Example: Converting Units
A Porsche can accelerate at 12 m / s 2 .
What is this in
km / h 2 ?
2
m 1km ⎛ 3600 s ⎞
km
12 2
⎟ = 155520 2
⎜
s 1000m ⎝ 1h ⎠
h
Vectors
Two definitions:
A = (A, θ) = (magnitude, angle)
^
^
A = (Ax,Ay) = Ax i + Ay j
Unit Vectors:
)
i)
j
= (1,0)
= (0,1)
Ax = A cos Θ
Ay = A sin Θ
Ax2 + Ay2 = A2
tan Θ = Ay/Ax
Initial position of the knight
^
^
(1,2) = 1i +2j
Position after second move
^
^
(4,3) = 4i +3j
y
x
Displacement
(4,3) - (1,2) = (3,1)
= 3 ^i +1 ^j
Magnitude = 32 + 12 = 10
−1
Angle θ = tan (1 / 3) = 18°
Grasshopper-Problem
(3,2)
(4,0)
(-2,-5)
(-3,2)
Displacement?
x
y
4
-3
3
-2
--------2
0
2
2
-5
--------1
Motion Along a Straight Line
In this chapter we will study kinematics, i.e., how objects move along a
straight line.
The following parameters will be defined:
Position
Displacement
Average velocity
Average speed
Instantaneous velocity
Average and instantaneous acceleration
For constant acceleration we will develop the equations that give us the
velocity and position at any time. In particular we will study the motion under
the influence of gravity close to the surface of the Earth.
Finally, we will study a graphical integration method that can be used to
analyze the motion when the acceleration is not constant.
Kinematics is the part of mechanics that describes the motion of physical
objects. We say that an object moves when its position as determined by an
observer changes with time.
We will study a restricted class of kinematics problems
Motion will be along a straight line.
We will assume that the moving objects are “particles,” i.e., we restrict our
discussion to the motion of objects for which all the points move in the same
way.
The causes of the motion will not be investigated. This will be done later in the
course.
Consider an object moving along a straight
line taken to be the x-axis. The object’s
position at any time t is described by its
coordinate x(t) defined with respect to the
origin O. The coordinate x can be positive or
negative depending whether the object is
located on the positive or the negative part of
the x-axis.
Displacement. If an object moves from position x1 to position x2 , the change
x2
x1
in position is described by the displacement
O
Δx = x2 − x1
.Δx .
.
x-axis
motion
For example if x1 = 5 m and x2 = 12 m then Δx = 12 – 5 = 7 m. The positive
sign of Δx indicates that the motion is along the positive x-direction.
If instead the object moves from x1 = 5 m and x2 = 1 m then Δx = 1 – 5 = -4 m.
The negative sign of Δx indicates that the motion is along the negative xdirection.
Displacement is a vector quantity that has both magnitude and direction. In this
restricted one-dimensional motion the direction is described by the algebraic
sign of Δx.
Note: The actual distance for a trip is
irrelevant as far as the displacement is
concerned.
Consider as an example the motion of an object from an initial position
x1 = 5 m to x = 200 m and then back to x2 = 5 m. Even though the total
distance covered is 390 m the displacement then is Δx = 0.
Important distinctions:
Position:
location of an object
Path, distance:
(scalar)
any connection between two points
Displacement Δ (vector):
shortest connection between two points
= final position - initial position
Velocity
Velocity is a vector that measures how fast and in what
direction something moves.
Speed is the magnitude of the velocity. It is a scalar.
Average Velocity
One method of describing the motion of an object is to plot its position x(t) as
a function of time t. In the left picture we plot x versus t for an object that is
stationary with respect to the chosen origin O. Notice that x is constant. In
the picture to the right we plot x versus t for a moving armadillo. We can get
an idea of “how fast” the armadillo moves from one position x1 at time t1 to a
new position x2 at time t2 by determining the average velocity between t1 and
t2.
vavg
x2 − x1 Δx
=
=
t2 − t1
Δt
Here x2 and x1 are the positions x(t2) and x(t1),
respectively.
The time interval Δt is defined as Δt = t2 – t1.
The units of vavg are m/s.
Note: For the calculation of vavg both t1 and t2
must be given.
