Module 1: Motion in One Dimension

Physics

Vocabulary and Major Points

Scalar: A physical measurement without directional information.

Vector: A physical measurement that includes directional information.

Displacement: the change in an objects position (a vector). In equa-

tions, represented by ‘x’.

Speed: The time rate of change of the distance traveled by an object

(how far did it go in a specified time). (a scalar)

Velocity: The time rate of change of an object’s position (a vector).

Velocity, represented by v, has direction.

Instantaneous velocity: The velocity of an object at one moment in

time.

Average velocity: The velocity of an object over an extended period of

time

Acceleration: the time rate of change of an object’s velocity (a vec-

tor).

Velocity (and acceleration) is relative based on where the person

measuring it is (frame of reference).

If velocity is zero, acceleration does not have to be zero.

If acceleration is zero, velocity does not have to be zero.

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 stops an

d

Ball falls b

ack to

starting p

ositio

n. T

hen

falls

30 m

ore feet to

the b

otto

m o

f the 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