Lecture 11: Linear and Rectilinear Kinematics - Velocity, Acceleration, and Displacement, Study Guides, Projects, Research of Kinematics

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Introduction & Rectilinear Kinematics:

Lecture 11

Statics: The study of

bodies in equilibrium.

Dynamics:

1. Kinematics – concerned with

the geometric aspects of motion

without any reference to the cause of

motion.

2. Kinetics - concerned with the

forces causing the motion

Mechanics: The study of how bodies

react to forces acting on them.

Scalars: are quantities which are fully described by a magnitude alone.

eg : distance , mass ,time ,volume etc

Vectors: are quantities which are fully described by both a magnitude and a direction.

eg: displacement , velocity , force ,acceleration etc.,

Introduction & Rectilinear Kinematics: Lecture 11

Linear motion : when a body moves either in a straight line or along a curved

path, then we say that it is executing linear motion.

1. when a body moves in a straight line then the linear motion is called

rectilinear motion.

eg ., an athlete running a 100 meter race along a straight track is said to be a

linear motion or rectilinear motion.

2. when a body moves along a curved in two or three dimensions path then the

linear motion is called cur vilinear motion.

eg., the earth revolving around the sun.

Rotatory motion: A body is said to be in rotatory motion when it stays at one

place and turns round and round about an axis.

example: a rotating fan, a rotating pulley about its axis.

Oscillatory motion : a body is said to be in oscillatory motion when it swings to

and fro about a mean position.

example: the pendulum of a clock, the swing etc.,

4

Rectilinear Kinematics: Continuous Motion Lecture 11

 In the picture, the car moves from ( point A to point B ). So it changes its position during a time interval ( t ).

Distance and Displacement

Displacement: refers to how far the object

travels, but also adds direction. This is a vector

quantity. For example: the car traveled (35km

to the east).

Distance: refers to how far an object travels.

This is a scalar quantity. For example: the car

traveled (35 km).

 We are only concerned with the length of travel we don’t distinguish between directions.

Distance = dx + dy = 4 + 3 = 7 km

36.8 o

Speed:

 Speed is a measure ofhow fast something moves.

 Speed is ascalar quantity, specified only by its magnitude.

 Speed is defined as thedistance covered perunit time:

speed = distance / time = x / t ( m/s, km/h, foot/min, … etc )

Aver age Speed: Lecture 11

 Average speed is thewhole distance covered divided bythe total time of travel. General

definition: Average speed = total distance covered / time interval

 Distinguish betweeninstantaneous speed andaverage speed:

• On most trips, we experience a variety of speeds, so the average speed and instantaneous speed are often quite different.

Instantaneous Speed:

 The speed at any instant is theinstantaneous speed.

 The speed registered by an automobile speedometer is

theinstantaneous spee d.

Example 1: If we travel 320 km in 4 hours, what is our average speed? If we drive at this

average speed for 5 hours, how far will we go?

Answer: vavg = 320 km/4 h = 80 km/h.

d =vavg x time = 80 km/h x 5 h = 400 km.

Example 2: A plane flies 600 km away from its base at 200 km/h, then flies back to its base

at 300 km/h. What is its average speed?

Answer: total distance traveled,d = 2 x 600 km = 1200 km;

total time spent ( for the round trip):

t = (600 km / 200 km/h) + (600 km / 300 km/h) = 3 h + 2 h = 5 h.

Average speed,vavg =d / t = 1200 km/5 h = 240 km/h.

Acceleration: Lecture 11

Acceleration is the rate of change in the velocity of a particle. It is a vector

quantity. Typical units are m/s^2 or ft/s 2.

