Why Study Motion?PositionTime

How do we know that the Earthinteracts with the man?

The man changes position with respectto the Earth as time passes.

Motion is the measure of that changein position with respect to time.

Speed = Change in position/time

v = x/t v = d/t

Types of Motion

Constant velocity.

Both speed and direction are constant.

Accelerated motion.

Speed or direction or both change.

Newton’s First Law

Newton’s Second Law

Speed

Instantaneous speed. Speedometer reading at any instant.

Average Speed. Total distance traveled divided by total time.

A car travels from Ewing Twp. To New York City, a distance of60 mi. The trip takes 1.5 hrs. Determine the instantaneous and average speeds for the trip.

vav = ∆x/∆t vav = 60 mi/1.5 hr vav = 40 mi/ hr

Since we do not have a time, we cannot answer this question.

Some Common Speeds v = d/t

v = 24,000 mi/24 hr

8,000 mi d = 2r

v ≈ 1,000 mi/ hr

93,000,000 mi v ≈ 67,000 mi/ hr

d = 2(93,000,000)mi

Don’t say you have not been around!!

t = (365)24 hr

The Ultimate Speed c = speed of light in a vacuum

(186,000 mi/s)/(24,000 mi/trip) =

8,000 mi c =

93,000,000 mi

186,000 mi/s

A light beam can make how many round trips around the Earth in 1s?

How long does it take light totravel from Sun to Earth?

c = d/t t = d/c

t = (93,000,000mi)/(186,000)mi/s)

t = (93,000,000/186,000)s

t = 500s t = 8.3 min

= 3.0 x 108 m/s

7.8 trips/s

Velocity Speed is a scalar. Speed has magnitude only.

Velocity is a vector. Velocity has magnitude and direction.

A car travels 5 mi, stops, and then travels 5 mi more. How far is the car from its starting point?

Why do me need to deal with vectors?

10 mi 0 mi 7 mi

Constant Velocity

v = ∆x/∆t vav = vinst

x

t

v

t

v = slope v = constant

Constant Linear Acceleration

x

t

v

t

a

t

v = ∆x/∆t

∆x/∆t = slope of the line

a = ∆v/∆t

∆v/∆t = slope of the line

a = constant

Constant Linear Acceleration Equations

t

v

v0

v = v0 + at

v0 = initial velocity a = slope of the line

x = v0t + ½ at2 distance traveled in time t

Velocity at time t

v2 = v02 + 2 ax

If the object starts from rest, v0 = 0, then the equations become:

v = a t x = ½ a t2 v2 = 2 ax

Constant Acceleration

Graphs

x = x0 + v0 t + ½ at2

v = v0 + a t

a = ∆v/∆t = constant

v = ∆x/∆t

Truck and Car Problem

A car is at rest at a red traffic light. Just as the light turns green the car is passed by a truck traveling at a constant velocity of 15.0 m/s. The car immediately accelerates at the constant rate of 3.0 m/s2 in the same direction as the truck. Both the car and truck travel in the same straight line direction on a level road.

a) At what time t1 are the car and truck traveling at the same velocity?

b) At what time t2 does the car pass the truck?

c) How far has the car traveled when it passes the truck?

d) How fast is the car traveling when it passes the truck?

Truck and Car Problemv(m/s)

t(s)

vT = 15

xT = 15t2

Truck

Car

t1 t2

15

vC = a t = 3t1 3 t1 = 15 t1 = 5 s

xC = ½ 3 t22 ½ 3 t2

2 = 15t2 t2 = 10 s

xT = 15t2 xT = 15(10) xT = 150

t1 = truck & car at same speed

t2 = truck & car at same distance

vC = a t = 3t2 vC = 3(10) vC = 30 m/s

Acceleration Due To Gravity

v = v0 - g t y = v0t - ½ g t2 v2 = v02 - 2 gy

y = distance traveled

Up is + Down is vo = initial velocity

g = acceleration due to gravity

a = -g g = 10 m/s2 g = 1000 cm/s2

A ball is dropped from rest and falls for 3.0 seconds.How fast is it moving and how far has it fallen?

v = 0 – (10m/s2)(3.0s)

v = – 30 m/s

y = 0 - ½ (10m/s2)(3.0s)2

y = - 45 m

Acceleration Due To Gravity

Sample Problem 1

v = v0 - g tv2 = v02 - 2 gy

What is the maximum height reached by the ball?

vo = 20.0 m/s

How long did it take for the ball to reach that height?

g = 10 m/s2

v0

-v0

How long did the total flight last?

(0)2 = (20.0)2 - 2 (10)y

y = 20 m

0 = 20.0 – 10 t

t = 2.0 s

-20.0 = 20.0 – 10 t

v = v0 - g t

t = 4.0 s

t = 2 x 2.0 s

t = 4.0 s

Graph for Sample Problem 1

v0

-v0

v (m/s)

t (s)

20 (s)

-20 (s)

2 s 4 s0

Projectile Motion 1

In the horizontal (x) direction the speed is constant.

In the vertical (y) direction the acceleration is constant.

Projectile Motion 2

x = horizontal distancevx = horizontal velocity

vox = initial horizontal velocity

y = vertical distance

vy = vertical velocity

voy = initial vertical velocity

Horizontal

vx = constant

vx = vox vx = x/t

Verticaly = voyt – ½ gt2

vy = voy - gt

(vy )2 = (voy )2 – 2gy

Sample Problem 3

x = ___ m

t = ___ svox = 36 m/s

y = - 6.0 m

t = 1.1 s

x = vox t

y = voyt – ½ gt2

vox

y

x

- 6.0 m = 0 – ½ (10 m/s2)t2

x = (36 m/s)(1.1 s) x = 40 m

A ball is thrown horizontally with a velocity of 36.0 m/s over a level surface . It is at a height of 6.00 m when released. a) How long was the ball in flight?b) How far did the ball travel?

Sample Problem 4

vox = ___ mt = ___ s

x = 28 m y = - 2.0 m

t = .63 s

vox = x/t

y = voyt – ½ gt2

vox

y

x

- 2.00 m = 0 – ½ (10 m/s2)t2

vox = (28 m)/(.63 s) vox = 44 m/s

Is vox > 45 m/s (90 mph)?

A pitcher claims he can throw a baseball at a velocity greater than 90 mi/hr (45 m/s). He throws the ball horizontally over a level surface from a height of 2.0 m. The ball travels a horizontal distance of 28 m. Was the boy correct?

The boy was not correct.

Sample Problem 5

vox = constant velocity

vo

vox

voy

voy = constant acceleration, g

Maximum range occurs when θ = 45o

Explain the Picture