chapter 2 motion along a straight line 2.2 motion motion: change in position in relation with an...
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Chapter 2
Motion along a straight line
2.2 Motion
Motion: change in position in relation with an object of reference .
The study of motion is called kinematics.
Examples:
• The Earth orbits around the Sun
• A roadway moves with Earth’s rotation
Engineering Physics : Lecture 1, Pg 3
Position and DisplacementPosition and Displacement
Position: coordinate of position – distance to the origin (m)Position: coordinate of position – distance to the origin (m)
O
x
P
x
Engineering Physics : Lecture 1, Pg 4
Example PositionExample Position
4
0 x (cm)212 1
The position of the ball is cm 2x
The + indicates the direction to the right of the origin.
Engineering Physics : Lecture 1, Pg 5
Displacement Displacement ΔΔxx
Displacement: change in position x0 = original (initial) location x = final location Δx = x - x0
Example: x0 = 1m, x = 4m, Δx = 4m – 1m = 3m
1m 4m
Engineering Physics : Lecture 1, Pg 6
Displacement Displacement ΔΔxx
Displacement: change in position x0 = original (initial) location x = final location Δx = x - x0
Example: x0 = 4m, x = 1 m, Δx = 1m – 4m = - 3m
1m 4m
Engineering Physics : Lecture 1, Pg 7
Displacement, DistanceDisplacement, Distance
Distance traveled usually different from displacement.Distance traveled usually different from displacement. Distance always positive.Distance always positive. Previous example: always 3 m.Previous example: always 3 m.
Engineering Physics : Lecture 1, Pg 8
Distance: Scalar QuantityDistance: Scalar Quantity
Distance is the path length traveled from one location to another. It will vary depending on the path.
Distance is a scalar quantity—it is described only by a magnitude.
Engineering Physics : Lecture 1, Pg 9
What is the displacement in the situation depicted What is the displacement in the situation depicted bellow?bellow?
a) 3 m b) 6 m c) -3 m d) 0 ma) 3 m b) 6 m c) -3 m d) 0 m
1m 4m
Engineering Physics : Lecture 1, Pg 10
What is the distance traveled in the situation depicted What is the distance traveled in the situation depicted bellow?bellow?
a) 3 m b) 6 m c) -3 m d) 0 ma) 3 m b) 6 m c) -3 m d) 0 m
1m 4m
Engineering Physics : Lecture 1, Pg 11
Position, DisplacementPosition, Displacement
Engineering Physics : Lecture 1, Pg 12
Motion in 1 dimensionMotion in 1 dimension In 1-D, we usually write position as x(t1 ).
Since it’s in 1-D, all we need to indicate direction is + or .
Displacement in a time t = t2 - t1 is x = x(t2) - x(t1) = x2 - x1
t
x
t1 t2
x
t
x1
x2some particle’s trajectory
in 1-D
Engineering Physics : Lecture 1, Pg 13
1-D kinematics1-D kinematics
tx
tt)t(x)t(x
v12
12av
t
x
t1 t2
x
x1
x2trajectory
Velocity v is the “rate of change of position” Average velocity vav in the time t = t2 - t1 is:
t
Vav = slope of line connecting x1 and x2.
Engineering Physics : Lecture 1, Pg 14
Consider limit t1 t2
Instantaneous velocity v is defined as:
1-D kinematics...1-D kinematics...
dt)t(dx
)t(v
t
x
t1 t2
x
x1
x2
t
so v(t2) = slope of line tangent to path at t2.
