Norah Ali Al- Moneef 1 Chapter 1 Motion in a straight line 1-2 Displacement vs Distance Average...

Norah Ali Al- Moneef 1

Chapter 1Motion in a straight line

1-2 Displacement vs Distance Average Velocity1-4 Acceleration1.5 finding the motion of an object

Chapter 1Motion in a straight line

• Mechanics = Study of objects in motion.Position (x) – where you are locatedDistance (d ) – how far you have traveled, regardless of

direction Displacement (x) – where you are in relation to where you

started – 2 parts to mechanics.

• Kinematics = Description of HOW objects move.

• Dynamics = WHY objects move.

2Norah Ali Al- Moneef

1-2 Displacement vs DistanceAverage Velocity

• Displacement is a vector that points from an object’s initial position to its final position

• and has a magnitude that equals the shortest distance between the two positions.

_Only depends on the initial and final positions– Independent of actual paths between the initial and

final positions• Distance is a scalar

– Depends on the initial and final positions as well as the actual path between them


Displacement

position initial ox

position final x

ntdisplaceme oxxx

The displacement Δx is a vector that points from the initial position to the final position. SI Unit of Displacement: meter (m) 4Norah Ali Al- Moneef

Comparing Vector & Scalar ValuesDisplacement (a vector) versus distance (a scalar)

LAKE

A

B

We want to get from point A to point B. If we follow the road around the lake our direction is always changing. There is no specific

direction. The distance traveled on the road is a scalar quantity.

A straight line between A and B is the displacement. It has a specificdirection and is therefore a vector.

start

stop


6

Displacement

x (m)

t (s)4

3

-3

Displacement between t=1 s and t=5 s

x = 1.0 m - 2.0 m = -1.0 m

This type of x(t) plot shows the position of an object at any time, e.g.,

Position at t=3 s, x(3) = 1 m

Norah Ali Al- Moneef

Given the train’s initial position and its final position what is the displacement of the train?

What is the distance traveled by the train ?Displacement = f ix x x


Example:A boy travels from D to A,A to B .B to C.C to D

Displacement from D to D ( which are initial and final points ) = 0

Distance traveled = 8 +4+8+4 = 24 m


A student on her way to school walks four blocks south, five blocks west, and another four blocks south, as shown in the diagram. What is the distance she walks ?What is displacement from home to school ?

Distance = 4 blocks + 5 blocks +4 blocks = 13 blocks

Displacement = ( 82 + 5 2 ) ½ = 9.43 blocks


Example:

d5 km

It is easy to add displacements if they are perpendicular to each other.

E.g.

N

4 km

3 km

d 2 = 42 + 32

d = 5 km

= 37

tan = 3

4


Example:

Example :

Distance = 4 m + 3 m =7 m

Displacement = 5 m


= (3 + 4) km = 7 km

total distance

total displacement

= (3 + 4) km north

= 7 km north

N

4 km

7 km north

3 km

A car travels 4 km north then 3 km northA car travels 4 km north then 3 km north


Example:

A car travels 4 km north then 3 km southA car travels 4 km north then 3 km south

4 km

1 km north

3 km

total distance =

total displacement =

= (3 + 4) km = 7 km

= (3 + 4) km north

= 1 km north

N

example


A car travels 5 km north and 4 km east.

(a) Total distance travelled = ?

Total distance = 5 + 4 = 9 km

5 km

4 km

Example N


6.4 km5 km

4 km

(b) What is the displacement?

22 45

displacement = = 6.40 km

54

= 38.7tan =

Total displacement:

6.40 km 38.7° east of north

60o

1 m

example

A ball hung by a string swings from X to Y.

1 mX Y

What is the size of the displacement of the ball?

A /3 m

B 1 m

C 1 m towards the right


A girl cycles a circular track of diameter 70 m and stops at the starting point.(a) Distance travelled = ?

Distance travelled

= perimeter of track

= × 70 = 220 m

Example

70 m


(b)What is her displacement?

displacement = 0 m

Starting from origin, O a person walks 90-m east, then turns around and walks 40-m west.

