Motion in a straight line

Motion in a straight line

Chapter (2)

2.4

VELOCITY VECTOR, AVERAGE VELOCITY,

SPEED

2.7 2.2

POSITION VECTOR, DISPLACEMENT

VECTOR

2.8

ACCELARATION VECTOR

MOTION WITH CONSTANT ACCELERATION

Outline

2.3

FREE FALL

After studying this chapter, you will be able:

1. Determine the direction and magnitude of a particle’s position vector from its components, and vice versa.

2. Apply the relationship between particle’s displacement vector and its initial and final position vectors.

3. Indicate velocity vector in unit vector notation.

4. Determine the direction and magnitude of a particle’s velocity vector from its components, and vice versa.

5. Given a particle’s position vector as a function of time, determine its instantaneous velocity vector.

6. Determine the direction and magnitude of a particle’s acceleration vector from its components, and vice versa.

7. Determination of average acceleration vector in unit-vector notations.

8. Given a particle’s velocity vector as a function of time, determine its instantaneous acceleration vector.

9. Apply the constant acceleration equations to find acceleration, velocity, position, and time.

10. Understand the free fall equations.

Learning outcomes

What is a Vector ?

Physical quantities describe motion

Displacement vector

Acceleration vector

Velocity vector Position vector

Physical quantities describe motion

Position vector 2.2

 To locate an object means to find it’s position relative to a reference point origin ( or zero point ) of an axis.

 All position vectors are measured relative to the origin of the coordinate system.

 SI-unit “m” ( it’s a length unit)

 Position is a vector quantity: has a magnitude and a direction.

Direction on x-axis

Direction on y-axis

Positive if it is right to the reference point Negative if it is left to the reference point Positive if it is upward to the reference point Negative if it is downward to the reference point

Position vector 2.2

• Position vector is denoted with 𝑟

• Written in a unit vector notation in 3D as:

𝑟 = (𝑥) i + (𝑦) j + (𝑧) k

Position vector in three dimensions

Unit vector notation 𝑎 = 𝑎_𝑥 𝑖 + 𝑎_𝑦 𝑗 + 𝑎_𝑧 𝑘

Chapter-1

Position vector

𝑦 = 2𝑐𝑚

2.2

Position left to the reference point (x=-3cm)

Position right to the reference point (x=2cm)

On X-axis On y-axis

Left to Ref. Point (Negative direction)

Right to Ref. Point (Positive direction)

x = -3cm

Position vector in one dimension

Up the Ref. point (Positive direction)

Below the Ref. point

(Negative direction) 𝑦 = −4𝑐𝑚

0 origin

Position up the reference point (y=2cm)

Position below the reference point (y= -4cm)

𝑟 = (𝑥) i + (𝑦) j + (𝑧) k

𝑟 = −3 i

𝑟 = 2 i

𝑟 = (𝑥) i + (𝑦) j + (𝑧) k

𝑟 = 2 j

𝑟 = −4 j 𝑟 = 0 i + 2 j + 0 k

𝑟 = −3 i + 0 j + 0 k

Position vector 2.2

Position vector in one dimension

x = 2cm y = 2cm z = 2cm

𝑟 = 2 i + 0 j + 0 k 𝑟 = 0 i + 2 j + 0 k 𝑟 = 0 i + 0 j + 2 k

𝑟 = 2 i 𝑟 = 2 j 𝑟 = 2 k

Position vector 2.2

x = -3cm y = 2cm

𝑟 = (𝑥) i + (𝑦) j + (𝑧) k

𝑟 = −3 i + 2 j + 0 k

𝑟 = −3 i + 2 j

could be any two dimensions Position vector in two dimension

Position vector

11

2.2

Position vector in two dimension

x = 2cm y = 3cm z = 2cm

y = 2cm z = 2cm x = 2cm

𝑟 = 2 i + 2 j + 0 k 𝑟 = 0 i + 3 j + 2 k 𝑟 = 2 i + 0 j + 2 k 𝑟 = 2 i + 2 j 𝑟 = 3 j + 2 k 𝑟 = 2 i + 2 k