 Theinstantaneous acceleration is the time derivative of velocity.  Consider particle with velocity ( v ) at time (t ) and (v’ ) at ( t + ∆t ):

Instantaneous acceler ation

t

v

a

t ∆

∆ → 0

lim

 Acceleration can be positive ( speed increasing ) or negative ( speed decreasing ).

t

dt

dv

a

v t t

dt

d x

dt

dv

t

v

a

t

e.g. 12 3

lim

2

2

2

0

∆ →

 From the definition of a derivative:

 The derivative equations for velocity and acceleration can be manipulated to get:

dx

v

dt

dv

a

dt

2 2

d x

a

dt

dv dv dx dv

a v

dt dx dt dx

= = = a. dx = v dv

Var iable acceler ation

8

Constant Acceleration:

Lecture 11

The three kinematic equations can be integrated for the special case when

acceleration is constant ( a = a (^) c ) to obtain ver y useful equations. A common

example of constant acceler ation is gr avity; i.e., a body freely falling toward earth.

In this case, (a (^) c = g = 9.81 m/s 2 = 32.2 ft/s 2 ) downward. These equations are:

constant

0 0

v v a x x

a vdv a dx v v a x x

dx

dv

v

x

x

v

v

Example 1 : Continue Lecture 11

y = 20 + 10 t - 4.905 t^2

 At time (t = 1.019 sec. ) the corresponding altitude ( y ),

where the velocity equal to zero.

y = 20 + 10. (1.019) - 4.905. (1.019) 2

y = 25. 1 m

 Solve for ( t ) at which altitude equals zero and

evaluate corresponding velocity ( v ).

y = 20 + 10 t - 4.905 t^2 = 0

v = 10 - 9.81 t = 10 – 9.81 * 3.28 = - 22.2 m/s

t = - 1.243 sec. (meaningless)

t = 3.28 sec.

Example 2 :

Ball ( A ) is released from rest at a height of ( 40 ft ) at the same time a ball ( B ) is thrown upward, ( 5 ft ) from the ground. The balls pass one another at a height of ( 20 ft ). Find: The speed at which ball ( B ) was thrown upward.

Solution: Both balls experience a constant downward

acceleration of ( a c = 32.2 ft/s 2 ).

Example 2 : Continue Lecture 11

1) Fir st consider ball A. With the origin defined at the

ground, ball ( A ) is released from rest [ (vA ) (^) o = 0 ]

at a height of ( 40 ft ), [ (sA ) o = 40 ft ]. Calculatethe

time required for ball ( A ) to drop to ( 20 ft ), [ sA =

20 ft ] using a position equation.

sA = ( sA ) o + ( v a ) o. t + (1/2) a c. t 2

20 ft = 40 ft + (0). (t) + (1/2). (-32.2). (t 2 )

t = 1.115 s

2) Now consider ball B. It is throw upward from a height of ( 5 ft ) [ (sB ) (^) o = 5 ft ]. It must reach a height of ( 20 ft ) [ sB = 20 ft ] at the same time ball ( A ) reaches this height ( t = 1.115 s ). Apply the position equation again to ball ( B ) using ( t = 1.115 sec ).

sB = (sB) o + (v B) o. t + (1/2). a c. t 2

20 ft = 5 + (v B) o. (1.115) + (1/2). (-32.2). (1.115)^2

(v B) o = 31.4 ft/s

Example 4: Variable acceleration^ Lecture 11

Given that the velocity of a particle changes with time such that ( 𝑣 = 16 − 𝑡^2 ), find: A. Draw a velocity-time graph for the first ( 4 seconds ) of motion. B. Obtain the distance covered in the first ( 4 seconds ). C. The acceleration when (i) t = 1 seconds, (ii) t = 3 seconds.

Solution:

v

t

16 m/s

4

B) Distance travelled in the first 4 seconds, (Integration to find the distance).

v = dS / dt dS = v. dt

Distance covered ( S ) = ∫ ( 16 − 𝑡^2 ) 𝑑𝑡 = [ 16 t - t 3 / 3 ]

0

4

0

4

S = 42.67 m

a = dv / dt = ( 16 - t 2 ) = - 2 t

d dt

i. ) a = - 2 t

ii.) a = - 2 t

At t = 1 sec. a = - 2 m/s 2

At t = 3 sec. a = - 6 m/s 2

B) The accelaration can be find by differentiating the velocity.