What is the total walked distance?

What is the displacement?

Example:

130 m

A: 50-m, due east.


SpeedSpeed can be defined in a couple of ways:

How fast something is movingThe distance covered in a certain amount of timeThe rate of change of the position of an object

Units for speed are: miles / hour (mi/hr)kilometers / hour (km/hr)feet / second (ft/s)

This is the standard unit meters / second (m/s)


The speed has no direction and is always expressed as a positive number

Speed and VelocityThe average speed being the distance traveled divided by the time required to cover the distance:

How far does a jogger run in 1.5 hours (5400 s) if his average speed is 2.22 m /s?

Distance = 5400 s x 2.22 m / s = 11988 m



No

yes

Graphing Skills

Calculating Slope The slope of a straight line equals the vertical change divided by the horizontal change. Determine the slope of the blue line shown in the distance vs. time graph.

• Motion can be studied using a distance vs. time graph.– time (x-axis) = independent variable

– distance (y-axis) = dependent variable

• The slope of a distance vs. time graph equals speed.


Graphing Skills, continued1. Choose two points that you will use to

calculate the slope. Point 1: t = 1 s and d = 6 m Point 2: t = 4 s and d = 12 m

2. Calculate the vertical change and the horizontal change. vertical change = 12 m – 6 m = 6 m horizontal change = 4 s – 1 s = 3 s

3. Divide the vertical change by the horizontal change.

slope = 6 m /3 s = 2 m/s


basic formula

v xt

x f xit

Average velocity

Average velocity

•Can be positive or negative•Depends only on initial/final positions•e.g., if you return to original position, average velocity is zero


Speed and velocity are not the same. Velocity requires a directional component and is

therefore a vector quantity.Speed tells us how fast we are going but not which way.

Speed is a scalar (direction doesn’t count!)

example

Answer is d


Graphical Representation of Average Velocity

Between A and D , v is slope of blue line


A man walks from A to B at 1 km h–1, and returns at 2 km h–1.

A B1 km h–1

Example

2 km h–1

what is the average speed for the whole trip ?


Suppose AB = 1 km whole journey = 2 km

Time for whole trip = 11 h km 2

km 1h km 1

km 1

= 1 h + 0.5 h = 1.5 h

Ave. speed = distance / time2/1.5 =1.33 Km / s

A car travels 7 km north and then 3 km west in 10 minutes. Find

C B

A

3 km

7 km

Example

(a) average speed,

Ave. speed =

distance travelled

time taken

=(7 + 3) km

(10/60) h= 60 km h–1


ExampleA car travels 7 km north and then 3 km west in 10 minutes. Find

C B

A

3 km

7 km

(b) ave. velocity?

AC = 22 BCAB 22 37 = 7.62 km

tan = =23.2o3/7


ExampleA car travels 7 km north and then 3 km west in 10 minutes. Find

C B

A

3 km

7 km

AC = 7.62 km, =23.2o

Size of ave. velocity =

= 45.7 km h–1

displacementtime

7.62 km(10/60) h

=

Ave. velocity is 45.7 km h–1, 23.2° north of west.

(b) ave. velocity?



example

What is


example

From A to B

What is

A B


example


example

ExampleCarol starts at a position x(t=0) = 1.5 m.At t=2.0 s, Carol’s position is x(t=2 s)=4.5 mAt t=4.0 s, Carol’s position is x(t=4 s)=-2.5 m

a) What is Carol’s average velocity between t=0 and t=2 s?b) What is Carol’s average velocity between t=2 and t=4 s?c) What is Carol’s average velocity between t=0 and t=4 s?

a) 1.5 m/sb) -3.5 m/sc) -1.0 m/s


basic formula

v xt

x f xit

Instantaneous velocity

Let time interval approach zero

•Defined for every instance in time•Equals average velocity if v = constant•SPEED is absolute value of velocity


The velocity at any instant is called instantaneous velocity.

If a car moves at a constant velocity... its average and instantaneous velocities have the same value.