Position vector 2.2

Position vector in three dimension

𝑟 = (𝑥) i + (𝑦) j + (𝑧) k

𝑟 = 2 i + 2 j + 2 k

x = 2cm y = 2cm z = 2cm

x Y

Z

Position vector 2.2

Position vector in one dimension as a function in time

 If the object moves the position of an object changes as a function of time

 In terms of unit vector: 𝑟 t = x t î

Example:

x-component of a position vector as a function of time 𝑥(𝑡) = (3𝑡 + 5) Where x is the position and t is the time

Position vector 2.2

Position vector in three dimension as a function in time

𝑟(𝑡) = 𝑥(𝑡) i + 𝑦(𝑡) j + 𝑧 (𝑡) k

Example:

if we have 𝑥(𝑡) = (3𝑡 + 1), y(t)= 2t and z(t)= 𝑡² + 5, then the position vector is

𝑟(𝑡) = 3𝑡 + 1 i + 2𝑡 j + (𝑡² + 5) k 𝑟(𝑡) = 𝑥(𝑡) i + 𝑦(𝑡) j + 𝑧 (𝑡) k

Extra Example 2.2

Answer:

Displacement vector

■ displacement is the object’s overall change in position Is the difference between final position and initial position

■ SI-unit : m

■ It is a vector quantity

2.2

 ∆𝑥 = 𝑥 -𝑥

_o

 Direction: if x is positive  moving to the right if x is negative  moving to the left

 In terms of unit vector :

∆ 𝑟 = ∆𝑥 î

𝑥_𝑖 𝑥_𝑓

The displacement vector in one dimension

Displacement vector 2.2

In terms of unit vector

∆𝑟 = −4 î

On X-axis On y-axis

origin 0

In terms of unit vector

∆𝑟 = 7 𝑗

𝒙

_o

∆𝑥 = (−2) − (2)= - 4 m

∆𝒙 = 𝒙-𝒙_o

∆𝑦 = (3) − (−4)= 7cm

∆𝒚 = 𝒚-𝒚_o

Moving to the left

Direction on x-axis Direction on y-axis

Positive  moving to the right Negative  moving to the left Positive  moving upward Negative  moving downward

𝒚_o=-4cm 𝒚=3cm

Moving upward

𝒙

Displacement vector 2.2

Displacement vector in three dimensions

Δ 𝑟 = 𝑟 − 𝑟_°

Δ 𝑟 = 𝑥 − 𝑥_° i + y − 𝑦_° j + (𝑧 − 𝑧_°) k Δ 𝑟 = Δ𝑥 i + Δ𝑦 j + Δ𝑧 k

Where ∆𝑥 = 𝑥 − 𝑥_° , ∆𝑦 = y − 𝑦_° , ∆𝑧 = (𝑧 − 𝑧_°)

Extra Example 2.2

Answer:

Velocity vector 2.2

Average velocity Instantaneous velocity or velocity

- Is the ratio of displacement per time interval.

- SI unit: m/s.

- It is a vector quantity

- 𝒗

𝒂𝒗𝒈

- Is the velocity at a specific time - SI unit: m/s.

- It is a vector quantity.

-

𝑣

Velocity vector

Velocity Vector

■ Direction:

