Chapter 2 Motion in One Dimension
description
Transcript of Chapter 2 Motion in One Dimension
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Chapter 2 Motion in
One Dimension
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Motion is relative.
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An object can be moving with respect to one object and at the same time be at rest or moving at a different speed with respect to another.
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Frame of Reference
Is the point with which a motion is described.
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How fast are you moving
at this moment?
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Depends upon how you look at it?
If you look at it from the point of view with the room, most of you are not moving.
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If you look at it from the point of view from outer space, then you are moving as fast as the earth is rotating.
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Or from out side the solar system, the earth is moving
around the sun at a speed of
approximately 100,000 km/hr.
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Relative Motion Animation
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Vector A physical quantity that has both magnitude and direction.
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Ex:10 km, North
15 m/s, SW
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ScalarA physical quantity that has magnitude, but no direction.
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Ex: 55 km/hr
19 m
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Distance How far an object has moved.
No direction, therefore a scalar.
Ex: 20 km
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DisplacementThe change in position of an
object
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Difference b/n Distance & Displacement
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Displacement
=
Change in position
=Final position – Initial position
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x = xf – xi
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Note: Displacement is not always equal to distance moved.
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Displacement has a specific
direction, therefore it is a
vector.
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Displacement can be positive (+) or negative
(-).
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On the x-axis displacement to the right is (+) and displacement to the left is (-).
On the y-axis displacement upwards is (+) and displacement downwards is (-).
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Speed
Measure of how fast something
is moving.
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Is the distance covered per unit of time.
ex: 72 km/hr or 20 m/s
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Speed Unitsm/s, km/hr, or cm/s, same as velocity units.
Since speed has no direction, it is a scalar.
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The fast speed possible is the speed of light.
Which is 3 x 108 m/s (299,792,458 m/s)
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Instantaneous Speed
Is the speed of an object at any instant.
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Instantaneous Speed
Is the speed of an object at any instant.Ex: speedometer reading
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Average SpeedThe total distance divided by the time
interval during which the displacement occurred. (Vavg)
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Change in position
Vavg = ---------------------- Change in time
total distance = -----------------------
time interval
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Xvavg = ------ t
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xf – xi
vavg = --------- t
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The cheetah averages 70 m/s over 30 seconds. How far does it travel in those 30 seconds?
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VelocityIs the rate of change
of displacement.It is speed with a direction. (vector)
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The Vavg can
be (+) or (-), depending on the sign of the displacement.
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Three Ways to Change Velocity
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Three Ways to Change Velocity
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Ex 1: During a trip, a plane flies directly
East with an average velocity of 35 m/s. What distance does
the plane cover in 45 minutes?
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G: Vavg = 35 m/s, t= 45 min= 2700 s
U: X = ?
E: X = (Vavg )(t)S: X=(35m/s)(2700s)S: X = 94,500 m, E
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With your group, work together to solve practice problems: 2, 4, and 6 on page 44 (HP). Do these
problems in your notes. I will check for them in the
notebook check. Also use the GUESS method.
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2) 3.1 km4) 3 hr6a) 6.4 Hour6b) 77 km/hr South
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Velocity is not the same as
speed.
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Uniform MotionBoth velocity/speed and direction of the body/object remain
the same.
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Accelerated Motion
Is when the velocity/speed of
the object is changing.
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Acceleration (aavg)
Is the rate of change of velocity.
How fast you change your velocity
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How do you know your accelerating?
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How do you know your accelerating?
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aavg has direction and magnitude; therefore, it is a
vector.
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Change in Velocity
aavg = ---------------
Time interval
for change
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v vf – vi
aavg = --- = ---------
t t
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Units: m/s2 or cm/s2
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If acceleration is negative (-) it means the object is slowing down or decelerating
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Uniform Accelerated Motion
Constant acceleration,meaning the velocity changes by
the same amount each time interval.
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Ex 2: A car slows down with an acceleration of –1.5 m/s2. How long does it take for the car to stop
from 15.0 m/s to 0.0 m/s?
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G: aavg = -1.5 m/s2, vf = 0.0m/s vi = 15 m/s
U: t = ?
E: aavg = (vf – vi) / t
or t = (vf – vi) / aavg
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t = (0m/s – 15 m/s)
(-1.5 m/s2)
t = 10 s
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Do practice problems, on page 49 (HP), #2 and 4.
Work together with groups. These must be in
notes and you need to use the GUESS Method
to solve them.
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Displacement (x) depends upon: aavg, vi ,
and t.
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For an object moving with a uniform acceleration.
The vavg is the average of the vi and the vf.
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vi + vf
vavg = ------- 2
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By setting both vavg equations equal to each
other.
x vi + vf
----- = ------- t 2
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Multiply both side by t
x = ½ (vi + vf)t
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Ex 3: Adam, in his AMC Pacer, is moving at a velocity of 27 m/s, he applies the brakes and comes to a stop in 5.5 seconds. How far did
move before he came to a stop?