INSTANTANEOUS VELOCITY


Graphical Representation of Instantaneous Velocity

v(t=3.0) is slope of tangent (green line)


A particle moves along a straight line such that its position is defined by s = (t3 – 3 t2 + 2 ) m.

Determine the velocity of the particle when t = 4 s.

tt 63 2

At t = 4 s,

the velocity = 3 (4)2 – 6(4) = 24 m/s

dt

dxv


example

1 - 4 AccelerationAcceleration: is a rate at which a velocity is changing.

Instantaneous acceleration

= dv / dt = d2 x / d2 t



Acceleration is a change in velocity or a change in direction of velocity.

.

First car is accelerating because its velocity is increasing.Second car is accelerating because its direction is changing.Third car is accelerating because its velocity and direction are changing.

Positive/Negative Acceleration

Positive Acceleration –produces an increase in speed in the positive direction, or decrease of speed in the negative direction.Negative Acceleration – produces a decrease in speed in the positive direction or an increase of speed in the negative direction.


Acceleration (increasing speed) and deceleration (decreasing speed) should not be confused with the directions of velocity and acceleration :


Norah Ali Al- Moneef

Acceleration

There is a difference between negative acceleration and deceleration:

Negative acceleration is acceleration in the negative direction as defined by the coordinate system.

Deceleration occurs when the acceleration is opposite in direction to the velocity.

48

ExampleA car’s velocity at the top of a hill is 10 m/s. Two seconds later it

reaches the bottom of the hill with a velocity of 26 m/s. What is the acceleration of the car?

The car is increasing its velocity by 8 m/s for every second it is moving.



example


tma

tdt

dxdt

dv

t

va

t

2

0

/ 20sec0.210

10

lim

The Ferrari 348 can go from rest to 100 km h–1 in 5.6 s. What is its ave. acceleration (in m s–2)?

Example

Ave. acceleration

(100/3.6) m s–1

5.6 s= = 4.96 m / s2


•A jet starts at rest at the end of a runway and reaches a speed of 80 m/s in 20 s. What is its acceleration?

• a = 80 m/s - 0 m/s = 4 m/s2

20 s

Example


A skateboarder is moving in a straight line at a speed of 3 m/s and comes to a stop in 2 sec. What is his acceleration?

• a = 0 m/s - 3 m/s = -1.5 m/s2

2 m/s

Example


• A particle moves along a straight line such that its position is defined by X= (t3 – 12 t2 + 36 t -20 ) cm. Describe the motion of P during the time interval [0,9]

)4(6246 ttdt

dva

t02469

X-2012-4-2061

v360-12063

a-24-1201230

Total time = 9 secondsTotal distance = (32+32+81)= 145 meterDisplacement = form -20 to 61 = 81 meterAverage Velocity = 81/9= 9 m/s to the rightSpeed = 9 m/sAverage speed = 145/9 = 16.1 m/sAverage acceleration = 27/9= 3 m/s2 to the right


Example


Example

25 m


Example

1. Velocity & acceleration are both vectors. Are the velocity and the acceleration

always in the same direction?

NO!!

If the object is slowing down, the acceleration vector is in the opposite direction of the velocity vector!


2. Velocity & acceleration are vectors. Is it possible for an object to have a zero

acceleration and a non-zero velocity?

YES!!

If the object is moving at a constant velocity, the acceleration vector is zero!


• How long does it take a car going 30 m/sec to stop of it decelerates at 7 m/sec2?


example

sec3.4/7

/3002

sm

sm

aviv

t

atvv

at

f

if

if vv

Examples :1 ) What is the acceleration of a car that increased its speed from 10 m/s to 30 m/s in 4 seconds? a = (30 m/s – 10 m/s) ÷ 4s

= 20 m/s ÷ 4s= 5 m/s2

2)the same car now slows down back to 10 m/s in 5 seconds. What is his acceleration? a = (10 m/s – 30 m/s) ÷ 5s

= (- 20 m/s) ÷ 5s= - 4 m/s2

Means slowing down


deduce from the shape of a speed-time graph when a body is:(i) at rest(ii) moving with uniform speed(iii) moving with uniform acceleration(iv) moving with non-uniform acceleration

example


Velocity

(i)Time

(ii)