velocity is positive  to the right velocity is negative to the left

■ In terms of unit vectors : 𝑣_𝑎𝑣𝑔 = ( ^∆𝑥_∆𝑡 )î

2.3

In x-direction 𝒗_𝒂𝒗𝒈

𝑥 = ^Δ𝑥

Δ𝑡 = ^{𝑥−𝑥°}

𝑡−𝑡_°

Average velocity vector in one dimension

In y-direction 𝒗_{𝒂𝒗𝒈𝑦} = ^Δ𝑦_Δ𝑡 = ^𝑦−𝑦_𝑡−𝑡^°

°

velocity is positive  upward velocity is negative  downward

𝑡_°

𝑥_°

𝑡

𝑥

origin 0

, 𝑦_° 𝑦

In terms of unit vectors : 𝑣_𝑎𝑣𝑔 = ( ^∆𝑦_∆𝑡) 𝑗 𝑡

𝑡_°

Moving upward

Velocity Vector 2.3

Average velocity vector in three dimension

𝑣_𝑎𝑣𝑔=^∆𝒓

∆𝒕 =Δ𝑥 i+Δ𝑦 j+Δ𝑧 k

∆𝑡 = ^Δ𝑥

∆𝑡 i + ^Δ𝑦

∆𝑡 j + ^Δ𝑧

∆𝑡 k = 𝑣_{𝑎𝑣𝑔𝑥} i + 𝑣_{𝑎𝑣𝑔𝑦} j + 𝑣_{𝑎𝑣𝑔𝑧} k

Velocity vector 2.2

Average velocity Instantaneous velocity or velocity

- Is the ratio of displacement per time interval.

- SI unit: m/s.

- It is a vector quantity

- 𝒗

_𝒂𝒗𝒈

- Is the velocity at a specific time - SI unit: m/s.

- It is a vector quantity.

-

𝑣

Velocity vector

Velocity Vector 2.3

Instantaneous velocity in one dimension

In terms of unit vectors

𝑣 = 𝑣

_𝑥

𝑖 = (

^𝑑𝑥_𝑑𝑡

)î

In x-direction  𝑥(𝑡)

𝑣

_𝑥

= 𝑑𝑥 𝑑𝑡

In y-direction  𝑦(𝑡)

𝑣

_𝑦

= 𝑑𝑦 𝑑𝑡

In terms of unit vectors 𝑣 = 𝑣_𝑦 𝑗 = (𝑑𝑦

𝑑𝑡) 𝑗

Velocity Vector 2.3

Instantaneous velocity in three dimension

𝒗=^𝒅𝒓

𝒅𝒕 =𝑑𝑥 i+𝑑𝑦 j+𝑑𝑧 k

𝑑𝑡 = ^𝑑𝑥

𝑑𝑡 i + ^𝑑𝑦

𝑑𝑡 j + ^𝑑𝑧

𝑑𝑡 k = 𝑣_𝑥 i + 𝑣_𝑦 j + 𝑣_𝑧 k 𝑟 = 𝑥 i + 𝑦 j + 𝑧 k

𝒗 =

^𝒅𝒓_𝒅𝒕

Where 𝑣_𝑥 = ^𝑑𝑥_𝑑𝑡 𝑣_𝑦 = 𝑑𝑦

𝑑𝑡 𝑣^𝑧 = 𝑑𝑧 𝑑𝑡

Speed

■ Is The magnitude of instantaneous velocity.

■ Is the absolute value of the velocity vector.

■ A scalar quantity (has no direction)

■ Always positive

■ SI unit :m/s

2.3

Instantaneous Speeds (S)

Instantaneous speed = 𝐼𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦

Speed limits are always posted as positive numbers, and the radar monitors that measure the speed of passing cars also always display positive numbers

During the time interval 0 to 10 s, the position vector of a car on the road is given by:

𝑥 𝑡 = 17.2 − 10.1𝑡 + 1.1𝑡² (a) What is the position vector at t=0s?

(b)What is its instantaneous velocity vector at t= 6s?

(c) What is the car’s average velocity during this interval (from 0s to 10s)?

SOLUTION:

(a) 𝑟 𝑡 =(17.2 − 10.1𝑡 + 1.1𝑡²) î

At t=0  𝑟 0 = (17.2 − 10.1 0 + 1.1(0)) î 𝑟 0 =17.2 m î

(b) Take derivative to get the instantaneous velocity vector:

𝑣_𝑥 𝑡 = ^𝑑𝑥_𝑑𝑡 = _𝑑𝑡^𝑑 (17.2 − 10.1𝑡 + 1.1𝑡²)=−10.1 + 2.2𝑡 𝑣 𝑡 =(−10.1 + 2.2𝑡) î

At t=6 𝑣 𝑡 =(−10.1 + 2.2𝑡) î 𝑣 6 = (−10.1 + (2.2𝑥6)) î = 3.1^𝑚 𝑠 î

Example 2.1 (Page 44)

