distance you have traveled is 10.7 mi. The trip takes 30 minutes (with traffic and lights).What is your average speed?

10.7 miles

0.5 hourssav =

= 21.4 miles/hr

Average VelocityAverage velocity = displacement / elapsed time

If you return to your starting point, your average velocity is zero.

You and your dog go for a walk to the

park. On the way, your dog takes

many side trips to chase squirrels

or examine fire hydrants. When

you arrive at the park, you

affectionately pat your dog’s head.

Which statement correctly describes

your average speed and velocity

relative to that of your dog, on the

trip from home to the park?

a) Average speed and average velocity were both different

b) Average speed was the same, but average velocity was different

c) Average speed was different, but average velocity was the same

d) Average speed and average velocity were the same

Walking the Dog II

You and your dog go for a walk to the

park. On the way, your dog takes

many side trips to chase squirrels

or examine fire hydrants. When

you arrive at the park, you

affectionately pat your dog’s head.

Which statement correctly describes

your average speed and velocity

relative to that of your dog, on the

trip from home to the park?

a) Average speed and average velocity were both different

b) Average speed was the same, but average velocity was different

c) Average speed was different, but average velocity was the same

d) Average speed and average velocity were the same

Walking the Dog II

Your dog’s many side trips mean that he travelled more distance

than you in the same amount of time, so his average speed must

have been greater. However, his net displacement was the same

(starting and ending at the same place), so his average velocity must

have been the same.

Averaging Speed

Is the average speed of the red car a) 40.0 mi/h, b) more than 40.0 mi/h, or c) less than 40.0 mi/h?

Hint: is t1 = t2?

Position vs. Time

Consider this motion sequence, plotted here in one-dimension...

...and here as an x-t graph

average velocity

Instantaneous Velocity

The instantaneous velocity is tangent to the curve.

Evaluating the average velocity over a shorter and shorter period of time, one approaches the “instantaneous velocity”.

Graphical Interpretation of Average and Instantaneous Velocity

Position v. Time I

t

x

The graph of position versus

time for a car is given below.

What can you say about the

velocity of the car over time?

a) it speeds up all the time

b) it slows down all the time

c) it moves at constant velocity

d) sometimes it speeds up and

sometimes it slows down

e) not really sure

t

x

The graph of position versus

time for a car is given below.

What can you say about the

velocity of the car over time?

The car moves at a constant velocity

because the x vs. t plot shows a straight

line. The slope of a straight line is

constant. Remember that the slope of x

vs. t is the velocity!

a) it speeds up all the time

b) it slows down all the time

c) it moves at constant velocity

d) sometimes it speeds up and

sometimes it slows down

e) not really sure

Position v. Time I

t

x

a) it speeds up all the time

b) it slows down all the time

c) it moves at constant velocity

d) sometimes it speeds up and

sometimes it slows down

e) not really sure

The graph of position vs.

time for a car is given below.

What can you say about the

velocity of the car over

time?

Position v. Time II

a) it speeds up all the time

b) it slows down all the time

c) it moves at constant velocity

d) sometimes it speeds up and

sometimes it slows down

e) not really sure

x

The graph of position vs.

time for a car is given below.

What can you say about the

velocity of the car over

time?

The car slows down all the time because the slope of the x vs. t graph is diminishing as time goes on. Remember that the slope of x vs. t is the velocity! At large t, the value of the position x does not change, indicating that the car must be at rest.

t

Position v. Time II

Acceleration

Average acceleration:

Instantaneous acceleration:

Graphical Interpretation of Average and Instantaneous Acceleration

Sign of Acceleration

Speed increasingSpeed Decreasing

Does deceleration mean “negative acceleration”?

No. It means “decreasing speed”.

Acceleration

Acceleration (increasing speed) and deceleration (decreasing speed) should not be confused with the direction (or sign) of velocity and acceleration:

Graphical Interpretation of Average and Instantaneous Acceleration

Constant Acceleration

If the acceleration is constant, the velocity changes linearly with time:

Average velocity:

Propeller carv

ta:

v

tb:

v

tc:

v

td:

Which of the above

plots represents the v

vs. t graph for the

motion of the propeller

car after it was pushed?

e: Not Sure

Propeller carv

ta:

v

tb:

v

tc:

v

td:

Which of the above

plots represents the v

vs. t graph for the

motion of the propeller

car after it was pushed?

The car has a negative initial velocity, but a constant positive acceleration. So the plot should start at a negative value, and increase linearly with increasing time.

e: Not Sure

Motion with Constant AccelerationPosition as a function of time

Motion with Constant Acceleration

The relationship between position and time follows a characteristic curve.

My guess: a ~ 7 m/s2

Propeller car displacement

Which of the displayed plots represents the x vs. t graph for the motion of the propeller car after it was pushed?

x

t

Propeller car displacement

Which of the displayed plots represents the x vs. t graph for the motion of the propeller car after it was pushed?

The car starts at zero position, with a negative initial velocity, so the displacement is growing in the negative direction. The slope is changing linearly with time to become more positive. It should be a smooth change, go through a minimum, and then change more and more quickly with increasing timex = v0t + 1/2 at2

x

t

Motion with Constant AccelerationVelocity as a function of position

Stopping Distance

1/2 Stopping Distance

3/4 Stopping Distance

Tip for save driving: if you double your speed, what happens to your stopping distance?

v = 0.7 v0

deaccelerating with constant a, take final velocity = 0,

find (x-x0) = “stopping distance”

Hit the Brakes!

Freely Falling ObjectsFree fall from rest:

Free fall is the motion of an object subject only to the influence of gravity. The acceleration due to gravity is a constant, g.

g = 9.8 m/s2

For free falling objects, assuming your y axis is

pointing up, a = -g = -9.8 m/s2

a) 0.1 seconds

b) 0.2 seconds

c) 2 seconds

d) not enough information to work the problem

e) not really sure

Reaction Time

In the peculiar units used by Professor Norum, his reaction time is measured to be 20 cm. What is the corresponding reaction time, in seconds? [Use g=10 m/s2]

With v0=0 and x0 = 0, position as a function of time is:

20 cm = 0.2 m, so:

t = 0.2 seconds

1-D motion of a vertical projectile

1-D motion of a vertical projectilev

ta:

v

tb:

v

tc:

v

td:

- Clickers, everyday, in class.

- Assignment 1 on MasteringPhysics. Due Monday, August 29, at midnight!

- Reading, for next class (Chapter 2, 3.1-3.6)

- When you exit, please use the REAR doors!

a) 1 m

b) 1.4 m

c) 2 m

d) not enough information to solve the problem

e) not really sure

Air Track

The car undergoes constant acceleration on the tilted airtrack. Where will the velocity be twice that measured at 0.5m?

With v0=0 and x0 = 0, velocity as a function of position is:

We want v2 / v1 = 2:

x2 = 4x1 = 2 m