a summary textbook for physics of classical

A SUMMARY TEXTBOOK FOR PHYSICS OF CLASSICAL MECHANICS 101

Written by Waleed Sh. Abu Khader Reviewed by Saleem Abdo

University of Jordan School of Engineering First Edition 2015

Contents Chapter 2 : Motion in One Dimension 2.1 Distance and Displacement

2

2.2 Velocity

3

2.3 Acceleration

4

2.4 Time Relationships

5

2.5 Motion Description

14

2.6 Constant Acceleration

15

Problems

17

Chapter Two : Motion in One Dimension

2.1 Distance and Displacement Displacement Magnitude Direction

Distance Magnitude No direction

= 𝒙𝒇 − 𝒙𝒊 or 𝒚𝒇 − 𝒚𝒊

∑ 𝒅 = 𝒅𝟏 + 𝒅𝟐 + 𝒅𝟑 + ⋯ + 𝒅𝒏

∆𝒙 𝒐𝒓∆𝒚

𝒅

Example 2.1.1 : A car moves in the shown pattern calculate the distance and displacement when it moves from : 𝑅 +𝑦 a) A to B . b) A to the center to B back to A moving in a straight line .

𝐵

+𝑥

𝐴

a) Distance = 𝟐𝝅𝑹⁄𝟐 ‫نصف محيط الدائرة‬ Displacement = −𝟐𝑹 = 𝟐𝑹 (−𝒊̂ ) b) Distance = 𝟐(𝟐𝑹) = 𝟒𝑹 Displacement = 𝒛𝒆𝒓𝒐

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Remarks : -

When an object moves in a straight line and in one direction only , then this object has the same displacement and distance . ∆𝒙 = ∑ 𝒅 only in this case .

-

In general : ( ∆𝒙𝑨𝑩 = 𝒙𝒇 − 𝒙𝒊 → 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕 ) and ( ∑ 𝒅 = 𝒅𝟏 + 𝒅𝟐 → 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 ) as for the figure shown at the bottom : 𝑑2 𝑥𝑓

∆𝑥

𝑑1

𝑥𝑖

𝐴

𝐶

𝐵

1.2

Velocity

Average velocity : The total displacement moved by an object divided by the time passed to move this displacement . A quantity that has a magnitude and direction . ̅= 𝝑 -

∆𝒙 ∆𝒕

If the term is positive this indicates that the displacement is > 𝟎 . If the term is negative this indicates that the displacement is < 𝟎 .

Examlple 2.2.1 : A book has moved from point ( 𝑨 ) to point ( 𝑩 ) as shown in the figure during a period of time ( 𝟐. 𝟑 𝒔 ) , determine the average velocity of the book . +𝑦

⃗𝝑 =

1.8 𝑚 𝐴

𝐵

+𝑥

=

∆𝒙 ∆𝒕

−𝟏. 𝟖 𝟐. 𝟑

= −𝟎. 𝟕𝟖 𝒎/𝒔

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The negative sign indicates that the direction of the velocity is directed to the left ( in the direction of the negative x-axis ) as shown in the figure .

Example 2.2.2 : A car ( 𝑿 ) moves with ( − 𝟐𝟓 𝒎/𝒔 ) and another car ( 𝒀 ) moves with ( 𝟐𝟓 𝒎/𝒔 ) , explain whether both cars have the same velocity or not . 𝑨𝒏𝒔 : Both cars have different velocities , despite of the magnitude of their average velocities they have different directions . Furthermore we can say that they have the same speed which has the same magnitude .

2.3 Acceleration ⃗ = 𝒂

∆𝝑 ∆𝒕

Acceleration basically depends on the change of velocity . 𝒎 = 𝒔 = 𝒎/𝒔𝟐 𝒔

-

If the term is positive this indicates that the change in velocity is > 𝟎 . If the term is negative this indicates that the change in velocity is < 𝟎 . If the term equals zero this means that there is no change in velocity , in other words [ 𝝑𝒇 = 𝝑𝒊 ] .

Example 2.3.1 : A particle moves in the ( 𝒙𝒚 ) plane as the following path shows : from ( −𝟏, 𝟎 ) to ( 𝟓, 𝟎 ) then back to the origin , in a straight line , determine : I. II. III.