Engineering Physics : Lecture 1, Pg 15
1-D kinematics...1-D kinematics...
tv
tt)t(v)t(v
a12
12av
Acceleration a is the “rate of change of velocity” Average acceleration aav in the time t = t2 - t1 is:
And instantaneous acceleration a is defined as:
2
2
dt)t(xd
dt)t(dv
)t(a
dt)t(dx
)t(v using
Engineering Physics : Lecture 1, Pg 16
RecapRecap
If the position x is known as a function of time, then we can find both velocity v and acceleration a as a function of time!
adv
dt
d x
dt
2
2
vdx
dt
x x t ( )
x
a
vt
t
t
Engineering Physics : Lecture 1, Pg 17
More 1-D kinematicsMore 1-D kinematics
We saw that v = dx / dt In “calculus” language we would write dx = v dt, which we
can integrate to obtain:
2
1
t
t12 dttvtxtx )()()(
Graphically, this is adding up lots of small rectangles:
v(t)
t
+ +...+
= displacement
Engineering Physics : Lecture 1, Pg 18
High-school calculus:
Also recall that
Since a is constant, we can integrate this using the above rule to find:
Similarly, since we can integrate again to get:
1-D Motion with constant acceleration1-D Motion with constant acceleration
constt1n
1dtt 1nn
adv
dt
vdx
dt
0vatdtadtav
002
0 xtvat21
dt)vat(dtvx
Engineering Physics : Lecture 1, Pg 19
RecapRecap So for constant acceleration we find:
atvv 0
200 2
1attvxx
a const
x
a
v t
t
t
Engineering Physics : Lecture 1, Pg 20
Example: Displacement vs. TimeExample: Displacement vs. Time
8 m
b) What is the velocity at 5 s? Unable to determine
(no slope)c) What is the position after 10 s?
- 4 m
d) What is the velocity at 10 s?
0 m/s
e) What is the velocity during the second part of the trip?
sms
m
ss
mmvelocity /4
3
12
58
84
Engineering Physics : Lecture 1, Pg 21
f) What is the velocity during the forth part of the trip?
g) What is the velocity at 15 s?
h) What is the position after 15 s?
2 m. The displacement changes 1 m every 2 seconds, so the position at 15 s is one meter morethan the position at 13 s.
Engineering Physics : Lecture 1, Pg 22
Motion in One DimensionMotion in One Dimension
When throwing a ball straight up, which of the following is When throwing a ball straight up, which of the following is true about its velocity true about its velocity vv and its acceleration and its acceleration aa at the at the highest point in its path?highest point in its path?
(a)(a) BothBoth v = 0v = 0 andand a = 0a = 0..
(b)(b) v v 0 0, but , but a = 0a = 0..
(c) (c) v = 0v = 0, but , but a a 0 0..
y
Engineering Physics : Lecture 1, Pg 23
Solution Solution
x
a
vt
t
t
Going up the ball has positive velocity, while coming down Going up the ball has positive velocity, while coming down it has negative velocity. At the top the velocity is it has negative velocity. At the top the velocity is momentarily zero.momentarily zero.
Since the velocity is Since the velocity is
continually changing there mustcontinually changing there must
be some acceleration.be some acceleration. In fact the acceleration is caused In fact the acceleration is caused
by gravity ( by gravity (g = 9.81 g = 9.81 m/sm/s22).). (more on gravity in a few lectures)(more on gravity in a few lectures)
The answer is (c) The answer is (c) v = 0v = 0, but , but a a 0 0. .
Engineering Physics : Lecture 1, Pg 24
Galileo’s FormulaGalileo’s Formula
Plugging in for t:
atvv 0 200 at
21
tvxx
Solving for t:
avv
t 0
200
00 avv
a21
avv
vxx
)0
(220
2 xxavv
Engineering Physics : Lecture 1, Pg 25
Alternate (Calculus-based) DerivationAlternate (Calculus-based) Derivation
dt
d
d
d
dt
d x
x
vva
)0
(220
2 xxavv
(chain rule)
x
vvad
d vvxa dd
v
v
x
x
x
x 000
vvxa x a ddd
)vv(2
1)-(a 2
02
0 xx
(a = constant)
Engineering Physics : Lecture 1, Pg 26
Recap:Recap: For constant acceleration:
atvv 0
200 at
2
1tvxx
a const
+ Galileo and average velocity:
v)(v2
1v
)x2a(xvv
0av
02
02
Engineering Physics : Lecture 1, Pg 27
Problem 1Problem 1
A car is traveling with an initial velocity v0. At t = 0, the driver puts on the brakes, which slows the car at a rate of ab
x = 0, t = 0ab
vo
Engineering Physics : Lecture 1, Pg 28
Problem 1...Problem 1...