SOLUTION

(c) We know that:

 𝑥 𝑡 = 17.2 − 10.1𝑡 + 1.1𝑡²

 𝑣 𝑡 =(−10.1 + 2.2𝑡) î

■ At 𝑡₂=10s  𝑥₂=? 𝑥 𝑡 = 17.2 − 10.1𝑡 + 1.1𝑡²

𝑥 10 = 17.2 − 10.1(10) + 1.1(10)²=26.2m

■ At 𝑡₁=0s  𝑥₁=? 𝑥 0 = 17.2m (calculated from (a)) The average velocity 𝑣_𝑎𝑣𝑔

𝑥=^∆𝑥_∆𝑡 = ^26.2−17.2₁₀₋₀ =₁₀⁹ =0.9m/s In terms of unit vectors

The average velocity 𝑣_{𝑎𝑣𝑔𝑥}=^∆𝑥

∆𝑡 = ^{𝑥−𝑥°}

𝑡−𝑡_°

Example 2.1 (Page 44)

We need to calculate 𝑥₂ and 𝑥₁to find the

displacement

𝑣_𝑎𝑣𝑔 = 0.9 î

Acceleration vector 2.2

Acceleration vector

Average acceleration Instantaneous acceleration or Acceleration - is defined as the velocity change

per time interval.

- SI unit: m/𝒔^𝟐.

- It is a vector quantity

- 𝑎

𝑎𝑣𝑔

- is the acceleration at a specific time - SI unit: m/𝒔^𝟐.

- It is a vector quantity.

-

𝑎

The velocity changes could be in ( magnitude, or direction or both),to be able to say the particle undergoes an acceleration

Acceleration Vector

■ The average acceleration is defined as the velocity change per time interval

■ In term of unit vector notation: 𝑎_𝑎𝑣𝑔 = (^∆𝑣_∆𝑡^𝑥)î

2.4

𝑎_{𝑎𝑣𝑔𝑥} = Δ𝑣_𝑥

Δ𝑡 = 𝑣_𝑥 − 𝑣_𝑥°

𝑡 − 𝑡_°

acceleration vector in one dimension

Average acceleration

where 𝑎_{𝑎𝑣𝑔𝑥} = ^Δ𝑣_∆𝑡^𝑥 𝑎_{𝑎𝑣𝑔𝑦}= ^Δ𝑣_∆𝑡^𝑦 𝑎_{𝑎𝑣𝑔𝑧}= ^Δ𝑣_∆𝑡^𝑧 Average acceleration vector in three dimensions

𝑎_𝑎𝑣𝑔=^∆𝒗

∆𝒕 =^{Δ𝑣𝑥 i+Δ𝑣𝑦 j+Δ𝑣𝑧 k}

∆𝑡 = ^Δ𝑣𝑥

∆𝑡 i + ^Δ𝑣_∆𝑡^𝑦 j + ^Δ𝑣_∆𝑡^𝑧 k =𝑎_{𝑎𝑣𝑔𝑥} i + 𝑎_{𝑎𝑣𝑔𝑦} j + 𝑎_{𝑎𝑣𝑔𝑧} k

In x-direction ( 𝑚 𝑠²)

Acceleration vector 2.2

Acceleration vector

Average acceleration Instantaneous acceleration or Acceleration - is defined as the velocity change

per time interval.

- SI unit: m/𝒔^𝟐.

- It is a vector quantity

- 𝑎

𝑎𝑣𝑔

- is the acceleration at a specific time - SI unit: m/𝒔^𝟐.

- It is a vector quantity.