The total distance The displacement The average velocity

𝑑1 3 𝑠𝑒𝑐 −1

0

+𝑥 5 𝑠𝑒𝑐

5

𝑑2

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l. ∑ 𝒅 = 𝒅𝟏 + 𝒅𝟐 𝟔 + 𝟓 = 𝟏𝟏 𝒖𝒏𝒊𝒕𝒔 ll. ∆𝒙 = 𝒙𝒇 − 𝒙𝒇 𝟎 − (−𝟏) = +𝟏 𝒖𝒏𝒊𝒕 The positive sign indicates that the displacement is in the direction of the positive x-axis . ̅= lll. 𝝑

∆𝒙 ∆𝒕

From ( −𝟏 𝒕𝒐 𝟓 ) : 𝝑𝟏 =

∆𝒙𝟏 ∆𝒕𝟏

= From ( 𝟓 𝒕𝒐 𝒕𝒉𝒆 𝒐𝒓𝒊𝒈𝒊𝒏 ) : 𝝑𝟐 =

𝟓 − (−𝟏) = 𝟐 𝒖𝒏𝒊𝒕/𝒔𝒆𝒄 𝟑

∆𝒙𝟐 ∆𝒕𝟐

=

𝟎−𝟓 = −𝟏 𝒖𝒏𝒊𝒕/𝒔𝒆𝒄 𝟓

If asked to calculate the average speed : |𝒗 ̅| = =

∑𝒅 ∆𝒕

𝟏𝟏 = 𝟏. 𝟒 𝒖𝒏𝒊𝒕/𝒔𝒆𝒄 𝟑+𝟓

2.4 Time relationships In this section you will learn how to deal with graphs during a specific time interval . -

Position – Time Velocity – Time Acceleration – Time

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Starting with the *position time relationship plot . ‫بشكل عام عند التعامل مع الرسوم البيانية يجب النظر الى ما تمثله المحاور بعد ذلك عليك تحديد ميل المنحنى عند أي نقطة‬ .‫على الخط المستقيم و ما يمثله الميل من كمية فيزيائية‬

∆𝑥

Case #1 ) You can see that at ( 𝒕 = 𝟎 ) the moving object was at position ( 𝒙 = 𝟎 ) , which indicates that the object started from rest when it was at ( 𝒙 = 𝟎 ) .

∆𝑥∆𝑥

Now looking at the slope ∆𝒙⁄∆𝒕 And that physical quantity has been mentioned in section 2 in this chapter . furthermore if you take a tangent ( ‫ ) مماس‬at any point on the line and then calculate the slope of this tangent, you will see that at any point you take it would have the same slope which means same velocity . This condition is called a constant motion that has an acceleration equals to zero because the velocity is constant and does not change with respect to time , this will be explained more in 2.5 . Basically in a linear relationship the slope is constant to explain this phenomena look at the following mathematical derivation : 𝑦 = 𝑎𝑥 + 𝑏 ‫معادلة الخط المستقيم‬ a : represent the slope b: represent the y-intercept Assume that at ( 𝒕 = 𝟎 ) the y-intercept equals zero ( 𝒙 = 𝟎 ) ̅. 𝒕 + 𝟎 𝒙=𝝑 ̅= 𝝑

∆𝒙 ∆𝒕

𝑥(𝑚)

Example 2.4.1 : which of the following is true about each graph : I.

 𝝑𝑨 > 𝝑𝑩 > 𝝑𝑪  𝝑𝑨 < 𝝑𝑩 < 𝝑𝑪  𝝑𝑨 = 𝝑𝑩 = 𝝑𝑪

𝐵 𝐶 𝑡 (sec)

𝑨𝒏𝒔 ∶ Since each point has the same slope and happens to be located exactly on the line then they have the same velocity [ 𝝑𝑨 = 𝝑𝑩 = 𝝑𝑪 ] .

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𝐴

 𝝑𝑨 = 𝝑𝑩 = 𝝑𝑪  𝝑𝑨 > 𝝑𝑩 > 𝝑𝑪

𝐴

𝑥 (𝑚 )

ll.