(iii) (iv)

(i) at rest(ii) moving with uniform speed(iii) moving with uniform acceleration(iv) moving with non-uniform acceleration

Equations of Kinematics for Constant Acceleration

1.5 finding the motion of an object


x v0 (v0 at)

2t

x v0t 1

2at2

Substitute to eliminate vf

v f v0 at

x 1

2(v0 v f )t

v f v0 at

x 1

2(v0 v f )t

x (v f at) v f

2t

x v f t 1

2at2

Substitute to eliminate v0


Other Forms of Eq.s of Motion

x (v0 v f )

2

(v f v0 )

a

ax v f

2

2v0

2

2

Substitute to eliminate t

v f v0 at

x 1

2(v0 v f )t


Final List of 1-d Equations

Which one should I use?Each Eq. has 4 of the 5 variables: x, t, v0, v & aAsk yourself “Which variable am I not given and not interested in?” If that variable is t, use Eq. (5).

basic equations:

1) v v0 at

2) x 1

2(v0 v)t

3) x v0t 1

2at2

4) x v f t 1

2at2

5) ax v f2

2v0

2

2


Example• A car is traveling with an initial velocity v0. At t = 0, the driver puts on the

brakes, which slows the car at a rate of ab. At what time tf does the car stop, and how much farther xf does it travel?

x = xf , t = tf

v = 0

x = 0, t = 0ab

v0

• Above, we derived: v = v0 + at• Realize that a = -ab

• Also realizing that v = 0 at t = tf :find 0 = v0 - ab tf or

tf = v0 /ab70Norah Ali Al- Moneef

• To find stopping distance we use:

• In this case v = vf = 0, x0 = 0 and x = xf

fb2

0 x)a(2v b

20

f a2v

x

)x2a(xvv 02

02


• So we found that • Suppose that vo = 29 m/s

• Suppose also that ab = g = 9.81 m/s2

– Find that tf = 3 s and xf = 43 m

- A car starting from rest attains a speed of 28 m/sec in 20 sec. Find the acceleration of the car and the distance it travels in this time.


Example

m 280sec 20 0/ 282

1 t

2

1

/14sec20

0/28 2

smvvx

smsm

tvv

a

atvv

if

if

if

Example – be careful with signsA car moves from a position of +4.0 m to a position of –1.0 m in 2.0 sec. The initial velocity of the car is –4.0 m/s and the final velocity is –1 m/s.

(a) What is the displacement of the car?(b) What is the average velocity of the car?(c) What is the average acceleration of the car?

Answer: (a) x = xf – xi = –1.0 m – (+4.0 m) = – 5 m (b) vav = x/t = (– 5.0 m)/(2.0 s) = – 2.5 m/s (c) 2m/s5.1

s2

)/m4(/m1

ss

tt

vv

Δt

Δva

if

ifav

deceleration!73Norah Ali Al- Moneef


ecss

t

a

vvΔt

Δt

vv

Δt

Δva

if

ifav

s6.14961.0

)/m0(/4m1

C-


Example

Velocity /m/s

(i)

(ii)

(iii)

Time/s0 10 14 23

A bus stopped at a bus-stop for 10 seconds before accelerating to a velocity of 15 m/s in 4 seconds and then at a constant speed for the next 9 seconds. How does the graph look like?

15

How far did the bus go in this 23 seconds?


Example

• Distance travelled in first 10 seconds is zero!• Distance travelled in the next 4 seconds is = ½ x 4 x 15 = 30 m• Distance travelled in the final 9 seconds is = 9 x 15 = 135 m• Total distance travelled = 165 m


DecelerationDeceleration

• refers to decreasing speed

• is not the same as negative acceleration

• occurs when velocity and acceleration have opposite signs

Example: A ball thrown up in the air. The velocity is upward but the acceleration is downward. The ball is slowing down as it moves upward. (Once the ball reaches its highest point and starts to fall again, it is no longer decelerating.)

