Module 2: One Dimensional Motion Equations and Free Fall

Physics

Vocabulary and Major Points

Free Fall: The motion of an object when it is

falling solely under the influence of gravity.

More on Free Fall: When an object is thrown

upward in the presence of gravity, the object will

reach its maximum height when the object’s ve-

locity equals zero. When it returns to the height

from which it was thrown, its velocity will be

equal and opposite to its initial velocity.

Constant Acceleration. Acceleration remains at

the same pace.

Gravitational Acceleration: Objects falling near

the surface of the earth experience a constant ac-

celeration of 9.8 m/s2 straight down. The accel-

eration due to gravity is independent of the na-

ture of the object, as long as the object has mass.

Air Resistance: A force, caused by air, that af-

fects the rate of acceleration of objects.

Terminal Velocity: The velocity a falling object

has when, due to air resistance, its acceleration is

reduced to zero. This is the maximum velocity a

falling object to air resistance can achieve.

ov v at= +

2 2

0

2

0

2

1

2

v v a x

x v t a t

= + ∆

∆ = +

In a position-versus-time graph, time is on the x-axis, and position is on the y-axis.

The curve represents the object’s position at any given time.

The slope of the line represents the velocity.

The steeper the slope, the greater the speed or velocity.

If you make the time interval get smaller and smaller, you approach the instantaneous velocity.

A change in the sign of the slope indicates a change in direction.

Example 1.4 pg 19 Position-vs-Time Graph

In a velocity-versus-time graph, time is on the x-axis, and velocity is on the y-axis.

The curve represents the object’s speed at any given time (actually, velocity because of the

sign).

The slope indicate acceleration. The steeper the slope, the greater the acceleration.

Additionally, the area under the curve represents the object’s displacement.

The object is “speeding up” when the sign of both the velocity and slope (acceleration) are the

same.

Example 1.7 pg 28 Velocity-vs-Time Graph

Module 1: Motion in One Dimension

Physics

Extra Practice Problems

1) A person standing on the edge of a 30 foot cliff throws a ball straight up into the air. If the

ball travels 50 feet up and then falls to the bottom of the cliff, what would be the total distance

traveled? What would be the displacement of the ball?

Ball g

oes u

p 5

0 feet

befo

re it stop

s and

Ball falls b

ack to

starting p

ositio

n. T

hen

falls

30

mo

re feet to th

e bo

ttom

of th

e cliff.

Distance does not care about direction:

d = 50 ft + 50 ft + 30 ft. = 130 ft.

Displacement cares about direction. Let ‘up’ be positive.

x = + 50 ft - 50 ft - 30ft = - 30 feet. Or 30 feet down.

2) It takes an infant 3.5 minutes to crawl the 22 meters from the playpen to the

refrigerator and 11 meters back toward the playpen What is the infant’s av-

erage speed and average velocity?

Change in time is given as 3.5 minutes.

The total distance traveled is 22 meters + 11 meters = d = 33 meters

The total displacement, x, requires direction. Let toward the refrig be +

x= +22 m - 11m = 11 meters (toward refrig)

Speed = change in distance/change in time = 33m/3.5 min =

Velocity, v, = change in displacement/change in time = 11 m/ 3.5 min

Crib

Re-

frig Let toward refrig be positive

3) A runner’s average velocity over a race of 600 meters is 0.61 m/s. Since the race was run on an oval track, the

runner’s displacement was only 62 m. How much time did it take the runner to run the race?

X start

X finish

Total race: 600 m

Distance between stop & finish: 61m

Velocity depends on displacement; displacement doesn’t consider direction;

displacement = 62 m = d

Average velocity = 0.61 m/s

Solve for ∆t: ∆t = ∆x / v

∆t = 62 m

0.61m/s Units cancel & agree

Dumb Question: Written

just to test your knowledge

of ‘displacement’ and it’s

use in finding velocity.

4) While driving, a man slows his car from 72 mph to 65 mph in 1.8 seconds. What is his acceleration in miles/

s2?

Time units do not agree. Convert the

mph to miles per second:

Then solve for acceleration

Module 1: Motion in One Dimension

Physics

Extra Practice Problems (cont)

5) What would be the final velocity of a car that is initially moving at 62 m/s and accelerates at

–5.0m/s2 for 9.2 sec?

The car is slowing, but that’s still acceleration.

Must re-arrange equations to solve for vfinal.

Questions 6—10 Reference graphs in the book.

Pages 563 and 564

6) How many times does the car change directions and during what time intervals is the car moving the slowest?

To determine when the car changes direction, find where the slope changes sign.

@t =2; t=7.5;;t=16 so 3times.

To determine slowest, look for slopes of 0 (no change in position, as time marches on): @t=2; t=7.75; and

@ 13≤t≤16. At these locations, the car is either changing direction or no moving.

7) What is the instantaneous velocity at 5 seconds.?

You must find the slope at t=5 seconds. Look at the graph, and you can see the slope is negative.

Take an interval “around” it; the slope looks constant from t=4 to t=6

∆x = -.5m - .5m; ∆t=6s - 4s

v = -0.5m/s

8) Over what time intervals is the car speeding up?

This is a velocity-versus-time graph.

“Speeding up” means increasing acceleration, so we’re looking for places where the acceleration and the

the velocity share the same sign. Acceleration is the slope of the line. There are 2 places where the

signs are the same.

@ t=2, the velocity goes negative (look at the axis), and the slope is negative

And they both stay negative until about t=3.75.

@t=11, the slope (acceleration) turns positive, but the velocity doesn’t turn positive until t=14

So from t=14 until t=16.5 both acceleration and velocity have positive signs.

These 2 intervals are where the car is “speeding up”

9) During what time intervals is the acceleration equal to zero?

Look for places where the slope is 0:

From about t=4 to t=7

And about t=16.3 to t=16.7

10) What is the acceleration at 13.5 seconds?

You want the slope around 13.5 seconds. Slope looks constant from 13s to 14s

m=0 m/s—(-1)m/s = 1m/s2

1s