𝐶 𝐵

 𝝑𝑩 > 𝝑𝑨 > 𝝑𝑪

𝑡 (sec)

 𝝑𝑪 > 𝝑𝑨 > 𝝑𝑩

𝑨𝒏𝒔 ∶ Its clear that for each line has its own slope so its now time to compare each slope with the other , looking at each axis we can see that the slope here represent ( ∆𝒙⁄∆𝒕 ) ≡ 𝝑 . shown in the figure that slope ( C ) has the largest slope then comes ( A ) and finally ( B ) . [ 𝝑𝑪 > 𝝑𝑨 > 𝝑𝑩 ]

Case #2 ) For a non-linear relationship the changes at each point on the curve , since each point has its different slope as shown in the figure we get different in every time period .

slope

velocities

This leads us to a new definition which is called instantaneous velocity : At every certain time ( 𝒕 ) we get a certain velocity ( 𝝑𝒕 ) according to the slope . Mathematically : For the function shown in the figure ( 𝒙(𝒕) ) , as we take a tangent at ( 𝒕 ) we as will are deriving ( ‫ ) نشتق‬the function of ( 𝒙(𝒕) ) : (𝒅𝒙⁄𝒅𝒕)|

𝒕=𝒕𝟏

As you can see here if we replace ( 𝒅 ) by ( ∆ ) we get ( ∆𝒙/∆𝒕 ) which is the velocity at the mentioned time ( 𝒕𝟏 ) .

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Remarks : -

The slope of the tangent represents here ( 𝝑(𝒕) ) : instantaneous velocity ∆𝒙

[ 𝝑(𝒕) = 𝐥𝐢𝐦 ∆𝒕 ]. 𝒕→𝟎

-

̅ 𝑨𝑪 = If you take the average velocity in consideration here ( 𝝑

∆𝒙 ∆𝒕

=

𝒙𝑪 −𝒙𝑨 𝒕𝑪 −𝒕𝑨

). The

average velocity is always calculated in a time period since the position – time relationship is not linear , the slope of the intersection of the curve between two points is the average velocity .

Example 2.4.2 : An object moves according to the graph shown ( position – time ) relationship . Determine the following : 5

ll. The instantaneous velocity at ( 𝒕 = 𝟑 𝒔𝒆𝒄 )

4

x(m)

l. The instantaneous velocity at ( 𝒕 = 𝟐 𝒔𝒆𝒄 )

3 2 1

lll. The average velocity between ( 𝒕 = 𝟏 𝒔𝒆𝒄 𝒂𝒏𝒅 𝒕 = 𝟒 𝒔𝒆𝒄 ) .

0 0

2

4

6

8

6

8

t ( sec )

5

x(m)

4

l. Graphically :

3 2 1

Draw a tangent at ( 𝒕 = 𝟐 𝒔𝒆𝒄 )

0

The slope of the tangent represents the instantaneous velocity at 𝒕 = 𝟐.

0

2

4

t ( sec )

I prefer the mathematical way for the second degree equation which can be easily obtained from the graph . : The general formula for the second degree equation is : 𝒚 = 𝒂 (𝒙 − 𝒉)𝟐 + 𝒌 Where ( 𝒉, 𝒌 ) represents the head coordinates of the parabola . From the graph you can see that ( 𝒉, 𝒌 ) = ( 𝟑, 𝟒 ) . Take any point on the parabola [ 0,0 ] for [ x,y ]

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And find the variable [ a ] . the equation is therefore : 𝒚=−

𝟒 𝟒 (𝒙 − 𝟑)𝟐 + 𝟒 →→ 𝑿 = − (𝒕 − 𝟑)𝟐 + 𝟒 𝟗 𝟗

As you derive this you get ( 𝒗(𝒕) ) So (𝒅𝒙⁄𝒅𝒕)|

𝒕=𝟐

𝟒

= − 𝟗 × 𝟐(𝒕 − 𝟑)(𝟏) → 𝒕 = 𝟐 𝝑(𝟐) =

𝟖 𝒎/𝒔 𝟗

ll. The same as the last question but if you draw a tangent at the head of the parabola the slope will be zero therefore the velocity is zero at this moment . lll. Draw an intersect between these two points as show in the figure : The slope of the intersect gives the average

5 4

try it your self .

x(m)

Velocity .