If up is our convention for positive, then both when the ball is rising and falling, the acceleration is negative (during the instant of bounce, the acceleration is positive).


Freely Falling Objects

• A freely falling object is any object moving freely under the influence of gravity alone.

• It does not depend upon the initial motion of the object– Dropped – released from rest– Thrown downward– Thrown upward


1.6 the acceleration of gravity and falling objects

Objects thrown straight up

The acceleration of gravity (g) for objects in free fall at the earth's surface is 9.8 m/s2. ( down ward )

The acceleration of a falling object is due to the force of gravity between the object and the earth.

Galileo showed that falling objects accelerate equally, neglecting air resistance.

Galileo found that all things fall at the same rate.

On the surface of the earth, in a vacuum, all objects accelerate towards the surface of the earth at 9.8 m/s2.

g actually changes as we move to higher altitudes

1.6 the acceleration of gravity and falling objects


a = - g


Acceleration of Freely Falling Object• The acceleration of an object in free fall is directed

downward, regardless of the initial motion• The magnitude of free fall acceleration is g = 9.80 m/s2

– g decreases with increasing altitude– g varies with latitude– 9.80 m/s2 is the average at the Earth’s surface– The italicized g will be used for the acceleration due to gravity

• Not to be confused with g for grams


: The acceleration of gravity is the same for all objects near the surface of the Earth, regardless of mass

What about the mass of an object?

Acceleration of Free Fall• We will neglect air resistance• Free fall motion is constantly accelerated motion in one dimension• Let upward be positive• Use the kinematic equations with ay = -g = -9.80 m/s2


Free Fall – an object dropped• Initial velocity is zero• Let up be positive• Use the kinematic equations

– Generally use y instead of x since vertical• Acceleration is

ay = -g = -9.80 m/s2

vo= 0

a = -g


Free Fall – an object thrown downward• ay = -g = -9.80 m/s2

• Initial velocity 0– With upward being positive, initial

velocity will be negative

vo≠ 0

a = -g

Free Fall -- object thrown upward

• Initial velocity is upward, so positive• The instantaneous velocity at the

maximum height is zero• ay = -g = -9.80 m/s2 everywhere in

the motion

v = 0

vo≠ 0

a = -g


• The motion may be symmetrical– Then tup = tdown

– Then v = - vo

• The motion may not be symmetrical– Break the motion into various parts

• Generally up and down

v v0 gt

y v0t 1

2gt 2

v 2 v02 2gy

v v0 at

x v0t 1

2at 2

v 2 v02 2ax

Equations of Kinematics for Constant Acceleration

For free fall



example

A ball is dropped from a tall building and strikes the ground 4 seconds later. A ) what velocity does it strike the ground B ) what distance does it fall?


example

a=-g = -


example


example

How high can a human throw a ball if he can throw it with initial velocity 90 m / h?.


example


example

ExampleA man throws a brick upward from the top of a 50 m building. The brick has an initial upward velocity of 20 m/s.

a) How high above the building does the brick get before it falls?

b) How much time does the brick spend going upwards?

c) What is the velocity of the brick when it passes the man going downwards?

d) What is the velocity of the brick when it hits the ground?

e) At what time does the brick hit the ground?a) 20.4 mb) 2.04 sc) -20 m/sd) -37.2 m/se) 5.83 s



example

• A marble is dropped from a bridge and strikes the water in 5 seconds. Calculate the speed with which it strikes and the height of the bridge.

• (Vf = 49 m/s, Δ y = 122.5 m)

example

smv

ssmv

gtv

f

f

f

/49

)5)(/8.9( 2

my

ssmy

gty

5.122

)5)(/8.9(5.0

5.022

2


example• The observation deck of the World Trade Center is 420 m

above the street. Determine the time required for a penny to free fall from the deck to the street below.

25.0 gty ? /8.9 420 /0 2 tsmgmysmv

Given

i

sst

ts

tsm

m

tsmm

tsmm

26.97.85

7.85

/9.4

420

)/9.4(420

)/8.9(5.0420

2

22

22

22

22


example• With what speed must an object be thrown to reach

a height of 91.5 m? Assume negligible air resistance.

