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Motion in one dimension

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Motion in one dimension. Test 2. Kinematics. - the branch of physics that describes motion in terms of x, v, a, and t x- position (m) (sometimes d) v- velocity (m/s) a- acceleration (m/s²) t- time (s) Dynamics is kinematics + force exerted on an object. Displacement. - PowerPoint PPT Presentation
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Test 2
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Page 1: Motion in one dimension

Test 2

Page 2: Motion in one dimension

- the branch of physics that describes motion in terms of x, v, a, and t x- position (m) (sometimes d) v- velocity (m/s) a- acceleration (m/s²) t- time (s)

Dynamics is kinematics + force exerted on an object.

Page 3: Motion in one dimension

Defined as the change in position

f stands for final and i stands for initial Units are meters (m) in SI

Example: If Xf = 30m and Xi = 20m what is the displacement?

f ix x x

Page 4: Motion in one dimension

Distance is a scalar quantity which refers to "how much ground an object has covered" during its motion.

Displacement is a vector quantity which refers to "how far out of place an object is"; it is the object's overall change in position.

Page 5: Motion in one dimension

Distance traveled would be 2m +4m + 2m +4m = 12m

Displacement would be 0m, you end where you started.

Page 6: Motion in one dimension

Xi =120m

Xf = 5m

ΔX= Xf - Xi

ΔX= 5-120= -115Negative displacement moves left or down;

positive displacement moves right or up.

Page 7: Motion in one dimension

From A to B xi = 30 m xf = 52 m x = 22 m The displacement is

positive, indicating the motion was in the positive x direction

From C to F xi = 38 m xf = -53 m x = -91 m The displacement is

negative, indicating the motion was in the negative x direction

Page 8: Motion in one dimension
Page 9: Motion in one dimension

The average speed of an object is defined as the total distance traveled divided by the total time elapsed (scalar quanity)

SI units are m/s (meters per second) Always positive

total distanceAverage speed

total time

dv

t

Page 10: Motion in one dimension

Velocity is magnitude and direction 7 m/s East

If V is positive motion is to the right. If V is negative motion is to the left.

Speed is the magnitude of velocity (no direction)

Velocity is a vector quanity

Page 11: Motion in one dimension

It takes time for an object to undergo a displacement

The average velocity is the rate at which the displacement occurs

Velocity can be positive or negative t is always positive

fi

averagefi

x xxv

t t t

Page 12: Motion in one dimension

Cars on both paths have the same average velocity since they had the same displacement in the same time interval

The car on the blue path will have a greater average speed since the distance it traveled is larger

Page 13: Motion in one dimension

How fast is a train moving if it travels 25 m in 5 sec?

Given:

X=25m

t=5s

Solve:

V = x/t

V = 25m / 5 sec

V = 5 km/s

Page 14: Motion in one dimension

If a car travels 100 miles in 2 hours what is the velocity of the car?

________ mph

Convert this is meters/second

Page 15: Motion in one dimension

fi

averagefi

x xxv

t t t

A car travels 40 meters between the hours of 10am and 12am. What is the cars average velocity?

Page 16: Motion in one dimension

A runner runs 3 seconds at 4m/s, stops for one second, and begins running for 2 seconds and covers 3m.

What is the velocity from (a) A to B, (b) B to C, and (c) A to D

Page 17: Motion in one dimension

Constant Velocity• Constant velocity is when an object’svelocity remains the same for a givenamount of time.

• An objects current velocity would be equal to the average velocity.

Page 18: Motion in one dimension

Changing velocity means an acceleration is present

Acceleration is the rate of change of the velocity

Units are m/s²

fi

fi

v vva

t t t

Page 19: Motion in one dimension

Which line goes with which car?

Page 20: Motion in one dimension

What is the acceleration of a car that travels from 20m/s to 50m/s in a period of 5s?

Given:

20m/s = vi

50 m/s =vf

5s = t

Solve:a = vf – vi

t

a = 50m/s – 20m/s 5s

a = 6 m/s2

Page 21: Motion in one dimension
Page 22: Motion in one dimension

fi

fi

v vva

t t t

We know that

Vf= Vi + aΔt

Page 23: Motion in one dimension

Xo = Xi

X= Xf

Page 24: Motion in one dimension

1. V= Vo + at2. X – Xo= ½ (v + vo)t3. X – Xo= Vo t + ½ at2

4. V2 - Vo2 = 2a (X – Xo)

Page 25: Motion in one dimension

An object moving under the influence of gravity alone.

Regardless of its initial conditions.

Page 26: Motion in one dimension

Drop something Throw something and let it fall Throw something down

Objects thrown upward or downward and those released from rest are all falling freely once released.

Once in freefall motion all objects experience the same acceleration; downward and equal to the acceleration due to gravity. (Earths Gravity= 9.8 m/s²

Page 27: Motion in one dimension

1. V= Vo - gt2. y – yo= ½ (v + vo)t3. y – yo= Vo t - ½ gt2

4. V2 - Vo2 = - 2g (y – yo)

Page 28: Motion in one dimension
Page 29: Motion in one dimension

Air resistance is neglected. All objects regardless of there mass will move at the same rate if they have the same initial conditions

Free fall motion is symetrical

Page 30: Motion in one dimension

Given:vi = 0.0 m/s y = –8.52 ma = –9.8 m/s2

Find:

t = ??

y-y. = v.t + 1/2gt2

-8.52m = 0 + .5(-9.8m/s2)t2

-8.52m = -4.9m/s2 t2

-8.52m/-4.9m/s2 = t2

1.7s2 =t2

1.3 s = t

JD drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground.

Page 31: Motion in one dimension

Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height.

Given: vi = 26.2 m/s vf = 0 m/sa = –9.8 m/s2

y = ?? (0 m/s)2 = (26.2 m/s)2 + 2(-9.8m/s2)y 0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2)y0 m2/s2 -686.44 m2/s2 =(-19.6 m/s2)y-686.44 m2/s2 =(-19.6 m/s2)yy = (-686.44 m2/s2)/ (-19.6 m/s2)

y = 35.0 m

V2 - Vo2 = - 2g (y – yo)

Page 32: Motion in one dimension

The observation deck of a skyscraper is 420 m above the street. Determine the time required for a penny to free-fall from the deck to the street below.

y= -420m

g =9.8m/s2

y-y.= v.t - ½ gt2

-420m = 0.5 (9.8m/s2)t2

-420m = -4.9m/s2t2

-420m/-4.9m/s2= t2

85.7m2/s2= t2

9.26s = t

Page 33: Motion in one dimension

Terminal Velocity

• An object that falls will experience a force due to gravity and a force due to the air.

• These forces are opposite in direction.

• The object will reach a point where theforces balance, which means the object will no longer accelerate. It will continue with constant velocity.

• Terminal velocity is the point reached by an object where it will no longer accelerate.

Page 34: Motion in one dimension

A Ball Thrown UpConsider a ball that is thrown straight up.

• It will have maximum velocity the instant itleaves your hand and when it returns toyour hand. (The ball is caught at the sameheight.)

• On the way up, the ball will have negativeacceleration, therefore the ball’s velocitywill decrease.

Page 35: Motion in one dimension

Consider a rock thrown downward from a high altitude hot-air balloon. If the initial velocity is 15 m/s and air resistance is neglected, what is the acceleration of the rock after one second?

What is the velocity of the rock?

9.8 m/s2


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