3 2 1 0 0

2

4

6

t ( sec )

Example 2.4.3 : A particle moves according to this equation : [ 𝒙(𝒕) = 𝒕𝟐 − 𝟐𝒕 ] . l. Find the average velocity from [ 𝒕 = 𝟑 𝒕𝒐 𝒕 = 𝟓 ] seconds ll. Find the general equation of the velocity ( 𝝑(𝒕) ) . lll. Find the instantaneous velocity at ( 𝒕 = 𝟐 𝒔𝒆𝒄 ) .

̅= l. 𝝑

𝒙𝟏 −𝒙𝟐 𝒕𝟏 −𝒕𝟐

𝒙(𝟑) = 𝟑𝟐 − 𝟐 × 𝟑 =𝟑𝒎 𝒙(𝟓) = 𝟓𝟐 − 𝟐 × 𝟓 = 𝟏𝟓 𝒎

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8

̅= 𝝑

𝒙(𝟓) − 𝒙(𝟑) 𝟓−𝟑 = 𝟔 𝒎/𝒔

𝒅𝒙

ll. 𝒅𝒕 = 𝝑(𝒕) 𝝑(𝒕) = 𝟑𝒕 − 𝟐 lll. 𝝑(𝟐) 𝝑(𝟐) = 𝟑(𝟐) − 𝟐 = 𝟒 𝒎/𝒔

Remark : -

Notice from example 2.4.3 that it is not necessary that at 𝒕 = 𝟐 𝒔𝒆𝒄 the instanteouse velocity would equal the average velocity for ( ∆𝒕 = 𝟐 ) .

𝜗 (𝑚/𝑠

*Velocity – Time plot : -

𝜗′

At ( 𝒕 = 𝟎 ) is called the initial velocity . If 𝝑𝒊 = 𝟎 the moving particle is initially at rest . The slope of the line represents the acceleration . The area under the line represents the 𝜗𝑖 displacement . 𝑡𝑜 Slope = ∆𝝑⁄∆𝒕 = 𝒅𝝑⁄𝒅𝒕 = 𝒂(𝒕) . 𝒕′ Area = ∫𝒕 𝝑 𝒅𝒕 = 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕 𝒐𝒄𝒄𝒖𝒓𝒆𝒅 𝒅𝒖𝒓𝒊𝒏𝒈 [ 𝒕𝒐 − 𝒕′ ] .

‫وليد شاهر أبوخضر‬

𝑡′

𝒐

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𝑡 (𝑠)

Example 2.4.4 : A particle is moving according to this equation : 𝝑(𝒕) = 𝟑𝒕𝟐 − 𝟏𝟐 Answer the following : I. II. III.

I.

Find The general equation of the acceleration . Assuming that the particle started moving from ( 𝒙 = 𝟎. 𝟑 ) , calculate the displacement after ( 𝟐 𝒎𝒊𝒏 ) . Determine when does the object changes its moving direction .

⃗ (𝒕) = 𝒅𝝑⁄𝒅𝒕 𝒂 = 𝟑(𝟐)𝒕 𝒂(𝒕) = 𝟔𝒕

II.

𝒕

𝟐 ∫𝒕 𝝑 𝒅𝒕 = ∆𝒙 𝟏

𝒕𝟑 − 𝟏𝟐 𝒕 + 𝒄 𝟑 𝒕𝒐 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 (𝒄) 𝒂𝒕 𝒕 = 𝟎 → 𝒙 = 𝟎. 𝟑 =𝟑

𝒙(𝟎) = 𝟎 − 𝟏𝟐(𝟎) + 𝒄 = 𝟎. 𝟑 𝒄 = 𝟎. 𝟑 The general equation of the displacement : 𝒙(𝒕) = 𝒕𝟑 − 𝟏𝟐𝒕 + 𝟎. 𝟑 To calculate the displacement after 2 minutes you have two options : Method #1 : 𝟐

