PHYSICS I - Pia

U N I V E R S I DA D A U T Ó N O M A D E N U E VO L E Ó N P R E PA R AT O R I A N O . 9

B Y N ATA L I A D A E N N A GONZÁLEZ VIERA

GROUP 220

MAIN MOTION CHARACTERISTICS:

• POSITION • TRAJECTORY • DISTANCE • DISPLACEMENT • SPEED • ACCELERATION

ONE-DIMENSION MOTION: TWO-DIMENSION MOTION:

• UNIFORM LINEAR MOTION • UNIFORMLY ACCELERATED LINEAR MOTION • FREE FALL (VERTICAL THROW) • UPWARD VERTICAL THROW • HORIZONTAL MOTION • PROJECTILE/PARABOLIC MOTION

TRAJECTORY It is the path that a moving object follows through space.

It can be rectilinear, circular, parabolic, etc.

POSITION It is precisely where an object is located.

DISPLACEMENT It is defined to be the change in position of an object. It is the distance, measured in straight line, between the initial position and the final position.

If an object moves relative to a reference frame—for example, if a professor moves to the right relative to a whiteboard, or a passenger moves toward the rear of an airplane—then the object’s position changes. This change in position is known as displacement.

So position and displacement have a relation.

DISTANCE • It the length of the trajectory made by the moving object from the initial position to the final position.

The displacement and the distance travelled are two completely different concepts.

The trajectory and the distance travelled have a relation.

SPEED • It is defined as the rate of change of distance.

• It is a scalar quantity.

Speed =

!"#$%&'( $")(

VELOCITY • It is defined as the rate of change of displacement.

• It is the distance traveled in a specific direction.

• It is a vector quantity.

Velocity =

!"#*+%'()(&$ $")(

Simply, speed is the distance travelled per second‌

Velocity is the distance travelled per second in a specific direction‌

ACCELERATION

It is the name we give to any process where the velocity changes. Since velocity is a speed and a direction, there are only two ways for you to accelerate: change your speed or change your direction—or change both.

EXAMPLES D I S TA N C E & D I S P L AC E M E N T

Distance =

,!

-

= Ď€đ?‘&#x;

Distance = (3.14)(6)/2 3m

Distance = 9.42m

Displacement = 3m x 2 Displacement = 6 m

SPEED • A car travels between two cities 60 kilometer apart. For the first 40 kilometers it reaches 80 km/h on paved road. In the remaining kilometers it reaches 20 km/h on dirt road. s = d/t ------------------1.What is the total travel time t = d/v Speed = in hours? t = 40/80 distance/time 2.What is the speed? t = .5 hr = 60km/1.5hrs ---------------- = 40km/hr t = d/v t = 20/20 t = 1 hr ----------------Tt = 1.5 hrs

VELOCITY • A driver travels in his car 650km east for 7 hours, and then 325km to the west for 3.5 hours.

V = displacement / time V = 650-325km ---------------------------------------

7hr+3.5hr V = 325km ------------------------------

• What is the velocity expressed in kilometers per hour?

10.5hr V=30.952km/hr

UNIFORM LINEAR MOTION It is a motion that occurs in one dimension of space at a constant speed and direction. It is a key principle of physics, directly related to Newton's first law.

UNIFORMLY ACCELERATED LINEAR MOTION It is motion with a constant, uniform change in velocity.

This often, but does not always, include a change in speed.

EXAMPLES U N I F O R M AC C E L E R AT E D L I N E A R M OT I O N

FORMUL AS Vf=Vo+at à x unnecessary x=(Vf+Vo/2)t à a unnecessary x=Vot+1/2at2 à Vf unnecessary Vf2=Vo2+2ax à t unnecessary

ACCELERATION

a= ? x=1000m t=10sec Vo=0

• What acceleration does a car develop over a distance of 1000 meters if it 2 a=2(1000/102) x=Vot+1/2at goes over them in 10 2 a= 2000/100 x-Vot=1/2at seconds and start from rest?

2(x-Vot)=at2 a= 20m/s2

2(x-Vot/t2)=a

FREE FALL

EXAMPLES FREE FALL

FORMUL AS Vf=Vo+gt - y unnecesarry y=Vot+1/2gt2 - Vf unnecessary 2 2 Vf =Vo +2gy - t unnecessary

FREE FALL • A freely dropped body reaches the ground with a velocity of 29.4m/s. Determine: • The fall time (we don’t need y) g=9.8m/s2 Vo=0 Vf=29.4m/s

Vf=Vo+gt t=Vf-Vo / g t=29.4-0 / 9.8 t=3sec

FREE FALL •The height from which dropped

y=Vot+1/2gt2 y=(.5)(9.8)(3)2 y=44.1m

EARTH

y=Vot+1/2gt2 2 y=(.5)(1.6)(3) y=7.2m

MOON

FREE FALL • A bottle dropped from a balloon reaches the floor in 20 seconds.

Determine the height at which to find the balloon if it is at rest in the air.

y=Vot+1/2gt2 y=0(20)+.5(9.8)(20)2 y=+4.9(400) y=1960m

UPWARD VERTICAL THROW • When you throw an object upwards, it will eventually fall back to the ground under the earth's gravity. In fact, all objects near the earth's surface fall with a constant acceleration of about 9.8. This is called the acceleration due to gravity and is usually denoted by the symbol g.

