### v - Bloom High School

```Honors Physics
Glencoe Science, 2005
 Change in
• Dt = tf - ti
 Change in
velocity
time- ending time minus initial time
velocity- final velocity minus initial
• Dv = vf - vi
 Average
acceleration- change in velocity
between two distinct time intervals
• ā = Dv/Dt = (vf - vi)/(tf - ti)
 Instantaneous
(m/s2)
acceleration- change in velocity
at an instant in time
• Only found by determining the tangent of a curve on a
velocity-time graph
 Shows
an object slowing down, speeding up,
without motion, or at constant motion
 Represents
motion graphically
 Plots the velocity versus the time of the
object
 Sometimes
we have an initial velocity
• Like when we pull through a stop light as it turns
green
 Our
acceleration formula can be rearranged
• Dv= ā Dt  vf - vi = ā Dt
 If
we are looking for the final velocity, then
we multiply the acceleration by the time and
• vf = ā Dt + vi
A bus that is traveling at 30.0km/h speeds up at a
constant rate of 3.5m/s2. What velocity does it reach
6.8s later?
 Convert all to terms of m and s:

• (30.0km/h)(1000m/km)(1h/3600s)=vi

Define known & unknown:
• a=3.5m/s2
• Dt=6.8s

vi=8.33m/s
vf=?
Choose the appropriate equation
• vf=āDt + vi


Rearrange if necessary (it is not in this case)
Plug & Chug
• vf=(3.5m/s2)(6.8s)+(8.33m/s)
• vf=32.13m/s
 Looking
find:
at a position-time graph, we can
• Figure 3-9
• Velocity (slope)
• Specific positions at specific times
 From
our original velocity equation,
v=Dd/Dt, we can find that Dd=vDt
 The area under the line in a velocity-time
graph is vDt, which is the displacement!
• Figure 3-10
• Slope is v/Dt, which is acceleration!
 Position
(df) of an object under acceleration
can be found with:
• df=½at2
(m/s2)(s2)=m
• df=di+½at2
(m)+(m/s2)(s2)=m
 If
there is an initial distance that we need to
 If
there is an initial velocity, then we can also
include that term!
• df=di+vit+½at2
(m)+(m/s)(s)+(m/s2)(s2)=m
• This equation is only useful if time of travel is known
 Unlike
the prior equation, sometimes time is
not known, so we need to relate velocity to
distance traveled
• vf2=vi2+2a(df-di)
(m/s)2=(m/s)2+(m/s2) (m-m)
Equation
Variables
Initial Conditions
vf = ā Dt + vi
vf, ā, Dt
vi
df=di+vit+½at2
df, t, a
di, vi
vf2=vi2+2a(df-di)
vf, a, df
vi, di
A race car travels on a racetrack at 44m/s and slows
at a constant rate to a velocity of 22m/s over 11s. How
far does it move during this time?
 Define known & unknown:

• vi=44m/s
• Dt=11s
• a=?

vf=22m/s
Dd=?
Choose the appropriate equation
• We need to find ā first
• vf=āDt + vi

Rearrange if necessary
• ā=(-vi+vf)/Dt

Plug & Chug
• ā=(-44m/s+22m/s)/(11s)
• ā=-2m/s2
that we have ā, we can solve for df
 Define known & unknown:
 Now
• vi=44m/s
• Dt=11s
• Dd=?
vf=22m/s
a=-2m/s2
 Choose the appropriate
• df=di+vit+½at2
 Rearrange if necessary
equation
 Plug & Chug
• df=(0m)+(44m/s)(11s)+½(-2m/s2)(11s)2
• df=363m
A car accelerates at a constant rate from 15m/s to
25m/s while it travels a distance of 125m. How long
does it take to achieve this speed?
 Define known & unknown:

• vi=15m/s
• Dt=?
• Dd=125m

vf=25m/s
a=?
Choose the appropriate equation
• We need to find ā first, but don’t know time
• vf2=vi2+2a(df-di)

Rearrange if necessary
• a=(vf2-vi2)/(2Dd)

Plug & Chug
• a=((25m/s)2-(15m/s)2)/((2)(125m))
• a=1.6m/s2
that we have ā, we can solve for Dt
 Define known & unknown:
 Now
• vi=15m/s
• Dt=?
• Dd=125m
vf=25m/s
a=1.6m/s2
 Choose the appropriate equation
• vf = ā Dt + vi
 Rearrange if necessary
• (vf-vi)/ā=Dt
 Plug & Chug
• (25m/s-15m/s)/(1.6m/s2)=Dt
• Dt=6.3s
 Free
Fall- an object falling only under the
influence of gravity
 Acceleration due to gravity- an object
speeds up due to the Earth’s gravitational
pull
• g=9.8m/s2
• Gravity is a specific kind of acceleration: like a
quarter is a specific kind of money
 Gravity
always points toward the center of
the Earth
 As
gravity (g) is a kind of acceleration (a),
we can replace all of the “a”’s with “g”
 This can only be done if the object is in free
fall
Equation
Variables
Initial Conditions
vf =gDt + vi
vf, Dt
vi
df=di+vit+½gt2
df , t
di, vi
vf2=vi2+2g(df-di)
vf, df
vi, di
A
construction worker accidentally drops a
brick from a high scaffold. What is the velocity
of the brick after 4.0s?
 Define known & unknown:
• vi=0m/s
• Dt=4.0s
• Dd=?
vf=?
g=9.8m/s2
 Choose the appropriate equation
• vf =gDt + vi
 Rearrange if necessary (not necessary)
 Plug & Chug
• vf =(9.8m/s2)(4.0s)+(0m/s)
• vf =39.2m/s
A
construction worker accidentally drops a
brick from a high scaffold. How far does the
brick fall during this time?
 Define known & unknown:
• vi=0m/s
• Dt=4.0s
• Dd=?
vf=39.2m/s
g=9.8m/s2
 Choose the appropriate equation
• df=di+vit+½gt2
 Rearrange if necessary (not necessary)
 Plug & Chug
• df=(0m)+(0m/s)(4.0s)+½(9.8m/s2)(4.0s)2
• df=78.4m
A tennis ball is thrown straight up with an initial
speed of 22.5m/s. It is caught at the same distance
above the ground. How high does the ball rise?
 Define known & unknown:

• vi=22.5m/s
• Dt=?
• Dd=?

vf=0m/s
g=-9.8m/s2
Choose the appropriate equation
• vf2=vi2+2g(df-di)

Rearrange if necessary
• Dd=(vf2-vi2)/(2g)

Plug & Chug
• Dd=((0m/s)2-(22.5m/s)2)/(2(-9.8m/s2))
• Dd=25.82m
A tennis ball is thrown straight up with an initial
speed of 22.5m/s. It is caught at the same distance
above the ground. How long does the ball remain in
the air?
 Define known & unknown:

• vi=22.5m/s
• Dt=?
• Dd=25.82m

vf=0m/s
g=-9.8m/s2
Choose the appropriate equation
• vf =gDt + vi

Rearrange if necessary
• Dt=(vf-vi)/g

Plug & Chug
• Dt=(0m/s-22.5m/s)/(-9.8m/s2)
• Dt=2.30s
 As
an objects’ speed approaches
3.0x108m/s (c), the time as observed
from the outside of the ship changes
 So if you are travelling very fast,
```