∆𝒙 = ∫ 𝟑𝒕𝟐 − 𝟏𝟐 𝒅𝒙 𝟎

= −𝟏𝟔 𝒎

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Method #2 : ∆𝒙 = 𝒙𝒇 − 𝒙𝒊 𝒙𝒊 𝒊𝒔 𝒂𝒕 𝒕 = 𝟎 → 𝟎. 𝟑 𝒎 𝒙𝒇 𝒊𝒔 𝒂𝒕 𝒕 = 𝟐 → 𝒙(𝟐) = 𝟐𝟑 − 𝟏𝟐(𝟐) + 𝟎. 𝟑 → −𝟏𝟓. 𝟕𝒎

∆𝒙 = −𝟏𝟓. 𝟕 − 𝟎. 𝟑 = −𝟏𝟔 𝒎

To be able to change the direction of the particle it should stop first then move in other direction .

III.

𝝑(𝒕) = 𝟎 𝝑(𝒕) = 𝟑𝒕𝟐 − 𝟏𝟐 = 𝟎 𝒕 = 𝟑 𝒎𝒊𝒏

*Acceleration – Time plot : -

At any (𝒕) you can determine the acceleration . The area under the line represents the velocity . If the acceleration is constant that means it does not depend on time .

-

Area = ∫𝒕 𝒂 𝒅𝒕 = 𝒂𝒗𝒆𝒓𝒂𝒈𝒆 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒐𝒄𝒄𝒖𝒓𝒆𝒅 𝒇𝒓𝒐𝒎 [ 𝒕𝒐 − 𝒕′ ] .

‫وليد شاهر أبوخضر‬

𝒕′ 𝒐

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The previous graphs gives you an introduction to the next section , which shows you that when you derive the function of the graph you get the next physical term.

Example 2.4.5 : The acceleration – time graph for a car starting from rest , is given below . Determine the following : I. II. III.

I.

II.

III.

Draw the velocity – time graph . Calculate the speed of the car at ( 𝒕 = 𝟑 𝒔𝒆𝒄 ) . Calculate the displacement during the first two seconds .

Calculating the area for each interval which represents the velocity :

From the velocity time graph its clear that at 𝒕 = 𝟑 the velocity equals 𝟒 𝒎/𝒔 𝝑(𝟑) = 𝟒 𝒎/𝒔 The area under the velocity –time graph represents the displacement .

𝑨𝒓𝒆𝒂|𝒇𝒓𝒐𝒎 (𝟎−𝟐) = =

𝟏 𝒃𝒉 𝟐

𝟏 ×𝟐×𝟒 𝟐

∆𝒙 = 𝟒 𝒎

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Remarks : 𝒅⁄ 𝒅𝒕

-

𝒙→

-

𝒂→

𝒅⁄ 𝒅𝒕

𝝑→

∫ 𝒂 𝒅𝒕

𝒂

∫ 𝒗 𝒅𝒕

𝝑→

𝒙

1.5

Motion Description

To describe the motion for the velocity – time graph , you should determine : -

‫إشارة السرعة بحيث اذا كانت فوق محور الزمن تكون موجبة و تحته تكون سالبة‬ ‫ اذا كان ميل الخط موجب تكون إشارة التسارع موجبة اذا كان‬، ‫إشارة التسارع و تحدد من خالل ميل المستقيم‬ ‫الميل سالب تكون إشارة التسارع سالبة‬

Example 2.5.1 :

Time interval

Velocity

Acceleration

Motion

𝟎 − 𝟏𝟓

Positive

Positive

Accelerating

𝟏𝟓 − 𝟑𝟎

Positive

Zero

Constant

𝟑𝟎 − 𝟓𝟎

Positive

Negative

Deceleration

‫وليد شاهر أبوخضر‬

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12 10

velocity (m/min )

8 6 4 2 0 0

2

4

6

8

10

12

-2 -4 -6

Time ( min )

Remarks : -

If the velocity and the acceleration of the moving object has the same sign then the motion of the object is accelerating

-

If the velocity and the acceleration of the moving object has different signs then the motion of the object is decelerating

-

It is a common mistake that most of the students determine the motion of the object according to the sign of the acceleration only . It does not mean that if the sign of the acceleration is negative that the motion is decelerating .

