Lecture 2 - · PDF fileMotion represents a continuous change in the position of an object....
Transcript of Lecture 2 - · PDF fileMotion represents a continuous change in the position of an object....
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:
Particleβs Position
Particleβs 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 βx is defined as its change in position
in some time interval.
βπ₯ = π₯π β π₯π
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
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:
π΄π£πππππ π ππππ =π‘ππ‘ππ πππ π‘ππππ
π‘ππ‘ππ π‘πππ
SI Unit: meter per second (m/s)
Speed is a a scalar quantity.
The average Velocity during a time
interval t is the displacement Ξx
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).
πππ£π =βπ₯
βπ‘=π₯π β π₯π
π‘π β π‘π
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
πππ£π =βπ₯
βπ‘=π₯π β π₯π
π‘π β π‘π
πππ£π =0 β 0
5= 0 π/π ππ
The average velocity is zero meter/sec
Speed & VelocityAns. :
π΄π£πππππ πππππ =25π+25π
5=
50
5= 10 π/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
Example Find the displacement, average velocity, and average speed of
the car in the figure between positions A and F.
Total distance traveled:
From the carβs 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. :
The instantaneous velocity ππππ π‘ is the velocity right
now, at this particular moment.
If the velocity is constant :
ππππ π‘ = πππ£π
SI unit : meter per second (m/s)
The instantaneous velocity
Acceleration
Acceleration is a change of velocity divided by a change of time.
πππ£π =βπ
βπ‘=ππ β ππ
π‘π β π‘π
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.
ππππ π‘ = πππ£π
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. :
πππ£π =βπ
βπ‘=19 β 10
3 π ππ=9 π/π ππ
3π ππ= 3
π/π ππ
π ππ
πππ£π = 3π
π ππ/π ππ = 3 π/π ππ2
β’ When the objectβs velocity and acceleration are in the
same direction, the object is speeding up (other hind
the value for acceleration is positive).
β’ On the other hand, when the objectβs velocity and
acceleration are in opposite direction, the object is
slowing down. (also called deceleration)
Velocity vs Acceleration
The Four Kinematic Equations
π = ππ + ππ‘ π =π β ππ
π‘β π β ππ = ππ‘ ββ΄ π = ππ + ππ‘
π₯π = π₯π +1
2(π + ππ)π‘
π =π₯π β π₯π
π‘β π₯π β π₯π = ππ‘
β π =π + ππ2
π₯π β π₯π =(π + ππ)
2π‘
β΄ π₯π= π₯π +1
2(π + ππ)π‘
π₯π = π₯π + πππ‘ +1
2ππ‘2
π₯π = π₯π +(π + ππ)
2π‘
β π = ππ + ππ‘
π₯π = π₯π +(ππ + ππ‘ + ππ)
2π‘
π₯π = π₯π +(2ππ + ππ‘)
2π‘ ββ΄ π₯π + πππ‘ +
1
2ππ‘2
The Four Kinematic Equations
ππ2 = ππ
2 + 2π (π₯π β π₯π)
π₯π = π₯π +(π + ππ)
2π‘
β π =π β ππ
π‘β π‘ =
π β πππ
π₯π = π₯π +(π + ππ)
2
(π β ππ)
π= π₯π +
1
2ππ2 β ππ
2
π₯π β π₯π =1
2ππ2 β ππ
2 β 2π π₯π β π₯π = π2 β ππ2
β΄ ππ2 = ππ
2 + 2π (π₯π β π₯π)
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. :
ππ₯π = 0 + ππ₯π‘
ππ₯π = ππ₯π‘ = 2 π/π ππ 3π ππ
ππ₯π = 6m/sec
π₯π = π₯π + ππ₯ππ‘ +1
2ππ₯π‘
2
ππ₯π = ππ₯π + ππ₯π‘
π₯ =1
2ππ‘2
π₯ =1
22 π/π ππ2 (3π ππ)2
π₯ = 9 m
Homework
Problem 1:
The velocity of a particle moving along the x-axis varies
according to the expression vx = (40 β 5t 2) m/s, where t
is in seconds.
Find the average acceleration in the time interval
t = 0 to t = 2.0 sec.