-

𝑎

The velocity changes could be in ( magnitude, or direction or both),to be able to say the particle undergoes an acceleration

Acceleration Vector

■ The instantaneous acceleration describes the acceleration at a very specific time

2.3

acceleration vector in one dimension Instantaneous acceleration

Instantons acceleration vector in three dimensions

𝒂=^𝒅𝒗

𝒅𝒕 =^{𝑑𝑣𝑥 i+𝑑𝑣𝑦 j+𝑑𝑣𝑧 k}

𝑑𝑡 = ^𝑑𝑣𝑥

𝑑𝑡 i + ^𝑑𝑣𝑦

𝑑𝑡 j + ^𝑑𝑣𝑧

𝑑𝑡 k = 𝑎_𝑥 i + 𝑎_𝑦 j + 𝑎_𝑧 k

x-component 𝑎_𝑥 = 𝑑𝑣_𝑥 𝑑𝑡

In term of unit vector notation: 𝑎 = (^𝑑𝑣_𝑑𝑡^𝑥)î

𝒂 =

^𝒅𝒗

𝒅𝒕 𝒗 = 𝑣_𝑥 i + 𝑣_𝑦 j + 𝑣_𝑧 k

Acceleration Concepts 2.4

 Deceleration of an object: is a decrease in the speed of the object over time.

 If the velocity and acceleration are in the same direction, the object speeds up.

 If the velocity and acceleration are in opposite directions, the object slows down.

Velocity Acceleration Motion

+ +

Speeding up in the

positive direction

- -

Speeding up in the

negative direction

+ -

Slowing down in the

positive direction

- +

Slowing down in the

negative direction

Concept Check

■ When you’re driving a car along a straight road, you could be traveling in the positive or negative direction and you could have a positive

acceleration or a negative acceleration.

If you have negative velocity and positive acceleration you are:

A. slowing down in the positive direction.

B. speeding up in the negative direction.

C. speeding up in the positive direction.

D. slowing down in the negative direction.

Exercise 2.37 (page 62)

SOLUTION:

(a) The velocity is given by the time derivative of the position function

𝑣 𝑡 = _𝑑𝑡^𝑑x(t)= _𝑑𝑡^𝑑 2.1𝑡³ + (1)𝑡² + (−4.1)𝑡 + (3 ) = 3(2.1)𝑡² + 2(1)𝑡 +(-4.1) 𝑣 𝑡 =6.3𝑡² + 2𝑡 - 4.1

At t=10s  𝑣(10)=(6.3)(10²)+(2)(10)-4.1

= 645.9 m/s

Exercise 2.37 (page 62)

SOLUTION:

(b) To find the time when the object is at rest, set the velocity to zero, and solve for time; t.

𝑣 𝑡 = 6.3𝑡² + 2𝑡 + −4.1 0 = 6.3𝑡² + 2𝑡 − 4.1

Where, a = 6.3 b = 2 c = -4.1

𝑡 = ^{−𝑏± 𝑏}_2𝑎^{2−4𝑎𝑐} = ^{−(2)± (2)2}−4(6.3)(−4.1)

2(6.3) =

𝑡 = 0.664𝑠 OR, 𝑡 = −0.981𝑠

This is a quadratic equation of the form 𝑎𝑥² + bx + c = 0, whose solution is

𝑥 = −𝑏 ± 𝑏² − 4𝑎𝑐

6.3𝑡² + 2𝑡 − 4.1 = 0 2𝑎

X

Exercise 2.37 (page 62)

SOLUTION:

(c) The acceleration is given by the time derivative of the velocity

a 𝑡 = _𝑑𝑡^𝑑 v(t)= _𝑑𝑡^𝑑 6.3𝑡² + 2𝑡 − 4.1 = 12.6𝑡 + 2

At t=0.5s  a 0.5 =(12.6)(0.5)+2

= 8.3 m/𝑠²

Constant Acceleration 2.7

Many physical situations involve constant acceleration .

 Constant acceleration does not mean the velocity is constant.

 Constant acceleration means the velocity changes with constant rate.

 If v = constant  a=0.

 If v changes with constant rate  a= constant

Constant Acceleration

■ We can derive useful equations for the case of constant acceleration.

■ The Five Kinematic Equations

2.7

Constant Acceleration 2.7

 When the object starts from rest

 When the object stops

 .

