### Study Guide for Unit 1 Test

```Mrs. Ashley

1.
2.
3.
4.
Has magnitude but no
direction
Speed
Mass
Temperature
Magnitude of a vector
Scalar

1.
2.
Has magnitude and
direction
Velocity
acceleration
Vector

A baseball player runs 27.4 meters from the
batter's box to first base, overruns first base
by 3.0 meters, and then returns to first base.
Find the total distance traveled by the player.
What is the magnitude of the player's total
displacement from the batter's box?






Distance (total) = d1 + d2 + d3
Total displacement
Total Distance is: 27.4 meters + 3.0 m + 3.0
m = 33.4 m
Total displacement =(taking into account
direction) d1 + d2 - d3
27.4 m + 3.0 m -3.0 m
Total displacement is 27.4 m from the batters
box.





Speed = distance/time
Average Speed = total distance/time interval
Velocity = Δ displacement/ Δ time
Acceleration = change in velocity/time interval
(rate of change in direction or change in speed
or both is acceleration)
A curved path is always a change in direction,
so it is accelerating even if speed doesn’t
change

How long will it take an object to move 100
meters if the object is traveling with an
average speed of 0.5 meter per second?


t= d/v
t = 100m/.5 m/s = 200 s



Velocity = Δ displacement/ Δ time
Average acceleration= velocityf - velocity i
time f - time
i

A car is traveling along a straight highway
with a velocity of 30 m/s when the driver hits
the brakes when a car pulls out in front of it.
If it takes the car 5 seconds to slow down to a
velocity of 3.0 m/s, what is the car’s average
acceleration?








average acceleration = change in velocity
elapsed time
a = 3.0 m/s – 30 m/s
5s
a = 27 m/s
5s
a = -5.4 m/s2



Constant velocity means that an object is
moving the same distance in the same time in
the same direction
Instantaneous velocity =acceleration X time
v=at
Average velocity = beginning velocity +
ending velocity divided by 2, if acceleration
is constant
 Distance
traveled = ½ at2
velocity is zero to start)
(if

When we want to find final velocity when we
know initial velocity, acceleration, initial
distance and final distance:
 v2f

= v2i + 2a( df - di)

If we what to find the velocity of an object
after a certain amount of time and know the
objects constant acceleration we use this
formula:
 vf
= vi + at

If we want to find the final position and know
the original position, the velocity, time and
acceleration we use the following formula:
 df

= di + vit +
1/2
at2

If we want to find final distance and know the
original position, original velocity, final
velocity, and acceleration then we use:
 df


= di + v2f - v2i
2a
This formula can be used for braking
distance, for example

Formula for stopping distance of a car: (tr
is reaction time and notice that a is negative)
 df
= vitr - vi2/ 2a

A car slows down from 40 m/s to a stop in a
distance of 100 m. What was its acceleration,
assumed constant?
 v2f
= v2i + 2a( df - di)
 a = v2f - v2i

2 ( d f - d i)
 a=
0- (40 m/s )2

2(100m)
 a = -8 m/s2

If a car slows down to rest uniformly and
stops in 63.0 m and its original speed was 21
m/s How long did it take to come to a
complete stop?
 df
= di + average velocity times
 df
= di
t
+ vi + vf t
2
 t = 2 (df - di)

vi + vf

t = 5.5 s or 6 s
Determine the stopping distance for a car
with an initial velocity of 80 km/h and a
human reaction time of 1.0 s for an
acceleration of
-6.0m/s2


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The origin is the location of the car at the
beginning of the reaction time: You have
different units, so need to change everything
to m/s:
(80km/h) (1h/3600s) ( 1000m/km) =
22.2m/s
d = vt (22.2m/s) (1.0 s) = 22.2 m
df = di + v2f - v2i
2a


df = 22.2m + 0- 22.2m/s2 x -22.2m/s2
2(6.0 m/s2)

df

Df = 22.2m + 41.07 = 63.27 m= 63 m
=
22.2m+492.84/12 m
Regardless of mass, acceleration is at the
same rate
9.8 m/s2 This means that
acceleration is 9.8 meters per second in each
second of fall.
 Free fall acceleration is due to gravity so “g” is
used in the formulas
 Velocity = gravity/time
 Distance = vot+ ½ gt2


If you know the original velocity and the time,
you can figure the velocity at the time
elapsed by this formula:
 vf
= vo + gt



An object slows at 9.8 meters per second
squared
Instantaneous speed is same on the way up
as the way down at the same position.
Up is usually expressed as a positive and
down is expressed as a negative in terms of
acceleration due to gravity

If a ballplayer catches a ball 3 seconds after
throwing it vertically upward, with what speed
did he throw it and what height did it reach?

Upward is the positive direction, then d initial
= 0. The acceleration is -9.81 m/s2 The
displacement is 0, assuming the ball was
caught at the exact location from where it
was thrown. Use y as the height.
 Yf
= Yo + vot +
1/2
at2
 Yf
= Yo + vot +
 vo
= -1/2 at
 vo
=15 m/s
1/2
at2

df = di + v2f - v2o
2a
df = 0 + 0- (14.7 m/s)2
2(-9.81m/s2

df = 1 m



```