Chapter 2 Motion in One Dimension. Motion is relative

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, North15 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 (-).

SpeedMeasure 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 positionVavg = ----------------------

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: vf

2 =(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