1.6 1 2 3 4

‫وليد شاهر أبوخضر‬

Constant acceleration 𝝑𝒇 = 𝝑𝒊 + 𝒂 𝒕 𝝑𝒇 = 𝝑𝒊 𝟐 + 𝟐 𝒂 ∆𝒙 𝟏 ∆𝒙 = 𝝑𝒊 𝒕 + 𝒂 𝒕𝟐 𝟐 𝝑𝒊 + 𝝑𝒇 ∆𝒙 = [ ] ∆𝒕 𝟐 𝟐

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These equation are the basics of motion in one dimension with constant acceleration , the idea here to solving problems is that you simply determine what is given in the problem and what is required , to choose the equation to substitute in , the following examples will tell you how . The following clues may help you solve the problem : Initially at rest Stopped moving Started moving with

𝝑𝒊 = 𝟎 𝝑𝒇 = 𝟎 𝒈𝒊𝒗𝒆𝒏 𝝑

Example 2.6.1 : A car accelerates from ( 𝟏𝟎 𝒎/𝒔 ) to ( 𝟐𝟎 𝒎/𝒔 ) through ( 𝟐 𝒔𝒆𝒄 ) , calculate its acceleration : ⃗ = 𝒂

∆𝝑 𝝑𝒇 − 𝝑𝒊 = ∆𝒕 ∆𝒕

When re-arranging the equation you get equation ( # 1 ) : 𝝑𝒇 = 𝝑𝒊 + 𝒂 𝒕 𝟐𝟎 = 𝟏𝟎 + 𝒂 × 𝟐 𝒂 = +𝟓

𝒎 → 𝝑𝒇 > 𝝑𝒊 𝒔𝟐

Example 2.6.2 : A plane lands on a road with a velocity ( 𝟐𝟓𝟎 𝒌𝒎/𝒉 ) , having an acceleration of a magnitude ( 𝟒. 𝟖 𝒎/𝒔𝟐 ) , calculate : I. II.

I.

The distance traveled on the road till the plane stopped in meters . The time needed till it stops in seconds .

𝝑𝒇 𝟐 = 𝝑𝒊 𝟐 + 𝟐 𝒂 ∆𝒙 𝟐

𝟐𝟓𝟎 × 𝟏𝟎𝟑 𝟎=[ ] + 𝟐 (−𝟒. 𝟖) × ∆𝒙 𝟔𝟎 × 𝟔𝟎 ∆𝒙 = 𝟓𝟎𝟐. 𝟑 𝒎 ‫وليد شاهر أبوخضر‬

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II. 𝝑𝒇 = 𝝑 𝒊 + 𝒂 𝒕 𝟐𝟓𝟎 × 𝟏𝟎𝟑 𝟎=[ ] + (−𝟒. 𝟖) × ∆𝒕 𝟔𝟎 × 𝟔𝟎 ∆𝒕 = 𝟏𝟒. 𝟒𝟕 𝒔 Since 𝝑𝒇 = 𝟎 < 𝝑𝒊 → 𝒂 < 𝟎 to stop the plane the motion should be decelerating.

Problems -

Fundamentals and Basics :

2-1 An object starts moving in the ( 𝒙𝒚 ) plane from ( 𝟑, −𝟒 ) to ( 𝟎, 𝟔 ) back to the origin . calculate the following : a) The total distance moved . b) The displacement done by the object .

𝑨𝒏𝒔 ∶ 𝟗. 𝟔 𝒖𝒏𝒊𝒕𝒔 𝑨𝒏𝒔 ∶ 𝟓 𝒖𝒏𝒊𝒕𝒔

2-2 A cyclist maintains a constant velocity , determine the following according to the graph shown : a) The slope of the line and what does it represent. 𝑨𝒏𝒔 ∶ +𝟓

𝒎 𝒔

𝑨𝒏𝒔 ∶

, 𝒕𝒉𝒆 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒚𝒄𝒍𝒊𝒔𝒕

b) What is the cyclist’s positon from its starting positon at ( 𝟔𝟎 𝒔𝒆𝒄 ) ? 𝑨𝒏𝒔 ∶ 𝟓𝟓𝟎 𝒎 c) What is the displacement from the starting position at ( 𝟔𝟎 𝒔 ) ? 𝑨𝒏𝒔 ∶ 𝟑𝟎𝟎 𝒎

2-3 A particle moves in the ( 𝒙 ) axis where its position is given by the equation : .

