Lecture 2

Definitions

Mechanics:

The study of motion.

Kinematics:

The mathematical description of motion in 1-D and 2-D motion.

Dynamics:

The study of the forces that cause motion.

Consider in this chapter only motion in one dimension:

motion along a straight line.

First define position, displacement, velocity and

acceleration.

Then, using these concepts, study the motion of objects

traveling in 1-D with a constant acceleration.

Chapter Outline

Definition of Motion

Motion represents a continuous change in the position of an

object.

A car moving down a highway.

Example of motion:

Particle’s Position

Particle’s Position is the location of the particle with respect

to a chosen reference point that we can consider to be the

origin of a coordinate system.

Positions to the right of the origin are positive.

Positions to the left of the origin are negative.

Displacement

In other word; it is the difference between the final position xf

, and the initial position xi .

Displacement of a particle ∆x is defined as its change in position

in some time interval.

∆𝑥 = 𝑥𝑓 − 𝑥𝑖

A displacement to the right will be a positive displacement.

Displacement

For example

starting with xi = 60 m and ending at xf = 150 m, the

displacement is :

A displacement to the left will be a negative displacement.

Displacement

For example

starting with xi = 150 m and ending at xf = 60 m, the

displacement is :

Distance is the absolute value of the displacement.

Distance is always positive and tells how far something is from

something else but does not tell us whether it is to the right or to the

left.

Units are important in Physics (and in all of Science)

In the lab, we will usually measure distance or displacement in units

of meters (m).

Distance or displacement could also be measured in centimeters

(cm) or kilometers (km) or even miles (mi).

Distance and Displacement

Example

Find the displacement, and the distance

between A & D

Displacement

Ans.

Ans.

Total Distance = distance from A to B + distance from

B to C + distance from B to C

= 180 m + 140 m + 100 m

= 420 m.

Ans.

The average Speed of an object over

a given time interval is defined as the

total distance traveled divided by the

total time elapsed:

𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑠𝑝𝑒𝑒𝑑 =𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒

SI Unit: meter per second (m/s)

Speed is a a scalar quantity.

The average Velocity during a time

interval t is the displacement Δx

divided by the time t :

SI Unit: meter per second (m/s)

Velocity is a vector quantity.

Speed & Velocity

Speed & Velocity is also measured in km/h (and even in mi/hr).

𝜗𝑎𝑣𝑔 =∆𝑥

∆𝑡=𝑥𝑓 − 𝑥𝑖

𝑡𝑓 − 𝑡𝑖

Ans. :

question

If you run from x=0 m to x=25 m and back to you starting

point in a time interval of 5 sec .

Find the average velocity and average speed

Speed & Velocity

𝜗𝑎𝑣𝑔 =∆𝑥

∆𝑡=𝑥𝑓 − 𝑥𝑖

𝑡𝑓 − 𝑡𝑖

𝜗𝑎𝑣𝑔 =0 − 0

5= 0 𝑚/𝑠𝑒𝑐

The average velocity is zero meter/sec

Speed & VelocityAns. :

𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑 =25𝑚+25𝑚

5=

50

5= 10 𝑚/sec

The average speed is 10 meter/sec

Example The position versus time for a certain particle moving along the x axis

is shown in Figure bellow.

Find the average velocity in the time intervals :

a) 0 to 2 sec,

b) 2 s to 4 sec,

c) 0 to 8 sec.

POSITION-TIME GRAPH

a)

b)

c)

Ans. :𝜗𝑎𝑣𝑔 =

∆𝑥

∆𝑡=𝑥𝑓 − 𝑥𝑖

𝑡𝑓 − 𝑡𝑖

Example Find the displacement, average velocity, and average speed of

the car in the figure between positions A and F.

Ans. :

Displacement:

Average Velocity:

Total distance traveled:

From the car’s position described by the curve, we find that the

distance traveled from A to B is 22m plus 105m the distance traveled

from B to F for a total of 127m.

Ans. :

• Applying the Formula to find Total Distance:

Ans. :

The instantaneous velocity 𝜗𝑖𝑛𝑠𝑡 is the velocity right

now, at this particular moment.

If the velocity is constant :

𝜗𝑖𝑛𝑠𝑡 = 𝜗𝑎𝑣𝑔

SI unit : meter per second (m/s)

The instantaneous velocity

Acceleration

Acceleration is a change of velocity divided by a change of time.

𝑎𝑎𝑣𝑔 =∆𝜗

∆𝑡=𝜗𝑓 − 𝜗𝑖

𝑡𝑓 − 𝑡𝑖

This quantity has units of (meters/second)/second.

We will write this as m/sec2 (there are no "square seconds").

As with the velocity, we can describe the instantaneous acceleration, the

acceleration right now, at this particular moment. If the acceleration is

constant :

We are often interested in how fast the velocity is changing. This

is the acceleration.

