Chapter 2 Motion in
One Dimension
Motion is relative.
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.
Frame of Reference
Is the point with which a motion is described.
How fast are you moving
at this moment?
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.
If you look at it from the point of view from outer space, then you are moving as fast as the earth is rotating.
Or from out side the solar system, the earth is moving
around the sun at a speed of
approximately 100,000 km/hr.
Relative Motion Animation
Vector A physical quantity that has both magnitude and direction.
Ex:10 km, North
15 m/s, SW
ScalarA physical quantity that has magnitude, but no direction.
Ex: 55 km/hr
19 m
Distance How far an object has moved.
No direction, therefore a scalar.
Ex: 20 km
DisplacementThe change in position of an
object
Difference b/n Distance & Displacement
Displacement
=
Change in position
=Final position – Initial position
x = xf – xi
Note: Displacement is not always equal to distance moved.
Displacement has a specific
direction, therefore it is a
vector.
Displacement can be positive (+) or negative
(-).
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 (-).
Speed
Measure of how fast something
is moving.
Is the distance covered per unit of time.
ex: 72 km/hr or 20 m/s
Speed Unitsm/s, km/hr, or cm/s, same as velocity units.
Since speed has no direction, it is a scalar.
The fast speed possible is the speed of light.
Which is 3 x 108 m/s (299,792,458 m/s)
Instantaneous Speed
Is the speed of an object at any instant.
Instantaneous Speed
Is the speed of an object at any instant.Ex: speedometer reading
Average SpeedThe total distance divided by the time
interval during which the displacement occurred. (Vavg)
Change in position
Vavg = ---------------------- Change in time
total distance = -----------------------
time interval
Xvavg = ------ t
xf – xi
vavg = --------- t
The cheetah averages 70 m/s over 30 seconds. How far does it travel in those 30 seconds?
VelocityIs the rate of change
of displacement.It is speed with a direction. (vector)
The Vavg can
be (+) or (-), depending on the sign of the displacement.
Three Ways to Change Velocity
Three Ways to Change Velocity
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?
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
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.
2) 3.1 km4) 3 hr6a) 6.4 Hour6b) 77 km/hr South
Velocity is not the same as
speed.
Uniform MotionBoth velocity/speed and direction of the body/object remain
the same.
Accelerated Motion
Is when the velocity/speed of
the object is changing.
Acceleration (aavg)
Is the rate of change of velocity.
How fast you change your velocity
How do you know your accelerating?
How do you know your accelerating?
aavg has direction and magnitude; therefore, it is a
vector.
Change in Velocity
aavg = ---------------
Time interval
for change
v vf – vi
aavg = --- = ---------
t t
Units: m/s2 or cm/s2
If acceleration is negative (-) it means the object is slowing down or decelerating
Uniform Accelerated Motion
Constant acceleration,meaning the velocity changes by
the same amount each time interval.
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?
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
t = (0m/s – 15 m/s)
(-1.5 m/s2)
t = 10 s
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.
Displacement (x) depends upon: aavg, vi ,
and t.
For an object moving with a uniform acceleration.
The vavg is the average of the vi and the vf.
vi + vf
vavg = ------- 2
By setting both vavg equations equal to each
other.
x vi + vf
----- = ------- t 2
Multiply both side by t
x = ½ (vi + vf)t
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?
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
Final velocity (vf) depends upon: vi,
t, & aavg
From the aavg equation:
Multiple both sides by time
(t)
aavgt = vf - vi
Then add vi to both sides.
vf = vi + aavgt
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?
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
With our final speed equation, we can
substitute it into the x equation. This allows us to find x without
knowing vf.
We are going to substitute
vf = vi + aavgt into
x = ½ (vi + vf)t
Where vf is, replace it with
(vi + aavgt)x = ½ (vi + vi + aavgt)t
x = ½ (2vit + aavgt2)
x = vit + ½ aavgt2
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?
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
Open books to page 56 and read it. Try
and follow the algebra and
substitution for our last equation.
Final velocity after any displacement.
vf2 = vi
2 + 2aavgx
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.
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
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
Free Falling Objects
Galileo showed that a body falls with a
constant acceleration of 9.81 m/s2.
Acceleration of Gravity
g = – 9.81 m/s2
(For convenience we will use –
10 m/s2)
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
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.
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
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
Free falling bodies always have the same downward acceleration
Even though an object may be moving upwards, its acceleration is downwards.
The velocity is positive, but is decreasing.
When it reaches the peak, the velocity is
zero, but still accelerating downwards.
Then the object begins to fall with
a negative velocity.
A ball is thrown straight up with
an initial velocity of 30 m/s.
t (s) y (m) Vf (m/s) Aavg
(m/s2)0.001.002.003.004.005.006.00
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
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
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
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.
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
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
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