9/4 Acceleration

Text sections 2.1-3 and 1.5-6 HW “9/4 Airplane” due Friday 9/6

On web or in 213 Witmer for copying

For Thursday, look at text sections 2.7 and 3.1-2 Graphing and 2-D Motion

Suggested Problems: 2-25, 26, 29, 30

0m/s

Example ProblemA block slides from rest down a ramp, across a level section, then down another ramp of equal slope. Ignore friction.On the lever section the block moves with a constant velocity of 4m/s.

v = 4m/s left

What is the block’s average velocity on the upper ramp?

vf =

vi =

4m/s

The average of 0 and 4 is 2.vave = 2m/s down the

ramp

Average Velocity

Average velocity is the “middle” velocity as well as x/t.

Example:

An object slows down from 35m/s to 5m/s, what is the average velocity?

It took 6s to slow down, how far did the object move?

What is its speed at 3s, the “mid-time?”

Acceleration A ball rolls up and down a ramp as shown in

the strobe photograph. Which way does the acceleration point or does the acceleration = 0?

Turnaround pointBall rolling up the rampv

Pick a time interval, ti - tf and draw velocity vectors

titf

Draw velocity vectors tail to tailDraw v, (from i to f) which points the same direction as a.

vfvi

vi

vf

a points down the ramp.

a =vt

Acceleration and VelocityExample:

An object moving left slows down from 35m/s to 5m/s, what is the average velocity direction?

It took 6s to slow down, what is the object’s acceleration, magnitude and direction? (Always think about v.)

v = 30m/s to the right

a = 5m/s2 to the right

Acceleration at turnaround A ball rolls up and down a ramp as shown in

the strobe photograph. At the turnaround point, which way does the acceleration point or does the acceleration = 0 there?

Turnaround point

Acceleration at turnaround

Turnaround point

Ball rolling up the ramp v

Pick a time interval, ti - tf and draw velocity vectors

tf

Copy velocity vectors tail to tail

vi

vi

Turnaround point

Ball rolling down the ramp

ti

vfv

vf

Draw v, (from i to f) which points the same direction as a.

Even though v = 0, v is still changing and there is acceleration!!!!

Acceleration

a =vt

is an “operational definition” in that it defines a procedure for finding and using a.

Finding acceleration

Using Acceleration

“Change in Velocity” Vector, v

v = -4m/s leftv = 8m/sv = 4m/sv = 0m/sv = -4m/sv = -8m/sv = -12m/s

v = -4m/s leftv = -4m/s leftv = -4m/s leftv = -4m/s left

The “change in velocity” vector may point with or against the velocity vector.

Even though the object slows down, turns around, and speeds up in the opposite direction; v is constant!

Acceleration

a =vt

v = -4m/s leftv = 8m/sv = 4m/sv = 0m/sv = -4m/sv = -8m/sv = -12m/s

v = -4m/s leftv = -4m/s leftv = -4m/s leftv = -4m/s left

v and a point opposite,slowing down

v and a point the same direction,speeding up

Acceleration is a vector that points in the same direction as the “change in velocity” vector. In this case, a = 4m/s/s left.

In concept, it is “the amount and direction the velocity changes each second.”

Concepts so far-

Displacement, x (distance moved)

Instantaneous Velocity, v (at a particular time)

Average Velocity, vave (average over time)

Change in Velocity, v (speeding up or slowing down)

Acceleration, a (how much the velocity changes each second)

Problem:

An object goes from a velocity of 15 m/s right to 6 m/s right in 3 seconds. Find the acceleration, both its size (magnitude) and its direction, (left or right).

How do the directions of the velocity and acceleration compare? What is the object doing during these 3 seconds?

How far did the object travel during these three seconds? Hint: What is the average velocity?

What will the objects velocity be in three more seconds if the acceleration stays the same?

Problem:

A bullet exits a rifle at 85m/s. The barrel is 0.75m long.

What is the acceleration of the bullet?

Don’t use text equations, just the relationships between displacement, time, velocity and acceleration

Finding acceleration

vi = 10m/s

vf = 40m/s

v = 30m/s right

Return

a =vt

t = 6s

=306 = 5m/s/s right

Problem:

A bear is running 4 m/s north. The acceleration of the bear is 3m/s2 north. What is the bear’s velocity 2 seconds later?

v = 10 m/s north

What is the bear’s average velocity? How far did the bear run during this time?

vave = 7 m/s north

x = 14 m northReturn