𝑎𝑖𝑛𝑠𝑡 = 𝑎𝑎𝑣𝑔

AccelerationIf there is no change in speed, but there is a change in direction

the acceleration has occured

Example: Find the acceleration from the figure below

Acceleration

f

Ans. :

𝑎𝑎𝑣𝑔 =∆𝜗

∆𝑡=19 − 10

3 𝑠𝑒𝑐=9 𝑚/𝑠𝑒𝑐

3𝑠𝑒𝑐= 3

𝑚/𝑠𝑒𝑐

𝑠𝑒𝑐

𝑎𝑎𝑣𝑔 = 3𝑚

𝑠𝑒𝑐/𝑠𝑒𝑐 = 3 𝑚/𝑠𝑒𝑐2

• When the object’s velocity and acceleration are in the

same direction, the object is speeding up (other hind

the value for acceleration is positive).

• On the other hand, when the object’s velocity and

acceleration are in opposite direction, the object is

slowing down. (also called deceleration)

Velocity vs Acceleration

MOTION DIAGRAMS

***

The Four Kinematic Equations

The Four Kinematic Equations

𝜗 = 𝜗𝑜 + 𝑎𝑡 𝑎 =𝜗 − 𝜗𝑜

𝑡→ 𝜗 − 𝜗𝑜 = 𝑎𝑡 →∴ 𝜗 = 𝜗𝑜 + 𝑎𝑡

𝑥𝑓 = 𝑥𝑖 +1

2(𝜗 + 𝜗𝑜)𝑡

𝜗 =𝑥𝑓 − 𝑥𝑖

𝑡→ 𝑥𝑓 − 𝑥𝑖 = 𝜗𝑡

→ 𝜗 =𝜗 + 𝜗𝑜2

𝑥𝑓 − 𝑥𝑖 =(𝜗 + 𝜗𝑜)

2𝑡

∴ 𝑥𝑓= 𝑥𝑖 +1

2(𝜗 + 𝜗𝑜)𝑡

𝑥𝑓 = 𝑥𝑖 + 𝜗𝑜𝑡 +1

2𝑎𝑡2

𝑥𝑓 = 𝑥𝑖 +(𝜗 + 𝜗𝑜)

2𝑡

→ 𝜗 = 𝜗𝑜 + 𝑎𝑡

𝑥𝑓 = 𝑥𝑖 +(𝜗𝑜 + 𝑎𝑡 + 𝜗𝑜)

2𝑡

𝑥𝑓 = 𝑥𝑖 +(2𝜗𝑜 + 𝑎𝑡)

2𝑡 →∴ 𝑥𝑖 + 𝜗𝑜𝑡 +

1

2𝑎𝑡2

The Four Kinematic Equations

𝜗𝑓2 = 𝜗𝑜

2 + 2𝑎 (𝑥𝑓 − 𝑥𝑖)

𝑥𝑓 = 𝑥𝑖 +(𝜗 + 𝜗𝑜)

2𝑡

→ 𝑎 =𝜗 − 𝜗𝑜

𝑡→ 𝑡 =

𝜗 − 𝜗𝑜𝑎

𝑥𝑓 = 𝑥𝑖 +(𝜗 + 𝜗𝑜)

2

(𝜗 − 𝜗𝑜)

𝑎= 𝑥𝑖 +

1

2𝑎𝜗2 − 𝜗𝑜

2

𝑥𝑓 − 𝑥𝑖 =1

2𝑎𝜗2 − 𝜗𝑜

2 → 2𝑎 𝑥𝑓 − 𝑥𝑖 = 𝜗2 − 𝜗𝑜2

∴ 𝜗𝑓2 = 𝜗𝑜

2 + 2𝑎 (𝑥𝑓 − 𝑥𝑖)

Example: Consider a car that starts at rest and accelerates at 2

m/s2 for 3 seconds.

At that time, t = 3 s, how fast is it going? and how far has it gone?

Ans. :

𝜗𝑥𝑓 = 0 + 𝑎𝑥𝑡

𝜗𝑥𝑓 = 𝑎𝑥𝑡 = 2 𝑚/𝑠𝑒𝑐 3𝑠𝑒𝑐

𝜗𝑥𝑓 = 6m/sec

𝑥𝑓 = 𝑥𝑖 + 𝜗𝑥𝑖𝑡 +1

2𝑎𝑥𝑡

2

𝜗𝑥𝑓 = 𝜗𝑥𝑖 + 𝑎𝑥𝑡

𝑥 =1

2𝑎𝑡2

𝑥 =1

22 𝑚/𝑠𝑒𝑐2 (3𝑠𝑒𝑐)2

𝑥 = 9 m

Homework

Problem 1:

The velocity of a particle moving along the x-axis varies

according to the expression vx = (40 – 5t 2) m/s, where t

is in seconds.

Find the average acceleration in the time interval

t = 0 to t = 2.0 sec.

Problem 2:

Carrier Landing

A jet lands on an aircraft carrier at 140 mi /h (≈ 63 m/s).

a) What is its acceleration (assumed constant) if it stops in 2.0 sec ?

b) If the plane touches down at position xi = 0, what is the final

position of the plane ?

Homework