𝒙(𝒕) = √𝟏 + 𝟒𝒕 , where 𝒕 ≥ 𝟎 . Calculate the following : a) Time passed till the particle moves a distance of ( 𝟕 𝒎 ) from where its original position . 𝑨𝒏𝒔 ∶ 𝟏𝟓. 𝟕𝟓 𝒔 b) The instantaneous velocity at ( 𝒕 = 𝟔 𝒔𝒆𝒄 ) .

‫وليد شاهر أبوخضر‬

𝟐𝒎

𝑨𝒏𝒔 ∶ 𝟓

𝒔

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2-4 The velocity – time graph of a moving body is shown in the figure , The acceleration of the in the three time intervals shown is : 𝑨𝒏𝒔 ∶ 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 , 𝒅𝒆𝒄𝒓𝒆𝒂𝒔𝒊𝒏𝒈 , 𝒛𝒆𝒓𝒐

2-5 A student drives a bike with a constant acceleration in a straight line , starting from rest to a velocity of ( 𝟏𝟐. 𝟓 𝒎/𝒔 ) in ( 𝟐. 𝟓 𝒔 ) . Calculate : a) Its constant acceleration . b) Its total displacement . c) Its average velocity .

-

𝑨𝒏𝒔 ∶ +𝟓 𝒎/𝒔𝟐 𝑨𝒏𝒔 ∶ +𝟏𝟓. 𝟔 𝒎 𝑨𝒏𝒔 ∶ 𝟔. 𝟑 𝒎/𝒔

Higher Level :

2.1 Two cars whose opposite to each other are moving in straight line as shown , and separated ( 1000 m ) from each other , after ( 𝐭 𝐨 ) they meet at point ( 𝐨 ). ( Known the constant velocities of both cars ) calculate ( 𝒕𝒐 ) in seconds .

2.2 A road concrete engineer wants to construct a rail-way such that a ( 𝟔𝟗. 𝟒 𝒎/𝒔 ) plane could travel on the terminal having ( 𝟒. 𝟖 𝒎/𝒔𝟐 ) constant acceleration and take off after ( 𝟏𝟓 𝒔 ) . Calculate The minimum length of the road needed so that the plane fly safely having the least cost.

2.3 An object moves along has a position given by [ 𝒙 = 𝟔𝒕𝟐 − 𝒕𝟑 ] , where ( 𝒕 : is measured in seconds ) , find the following : a) What is the positon of the object when it reaches its maximum speed in meters ? b) What is the average velocity during the time interval [ 𝟏 𝒕𝒐 𝟓 ] seconds ? c) At what time would the acceleration of the object equals its velocity in magnitude ?

‫وليد شاهر أبوخضر‬

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2.4 Starting from rest , a car travels ( 𝟏. 𝟑𝟓 𝒌𝒎 ) in one minute , it accelerated at ( 𝟏 𝒎𝟐 /𝒔 ) until it reached its cruising speed . Then it drove the remaining distance at a constant velocity , what was its cruising speed in ( 𝒎/𝒔 ) ?

Finally : Some figures , graphs , problems and paragraphs are copied from some references and books : -

Physics for Scientists and Engineers with Modern Physics - 8th /6th Edition - Serway_Jewett Physics – Harcourt –geoprojects Free high school science texts – FHSST authors

Copying this textbook in other author’s or person’s name without a personal approval from the writer himself is considered a crime and will be severely punished according to law . TM

This work is published on www.academia.edu , for the latest work , papers , researches and summaries follow the author on : https://djasljsa.academia.edu/Kunka/Physics-(-Classical-Mechanics-) Waleed Shaher Abu Khader .

‫وليد شاهر أبوخضر‬

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