Example 5: Constant acceleration^ Lecture 11

Two cars ( A & B ) start simultaneously from point ( O ) and move in a straight line. One with a velocity of ( 66 m/s ), and an acceleration of ( 2 m/s^2 ), the other car with a velocity of ( 132

m/s ) and with a retardation of ( 8 m/s^2 ). Find the time which the velocities of the cars are

same and the distance from point ( O ) where they meet again. (^) a

t

2 m/s^2

8 m/s^2

O

Solution:

For Car ( A ):

A

B

a = dv / dt dv = a. dt dv = 2 dt

vA = 66 + 2. t

For Car ( B):

vB = 132 - 8. t

 When car ( A ) and car ( B ) have the same velocity:

vA = vB

66 + 2. t = 132 - 8. t

66 = 10. t

t = 6.6 sec

 So, after ( 6.6 sec. ), the velocities of car ( A ) and car ( B ) have the same velocity.

A

132 m/s

O

66 m/s B

6.6 s^ t

Lecture 12

Curvilinear Motion:

 When a Par ticle moving along a curve other than a str aight line, we say that the par ticle is in cur vilinear motion.

Coordinates Used for Describing the Plane Curvilinear Motion:

The motion of a particle can be described by using three coordinates, these are:

1. Rectangular coordinates.
2. Normal – Tangential coordinates.
3. Polar coordinates.

Rectangular coordinates

Normal-Tangential coordinates

Polar coordinates

y

x

P

t

n

PA

PB

PC

t

n

t

n

y

x

P

r

θ r

Path

Path

Path

O^ O

Vx

Vy^ V^ V

Plane Curvilinear Motion – without Specifying any

Coordinates:

(Displacement)

Lecture 12

Position:

 The position of the particle measured from a fixed point ( O ) will be designated by the

position vector [r = r (t ) ].

 This vector is a function of time (t ), since in general both its magnitude and direction

change as the particle moves along the curve path.

 Suppose that particle which

occupies position ( P ) defined by

(r ) at time (t) moving along a curve

a distance ( ∆S ) to new position (P’)

defined by (r ’ = r + ∆r ) at a time

(t + ∆t ).

 (∆r = r ’ – r = displacement).

Note: Since, here, the particle

motion is described by two

coordinates components, both the

magnitude and the direction of

the position, the velocity, and the

acceleration have to be specified.

P at time t

r (t) ∆ r^ P^ at time t+∆t

r (t+∆t)

O

∆s

Actual distance traveled by the particle (it is s scalar)

The vector displacement of the particle

r ( t )

r ( t + ∆ t )

∆ r

∆ S P at time (t + ∆ t)

P at time (t)

Path

4

Curvilinear Motion: Lecture 12

V

V’

∆V

Average Acceleration ( a av ):

_

Note: a av has the direction of ( ∆ V ) and its

magnitude is the magnitude of ( ∆ V ) divided by ( ∆t ).

_

Instantaneous Acceleration ( a ):

_

As ( ∆ t ) approaches zero:

Note: In general, the acceleration vector ( a ) is neither

tangent nor normal to the path. However, ( a ) is tangent

to the hodograph.

_

_

P

P

V 1

V 2

_

_

C

Hodograph V 1

V 2

a 2 a 1

_

_

_ _

A plot of the locus of points defined by the arrowhead of the velocity vector is called a hodograph. The acceleration vector is tangent to the hodograph, but not, in general, tangent to the path function.

Motion of a Projectile:^ Lecture 12

Projectile motion can be treated as two rectilinear motions , one in the horizontal

direction ( constant velocity ) experiencing zero acceler ation and the other in the vertical direction experiencingconstant acceler ation (i.e., gr avity).