Note: The upward direction is taken as positive. The velocity decreases uniformly, and it becomes zero when the throw object attains its maximum height.

Then the velocity changes its sign when the object accelerates uniformly downwards.

EXAMPLES UPWARD VERTICAL THROW

UPWARD VERTICAL THROW • A stone is thrown straight upward by a bag and it reaches a height of 12m. Calculate: • The time to reach the highest point. g= -9.8m/s2 y=12m Vf=0

y=Vot+1/2gt2 12=(.5)(9.8)t2 12=4.9t2 12/4.9=t2 t2=2.448 t=1.56 sec

•Its velocity when it reaches the ground.

Vf=Vo+gt Vf=(9.8)(1.56) Vf=15.28m/s

•Its position after the first second. ↑y=Vot+1/2gt2 y=(15.28)(1)+(.5)(-9.8)(1)2 y=15.28 - 4.90 y=10.38m

UPWARD VERTICAL THROW • A baseball is thrown straight up recovers 9 seconds later by the catcher.

• Find its maximum height reached

↓y=Vot+1/2gt2 2 y=(.5)(9.8)(4.5) y=99.225m

•its velocity after it was hit by a bat.

↑Vf=Vo+gt Vo=Vf-gt Vo= -(-9.8)(4.5) Vo=44.1m/s

HORIZONTAL MOTION • This type of projectile motion is called horizontal projectile motion. This motion generally occurs when the projectile is shot straight without forming any angle with the horizontal surface and the projectile falls downward until it hits the ground.

As shown in the figure below, the initial component of the vertical components of the velocity is zero. Horizontal velocity component of the projectile remains constant as the gravity does not affect it. Direction of the vertical component of the velocity is in downward direction during the trajectory. The magnitude of the vertical component of the velocity increases as the projectile moves downward, the force of gravity acts on it, results in acceleration of the projectile.

What is important to remember is that the motion along the horizontal direction does not affect the motion along the vertical direction and vice versa. Horizontal motion and vertical motion are totally independent of each other.

EXAMPLES HORIZONTAL MOTION

HORIZONTAL MOTION • A stone is launched horizontally with a velocity of 25 m/s from a height of 60 meters, calculate a) the time with it reaches the ground

Vx=25m/s Y=60m t=?

Y=1/2gt2 2y=gt2 2y/g=t2 t=

-3 4

t=

t=3.49sec

- 67 8.:

• b) the vertical velocity at 2 seconds

Vy=gt Vy=(9.8)(2sec) Vy=19.6m/s • c) the distance when it lands (the range)

X=Vox * t X=(25)(3.49) X=87.25m

HORIZONTAL MOTION • A ball rolls over the edge of a table with a velocity of 8 m/s; if it reaches 10 meters from the point where the table ended, a) from what height was launched?

2

Vx=8 m/s X=10m y=? t=?

X=Vox*t x/Vox=t 10/8m/s=t t=1.25sec

y=1/2gt y=(.5)(9.8)(1.25)2 y=7.65m

We need to obtain the time to be able of obtain the height using the y=1/2gt2.

• b) what’s the velocity when the ball is on the To obtain the vertical velocity if we already obtained the ground? time it’s so simple, using the formula: Vy=gt

V= �� - + �� V= 8- + 12.25V=14.63m/s

Vy=gt Vy=(9.8)(1.25) Vy=12.25m/s

If we also want to obtain the angle of the motion, you need the formula: đ?œƒ=tan-1 Vy/Vx

đ?œƒ=tan-1(12.25/8) đ?œƒ=tan-1(1.53) đ?œƒ=56.83Âş

PROJECTILE MOTION

• Projectile motion is a two dimensional concept and it follows the two dimensional kinematics. • A projectile has both the horizontal and the vertical components of motion.

P R O J E C T I L E M OT I O N

TRAJECTORY ---------------PARABOLIC PATH

• If we ignore the effects of air resistance, the horizontal velocity is constant and the vertical velocity changes with a uniform acceleration.

The path that the body follows is a parabola as can be seen from the proof below. • The shape of the trajectory can be found by combining the equations for vertical and horizontal velocity.

• Taking the vertical displacement as y and the horizontal displacement as x, we have.

Vx=Vcosđ?œƒ Vy=Vsinđ?œƒ • It also helps us to obtain another unknown values as like the vertical initial velocity (Vox)

x=Vox+t * The range can be obtained with the formula:

EXAMPLES PROJECTILE MOTION

PARABOLIC MOTION • A baseball player hit a baseball with an angle of 37º, respect to the horizontal and with an initial velocity of 15m/s.

Calculate: • a) its range

X=?

X=Vox+t X=11.97(1.84) X=22.02m

Vox=Vcosđ?œƒ Vox=15cosđ?œƒ Vox=11.97

•b) Its time to flight

Tt=?

Tt=0.92x2 Tt=1.84sec

Voy=Vsenđ?œƒ=15sen37=9.02 Vy=Voy+gt Vy/g=t 9.02/9.8=t t = 0.92 sec

•C) Its height Y=?

2 Y=Voyt+1/2gt Y=9.8(0.922)/2 Y=0.5(9.8)(0.922) Y=4.147m

THANK YOU N ATA L I A D A E N N A G O N Z Á L E Z V I E R A GROUP 220 U N I V E R S I DA D A U T Ó N O M A D E N U E VO LEÓN P R E PA R AT O R I A N O . 9