9/5 Graphing Motion

Text sections 2.7 and 3.1-2 HW “9/5 Graphing” due Monday 9/9

On web or in 213 Witmer for copying

For Friday, look at text sections 1.6-8 Vector Math (what fun!)

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

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

Describe the Motion

v

t

Constant speed then slowing down to rest.

Describe the Motion

v

t

Constant speed then slowing down to rest (for an instant), turning around and speeding up in the opposite direction.

Find the Displacement

v (m/s)

t (s)

How far did the object go in 9 seconds?10

10

What is the average velocity?

7 m/s

Displacement = 63 m

There are 31.5 shaded squares each representing 2 m of displacement. 2 x 31.5 = 63

Note that the displacement is the same as the shaded area.

Find the Displacement

v (m/s)

t (s)

At what time is the object back where it started from?10

10

What is its velocity at this time?

If it returns to its starting point it must turn around. At what time does it do that?-10

10s

6s

-8m/s

What is the displacement from 2s to 13s?

How many meters were put on the odometer from 2s to 13s?