0

 0 v

 0 v

mentioned else

something unless

0

 0

x

SOLUTION:

(a)

𝑎 = 4.3 𝑚 𝑠² Given 𝑣_o=0

𝑡 = 18𝑠

S = ? 𝑣 = ? 𝑨𝒕 𝒕 = 𝟏𝟖 𝒔

𝒗=𝒗_o + 𝒂𝒕 = 𝟎 + 𝟒. 𝟑 𝟏𝟖 = 𝟕𝟕. 𝟒 𝒎/𝒔 S= 𝟕𝟕. 𝟒 = 𝟕𝟕. 𝟒 𝒎/𝒔

Assuming a constant acceleration of a = 4.3 m/s², starting from rest:

(a) what is the speed of the airplane reached after 18 seconds?

(b) How far down the runway has this airplane moved by the time it takes off?

Solved problem 2.2 (page 46)

SOLUTION:

(b)

𝑎 = 4.3 𝑚 𝑠² Given 𝑣_°=0

𝑡 = 18𝑠 𝑥_°=0

𝑥 = ?

𝑨𝒕 𝒕 = 𝟏𝟖 𝒔

𝒙=𝒙_° + 𝒗_°𝒕 + ^𝟏 _𝟐𝒂𝒕^𝟐 = 𝟎 + 𝟎 𝟏𝟖 + ^𝟏_𝟐(𝟒. 𝟑)(𝟏𝟖^𝟐) = 𝟔𝟗𝟕𝒎

Assuming a constant acceleration of a = 4.3 m/s², starting from rest:

(a) what is the speed of the airplane reached after 18 seconds?

(b) How far down the runway has this airplane moved by the time it takes off?

Solved problem 2.2 (page 46)

Constant Acceleration (Free Fall) 2.7

■ Free fall is the motion of an object under influence of gravity and ignoring any other effects such as air resistance.

■ Free fall is an example of motion in one dimension with constant acceleration.

■ All objects in free fall accelerate downward at the same rate and is independent of the object’s mass, density or shape.

■ Near the surface of the Earth, the acceleration due to the force of gravity is constant and is always in the downward direction.

a = a_y = -g

downward s

m

g  9 . 8 /

²

Motion along x-axis motion is along y axis: x  y (Free Fall)

Constant Acceleration (Free Fall) 2.7

𝑣 = 𝑣_o + 𝑎𝑡 𝑥 − 𝑥_o = 𝑣_o𝑡 + 1

2 𝑎𝑡² 𝑣² = 𝑣_o² + 2𝑎 𝑥 − 𝑥_o 𝑥 − 𝑥_o = 1

2 𝑣^o + 𝑣 𝑡 𝑥 − 𝑥_o = 𝑣𝑡 − 1

2 𝑎𝑡²

𝑣 = 𝑣_o + 𝑎𝑡

y−𝑦_o = 𝑣_o𝑡 + ¹ ₂𝑎𝑡² 𝑣² = 𝑣_o² + 2𝑎 𝑦 − 𝑦_o y−𝑦_o = ¹ ₂ 𝑣_o + 𝑣 𝑡

y−𝑦_o = 𝑣𝑡 − ¹ ₂𝑎𝑡²

a= -g = -(9.8)

𝑣 = −24𝑚/s Given 𝑣_°=0

a = −g = −9.8 m/s² 𝑡 =?

From

𝑣 = 𝑣_o + 𝑎𝑡 𝑡 = ^𝑣−𝑣_𝑎 ⁰= ⁻²⁴⁻⁰

−9.8 = 2.45 seconds

Extra Exercise

SOLUTION:

■ At a construction site a pipe struck the ground with a velocity of -24 m/s.

How long was it falling ? (𝑣_°=0)

𝑣 = 𝑣_o + 𝑎𝑡 y−𝑦_o = 𝑣_o𝑡 + ¹ ₂𝑎𝑡² 𝑣² = 𝑣_o² + 2𝑎 𝑦 − 𝑦_o y−𝑦_o = ¹ ₂ 𝑣_o + 𝑣 𝑡

y−𝑦_o = 𝑣𝑡 − ¹ ₂𝑎𝑡²

The END OF

CHAPTER

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