Martha Casquete

Assignments: For next class: Finish reading Ch. 2, read Chapter 3

(Vectors) HW3 Set due next Wednesday, 9/11 HW3 will be in weebly.

Question/Observation Mondays

Research Q/O Wednesday with HW (due date

Wednesday)

Acceleration in Uniform Circular Motion

Projectile Motion

What today’s lesson has to do with drunk drivers?

Distance and displacement

Speed and Velocity

Acceleration

• v = d/t or v = Dd/Dt

Vector: displacement, velocity and speed

2-5

Rearrange this equation:

at = vf – vo

vf = vo + at (solved for final velocity)

Section 2.3

vf – vo t

• Remember that a =

If speed increases, there is an acceleration (both velocity and acceleration points in the same direction)

If speed decreases, there is a deceleration (the velocity and acceleration points in different direction)

• Units of acceleration = (m/s)/s = m/s2

Galileo Galilei (1564 – 1642)

Section 2.3

Distance is proportional to t2 ◦ (d = ½ gt2)

Velocity is proportional to t ◦ (vf = at )

Section 2.3

Free Fall: Objects under influence of gravity

2-8

d = ½ gt2

This equation will compute the distance (d) an object drops due to gravity (neglecting air resistance) in a given time (t).

Section 2.3

A ball is dropped from the top of a 75 m building. Does it reach the ground in 3.5s?

GIVEN: g = 9.80 m/s2, t = 3.5 s

Section 2.3

What is the speed of the ball in the previous example 3.5 s after it is dropped?

Use equation: vf = vo + at vo = 0

GIVEN: g = a = 9.80 m/s2, t = 3.5 s

SOLVE:

Section 2.3

Acceleration due to gravity occurs in BOTH directions. ◦ Going up (-)

◦ Coming down (+)

The ball returns to its starting point with the same speed it had initially. vo = vf

At the maximum height v= 0.

Section 2.3

Although an object in uniform circular motion has a constant speed, it is constantly changing directions and therefore its velocity is constantly changing. ◦ Since there is a change in direction there is an

acceleration.

What is the direction of this acceleration?

It is at right angles to the velocity, and generally points toward the center of the circle.

Section 2.4

Supplied by friction of the tires of a car

The car remains in a circular path as long as there is enough centripetal acceleration.

Section 2.4

This equation holds true for an object moving in a circle with radius (r) and constant speed (v).

From the equation we see that centripetal acceleration increases as the square of the speed.

We also can see that as the radius decreases, the centripetal acceleration increases.

Section 2.4

v2

r • ac =

Determine the magnitude of the centripetal acceleration of a car going 12 m/s on a circular cloverleaf with a radius of 50 m.

Section 2.4

Compute the centripetal acceleration in m/s2 of the Earth in its nearly circular orbit about the Sun.

GIVEN: r = 1.5 x 1011 m, v = 3.0 x 104 m/s

SOLVE:

Section 2.4

An object thrown horizontally will fall at the same rate as an object that is dropped.

Multiflash photograph of two balls Section 2.5

2-18 Section 2.5 Section 2.5

2-19

An object thrown horizontally combines both straight-line and vertical motion each of which act independently.

Neglecting air resistance, a horizontally projected object travels in a horizontal direction with a constant velocity while falling vertically due to gravity.

Section 2.5

Combined Horz/Vert.

Components

Vertical

Component

Horizontal

Component + =

Section 2.5

2-21 Section 2.5

2-22 Section 2.5

2-23 Section 2.5

2-24

Angle of release Spin on the ball Size/shape of ball/projectile Speed of wind Direction of wind Weather conditions (humidity, rain, snow,

etc) Field altitude (how much air is present) Initial horizontal velocity (in order to make

it out of the park before gravity brings it down, it must leave the bat at a high velocity)

Section 2.5

2-25

d = ½at2 (distance traveled, starting from rest)

d = ½gt2 (distance traveled, dropped object)

g = 9.80 m/s2 = 32 ft/s2 (acceleration, gravity)

vf = vo + at (final velocity with constant a)

ac = v2/r (centripetal acceleration)

Review

Dv

t

vf –vo t

• a = = (constant acceleration)

• v = d/t (average speed)