? ? 5.91 / /8.9

:2 tvmysomvsmg

Given

if

smsmv

smv

smv

msmvsm

gdvv

i

i

i

i

if

/3.42/1793

1793

/17930

)5.91)(/8.9(2/0

2

222

222

222

22


example• A 10kg block being held at rest above the ground is

released. The block begins to fall under only the effect of gravity. At the instant that the block is 2.0 meters above the ground, the speed of the block is 2.5m/sec. The block was initially released at a height of how many meters.

smgkgmsmvmyv /8.9 10 /5.2 2 00

my

mssm

smy

yg

vy

yyg

v

yygv

v

yygvv

3.2

2//8.9

)/5.2(5.0

5.0

2

)(2

0

)(2

0

2

0

2

0

0

2

02

0

020

2


examplea-How long does it take a ball to fall from a roof to the ground 7.0 m below? b - With what speed does it strike the ground?

Now we can see we need a relationship between ∆ y, a, vi, and t∆ y = vit + (0.5)at2 (7.0 m) = (0)t + (0.5)(9.8)t2 t = 1.20 s b. We need a relationship between ∆ y , a, vi, and vf

vf

2 = vi2 + 2a ∆ y

vf

2 = 02 + 2(9.8)(7.0) vf = 11.7 m/s


example2. How long does it take a ball to reach the ground 7.0 m below, if it is thrown straight up with an initial speed of 2.00 m/s?

Now we can see we need a relationship between ∆ y , a, vi, and t∆ y = vit + (0.5)at2

(7.0 m) = (-2.00)t + (0.5)(9.8)t2

t = {1.42, -1.01}

Since t < 0 has no meaning,

t = 1.42 s


example

1. How fast should you throw a ball straight down from 40 m up so that its impact speed would be the same as a mango’s dropped from 60 m?

2. A dune buggy accelerates uniformly at 1.5 m/s2 from rest to 22 m/s. Then the brakes are applied and it stops 2.5 s later. Find the total distance traveled.

19.8 m/s

188.83 m

Answer:

Answer:107Norah Ali Al- Moneef

A rubber chicken is launched straight up at speed v from ground level. Find each of the following if the launch speed is tripled (in terms of any constants and v).

a. max heightb. hang time c. impact speed 3 v

9 v 2 / 2 g

6 v / g

Answers

example


Sample Problems1. You’re riding a unicorn at 25 m/s and come to a

uniform stop at a red light 20 m away. What’s your acceleration? (15.62m/s2)

2. A brick is dropped from 100 m up. Find its impact velocity and air time. ( V= -44.27m/s)

3. An arrow is shot straight up from a pit 12 m below ground at 38 m/s.

a. Find its max height above ground. (73.7m)b. At what times is it at ground level? (1.94s)


A model rocket is launched straight upward with an initial speed of 50.0 m/s. It accelerates with a constant upward acceleration of 2.00 m/s2 until its engines stop at an altitude of 150 m. (a) What can you say about the motion of the rocket

after its engines stop? (b) What is the maximum height reached by the rocket? (c) How long after lift-off does the rocket reach its maximum height? (d) How long is the rocket in the air?

a) What can be said about the motion of the rocket after theengines stop?

The rocket will continue upward, but start to decelerate due to the Earth’s gravitational field until the upward velocity reaches zero. The rocket then begins to fall back to the ground with an acceleration equal to the Earth’s surface gravity


b ) the max height is

Step ( 1)

=y 2


Step ( 2 )


d) How long is the rocket in the air?


• If the direction of motion is upward, then y is positive; if the direction of motion is downward, then y is negative.

• For an object thrown upward:– Final velocity is 0 m/s (at the highest point).– Acceleration: g = 9.8 m/s2 downward.– Initial velocity is greater than 0 m/s.– Maximum height will be equal to y.

