v - Bloom High School
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Transcript 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
add the initial velocity
• 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
add, then:
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,