A car is traveling with an initial velocity v0. At t = 0, the driver puts on the brakes, which slows the car at a rate of ab. At what time tf does the car stop, and how much farther xf does it travel?
x = xf , t = tf
v = 0
x = 0, t = 0ab
v0
Engineering Physics : Lecture 1, Pg 29
Problem 1...Problem 1...
Above, we derived: v = v0 + at
Realize that a = -ab
Also realizing that v = 0 at t = tf :
find 0 = v0 - ab tf or
tf = v0 /ab
Engineering Physics : Lecture 1, Pg 30
Problem 1...Problem 1...
To find stopping distance we use:
In this case v = vf = 0, x0 = 0 and x = xf
fb2
0 x)a(2v
b
20
f a2v
x
)x2a(xvv 02
02
Engineering Physics : Lecture 1, Pg 31
Problem 1...Problem 1...
So we found that
Suppose that vo = 29 m/s Suppose also that ab = g = 9.81 m/s2
Find that tf = 3 s and xf = 43 m
b
20
fb
0f a
v
2
1x ,
a
vt
Engineering Physics : Lecture 1, Pg 32
Tips:Tips:
Read !Before you start work on a problem, read the problem
statement thoroughly. Make sure you understand what information is given, what is asked for, and the meaning of all the terms used in stating the problem.
Watch your units !Always check the units of your answer, and carry the units
along with your numbers during the calculation.
Understand the limits !Many equations we use are special cases of more general
laws. Understanding how they are derived will help you recognize their limitations (for example, constant acceleration).
Engineering Physics : Lecture 1, Pg 33
Problem 2Problem 2
kmkmtot ttt 32
avav v
xttvxSince
min61.0/20
22 h
hkm
kmt km
It normally takes you 10 min to travel 5 km to school. You leave class 15 min before class. Delays caused by traffic slows you down to 20 km/h for the first 2 km of the trip, will you be late to class?
usualkm v
kmt
33
min/5.0min10
5km
km
t
xv
usual
totusual
min6min/5.0
33
km
kmt km
min12min6min6 tott Not late!
Engineering Physics : Lecture 1, Pg 34
34
Example: A train of mass 55,200 kg is traveling along a straight, level track at 26.8 m/s. Suddenly the engineer sees a truck stalled on the tracks 184 m ahead. If the maximum possible braking acceleration has magnitude 1.52 m/s2, can the train be stopped in time?
m 236
m/s 52.12
m/s 8.26
2
02
2
22
22
a
vx
xavv
o
o
Know: a = 1.52 m/s2, vo = 26.8 m/s, v = 0
Using the given acceleration, compute the distance traveled by the train before it comes to rest.
The train cannot be stopped in time.
Engineering Physics : Lecture 1, Pg 35
A train is moving parallel and adjacent to a highway with a constant speed of 35 m/s. Initially a car is 44 m behind the train, travelingin the same direction as the train at 47 m/s, and accelerating at 3 m/s2.What is the speed of the car just as it passes the train?
Example MeetingExample Meeting
tvxx tot 2
2
1attvtvx octo 22 )/3(
2
1)/47()/35(44 tsmtsmtsmm
to = 0 s, initial timet = final timexo = 44 mvt = 35 m/s, constantvoc = 47 m/s, initial speed of carvc = final speed of car
atvv occ
2
2
1attvx occ
Car’s equation of motion
Train’s equation of motion:
To meet: xc = xt
25.112440 tt t = 2.733 s
smssmsmvc /199.55733.2)/3(/47 2
Oxo xt =
xc
Engineering Physics : Lecture 1, Pg 36
Recap Recap Scope of this courseScope of this course Measurement and Units Measurement and Units (Chapter 1)(Chapter 1)
Systems of unitsConverting between systems of unitsDimensional Analysis
1-D Kinematics 1-D Kinematics (Chapter 2)Average & instantaneous velocity
and acceleration Motion with constant acceleration
Example car problemExample car problem