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G: vi =27 m/s vf =0 m/s t= 5.5 s
U: x E: x = ½ (vi + vf)tS: x = ½(27 m/s+
0m/s)5.5sS: x = 74.3 m
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Final velocity (vf) depends upon: vi,
t, & aavg
From the aavg equation:
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Multiple both sides by time
(t)
aavgt = vf - vi
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Then add vi to both sides.
vf = vi + aavgt
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Ex 4: A pilot flying at 60 m/s opens the throttles
to the engines, uniformly accelerating the jet at a rate of 0.75
m/s2 for 8 seconds, what is his final speed?
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G: aavg= 0.75 m/s2, t=8s, vi= 60 m/s
U: vf = ?
E: vf = vi + aavg t
S: vf =60 m/s +
(0.75m/s2 x 8 s)S: vf = 66 m/s
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With our final speed equation, we can
substitute it into the x equation. This allows us to find x without
knowing vf.
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We are going to substitute
vf = vi + aavgt into
x = ½ (vi + vf)t
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Where vf is, replace it with
(vi + aavgt)x = ½ (vi + vi + aavgt)t
x = ½ (2vit + aavgt2)
x = vit + ½ aavgt2
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Ex 5: A plane flying at 80 m/s is uniformly
accelerated at a rate of 2 m/s2. What distance will it travel during a 10 second interval after acceleration
begins?
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G: t= 10 s, a =2m/s2, vi = 80 m/s,
U: x = ?
E: x = vit + ½ a t2
S: x = (80 m/s)(10 s) + ½(2 m/s2)(10 s)2
S: x = 900 m
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Open books to page 56 and read it. Try
and follow the algebra and
substitution for our last equation.
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Final velocity after any displacement.
vf2 = vi
2 + 2aavgx
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Ex 6: A bullet leaves the barrel of a gun, 0.5 m long, with a muzzle
velocity of 500 m/s. Find (a) its acceleration and
(b) the time it was in the barrel.
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G: x=0.5 m, vf =500m/s, vi = 0 m/s
U: aavg = ?
E: vf2 – vi
2 = 2aavgx or aavg = (vf
2 – vi2) / 2x
S: aavg =(500 2 – 0 2)/(2 x 0.5)
S: aavg = 2.5 x 105 m/s2
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G: x =0.5 m, vf =500m/s vi = 0 m/s
U: t = ?
E: x = vit + ½ aavgt2 or t2 = 2 x /aavg
S: t2 = 2(0.5) / 2.5 x 105 S: t = 0.002 s
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Free Falling Objects
Galileo showed that a body falls with a
constant acceleration of 9.81 m/s2.
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Acceleration of Gravity
g = – 9.81 m/s2
(For convenience we will use –
10 m/s2)
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That means after 1 sec the object will have increased its
speed by 10 m/s.So if starting from rest:
After 1 sec – 10 m/sAfter 2 sec – 20 m/sAfter 3 sec – 30 m/s
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Ex 7: A stone dropped from a cliff hits the
ground 3 seconds later. Find (a) the speed with which the stone hits the
ground, and (b) the distance it fell.
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G: t =3 sec,vi =0 m/s aavg = g = - 10 m/s2
U: vf = ?
E: vf = vi + aavgt
S:vf=0 m/s+(- 10 m/s2)(3s)
S: vf = - 30 m/s
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G: t =3 sec, vi =0 m/s aavg = g = –9.81 m/s2
U: x= ?
E: x = vi t + ½ aavgt 2
S: x = (0 m/s)(3 s) + ½ (– 10 m/s2)(3 s)2
S: x = – 45 m
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Free falling bodies always have the same downward acceleration
Even though an object may be moving upwards, its acceleration is downwards.
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The velocity is positive, but is decreasing.
When it reaches the peak, the velocity is
zero, but still accelerating downwards.
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Then the object begins to fall with
a negative velocity.
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A ball is thrown straight up with
an initial velocity of 30 m/s.
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t (s) y (m) Vf (m/s) Aavg
(m/s2)0.001.002.003.004.005.006.00
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t (s) y (m) Vf (m/s) Aavg
(m/s2)0.00 -101.00 -102.00 -103.00 -104.00 -105.00 -106.00 -10
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t (s) y (m) Vf (m/s) Aavg
(m/s2)
0.00 30 -101.00 20 -102.00 10 -103.00 0 -104.00 -10 -105.00 -20 -106.00 -30 -10
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t (s) y (m) Vf (m/s) Aavg
(m/s2)
0.00 0 30 -101.00 25 20 -102.00 40 10 -103.00 45 0 -104.00 40 -10 -105.00 25 -20 -106.00 0 -30 -10
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Ex 8: Amber hits a volleyball, so that it moves with an initial
velocity of 6m/s straight upward. If the ball starts from 2 m off the floor. How long will it remain in the air before hitting
the floor? Assume she is the last person to touch it.
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G: Vi = 6 m/s, x = -2 m, aavg= -10 m/s2
U: t :There is no easy equation to use so we need to find t
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So we need to find the Vf first.
E: vf2 = vi
2 + 2aavgx
S: vf2 =(6m/s)2 +
2(-10 m/s2)(- 2m)S: vf = +/- 8.7 m/s, since its
moving downwards its – 8.7 m/s
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Now we can find the time t.
E: aavg = (vf – vi) / t or
t = (vf – vi) / aavg
S: t = (- 8.7 m/s – 6m/s) / (-10 m/s2)
S: t = 1.47 s