• For an object that is falling downward:– Initial velocity is 0 m/s if the object is dropped.– Initial velocity is greater than 0 m/s and is negative if the object is thrown downward.– Acceleration: g = 9.8 m/s2 downward.– Distance object falls is negative and is equal to y.– The final velocity when the object first contacts the ground will be negative and will be greater

than 0 m/s. Vf is the velocity the object has as it first makes contact with the surface and it is NOT 0 m/s. The object will eventually slow down from vf to 0 m/s as it interacts with the surface, but the object first strikes the surface with velocity vf.

• Velocities in the upward direction are positive.• Velocities in the downward direction are negative.


Summary about free fall

• For an object that is falling downward, the y is negative. When using the vf2 =

vi2 – (2g·y) equation, the final velocity will be positive because you cannot take the square

root of a negative number. You will have to add the negative sign to show that the object is moving downward.

• For an object that is thrown upward from a point and then lands at that point:– vi = - vf time up = time down– Total time = time up + time down– Distance up = distance down

– All of this is true because the object slows down (deceleration) on the way up at the same rate as it speeds up on the way down. Both the deceleration and the acceleration are constant and due to the force of gravity pulling on the object.

• For Problems Involving Ascending (Rising) or Descending (Moving Downward) Objects:– These types of problems have an item being released or falling off of an object (I will

use a hot air balloon as an example) that is either rising or coming down with some speed. The key to solving these problems is to realize that the balloon and any object in the balloon or attached to the balloon is traveling at the same speed at the balloon.


For objects that are rising: – If a balloon is rising at 10 m/s and an object is released from the balloon. The

released object will have an upward velocity of 10 m/s and will be decelerated by gravity as it travels upward. The final velocity on the way up is 0 m/s. The upward distance y traveled by the object will be positive.

– Both the upward distance y and the time up can be determined from this information.

2f

i

m/s9.8g m/s 0 v

velocityballoonv

– After the object reaches its highest point, gravity will begin to accelerate the object down toward the ground. The initial velocity is 0 m/s and the distance the object falls will be the distance above the ground at which the object left the balloon plus the upward distance y traveled by the object to the highest point. The y down will be negative to indicate that the motion is downward.

upΔyreleaseofheightdownΔy

m/s9.8gm/s0v 2i

From this information, you can determine the time down and the velocity at which the object will strike the surface (vf). The final velocity should be negative to indicate that the direction of motion is downward.

The total time in the air will be equal to the time up + the time down. The maximum height will be equal to the height of release + y up.

•For objects that are descending (moving downwards) when the object is released:

If a balloon is descending at 10 m/s and an object is released from the balloon. The released object will have a downward velocity of -10 m/s and will be accelerated by gravity as it travels downward. The initial velocity vi and the y down will be negative to indicate that the motion is downward.

From this information, you can determine the time down and the velocity at which the object will strike the surface (vf). The final velocity should be negative to indicate that the direction of motion is downward.


•For objects that are hovering (staying on one place; not rising or moving downward):

–The initial velocity will be zero and you can solve this problem as you would for an object that is dropped downward. The y down will be negative to indicate that the motion is downward.

–From this information, you can determine the time down and the velocity at which the object will strike the surface (vf). The final velocity should be negative to indicate that the direction of motion is downward.

releaseofheightdownΔy

m/s9.8gm/s0v 2i

Speed, Velocity, & Acceleration

• Speed (v) – how fast you go• Velocity (v) – how fast and which way;

the rate at which position changes• Average speed ( v ) – distance / time• Acceleration (a) – how fast you speed

up, slow down, or change direction; the rate at which velocity changes


Speed vs. Velocity

• Speed is a scalar (how fast something is moving regardless of its direction). Ex: v = 20 mph

• Speed is the magnitude of velocity.• Velocity is a combination of speed and direction.

Ex: v = 20 mph at 15 south of west

• The symbol for speed is v.• The symbol for velocity is type written in bold: v or

hand written with an arrow: v


Notice in free fallNotice in free fall


Norah Ali Al- Moneef 1 Chapter 1 Motion in a straight line 1-2 Displacement vs Distance Average...

Documents

Transcript of Norah Ali Al- Moneef 1 Chapter 1 Motion in a straight line 1-2 Displacement vs Distance Average...