Graphical Determination of vavg
On an x versus t plot we can determine vavg from the slope of the straight line
that connects point ( t1 , x1) with point ( t2 , x2 ). In the plot below, t1=1 s and
t2 = 4 s. The corresponding positions are: x1 = - 4 m and x2 = 2 m.
vavg
x2 − x1 2 − (−4) 6 m
=
=
=
= 2 m/s
4 −1
3s
t2 − t1
Average Speed savg
The average speed is defined in terms of the total distance traveled in a time
interval Δt (and not the displacement Δx as in the case of vavg).
savg
total distance
=
Δt
Note: The average velocity and the average speed
for the same time interval Δt can be quite different.
On a graph of position versus time, the average velocity is
represented by the slope of a chord.
x (m)
x2
x1
t1
t2
Average velocity = vav , x
t (sec)
x2 − x1
=
t 2 − t1
Δr
Instantaneous velocity = v = lim
Δt →0 Δt
This is represented by the slope of a line tangent to the curve
on the graph of an object’s position versus time.
x (m)
t (sec)
Instantaneous Velocity
The average velocity vavg determined between times t1 and t2 provides a useful
description of “how fast” an object is moving between these two times. It is in
reality a “summary” of its motion. In order to describe how fast an object moves
at any time t we introduce the notion of instantaneous velocity v (or simply
velocity). Instantaneous velocity is defined as the limit of the average velocity
determined for a time interval Δt as we let Δt → 0.
Δx dx
=
v = lim
Δt dt
Δt → 0
From its definition instantaneous velocity is the
first derivative of the position coordinate x with
respect to time. It is thus equal to the slope of
the x versus t plot.
Speed
We define speed as the magnitude of an
object’s velocity vector.
The area under a velocity versus time graph (between the
curve and the time axis) gives the displacement in a given
interval of time.
v(m/s)
t (sec)
Margaret walks to the store using the following path: 0.500 miles west, 0.200
miles north, 0.300 miles east. If the first leg of her walk takes 10 minutes, the
second takes 8 minutes, and the third 7 minutes, compute her average velocity
and average speed during each leg and for the overall trip.
y
Take north to be in
the +y direction and
east to be along +x.
r3
Δr
r2
x
r1
Leg
Distance
(miles)
Δt
(hours)
vav
(miles/hour)
Average
speed
(miles/hour)
1
0.5
0.167
3.00 (west)
3.00
2
0.2
0.133
1.50 (north)
1.50
3
0.3
0.117
2.56 (east)
2.56
Total trip
(0.2,0.2)
0.28
0.417
0.679
(45° N of W)
2.40
Example: The position x(t) of a particle
Is given by
x(t) = 8 - 6 t + t2 [m]
x = tn dx/dt = n tn-1
v(t) = 0 - 6 + 2 t
[m/s]
For t=0 x(0)=8 and v(0) = -6
x=0 when t =2 and t =4
v= 0 when t = 3
Average Acceleration
We define the average acceleration aavg between
t1 and t2 as:
aavg =
v2 − v1 Δv
=
t2 − t1 Δt
Units: m/s2
Instantaneous Acceleration
If we take the limit of aavg as Δt → 0 we get the
instantaneous acceleration a, which describes
how fast the velocity is changing at any time t.
Δv dv
a = lim
= , a=
Δt dt
Δt → 0
dv d ⎛ dx ⎞ d 2 x
= ⎜ ⎟= 2
dt dt ⎝ dt ⎠ dt
The acceleration is the slope of the v versus t plot.
Note: The human body does not react to velocity
but it does react to acceleration.
(2-7)
Motion with Constant Acceleration
Motion with a = 0 is a special case but it is rather common, so we will develop
the equations that describe it.
a=
dv
→ dv = adt. If we integrate both sides of the equation we get:
dt
∫ dv = ∫ adt = a ∫ dt → v = at + C.
Here C is the integration constant.
C can be determined if we know the velocity v0 = v(0) at t = 0:
v(0) = v0 = (a )(0) + C → C = v0 → v = v0 + at
(eq. 1)
dx
→ dx = vdt = ( v0 + at ) dt = v0 dt + atdt. If we integrate both sides we get:
dt
at 2
∫ dx =∫ v0 dt +a ∫ tdt → x = v0t + 2 + C ′. Here C ′ is the integration constant.
C ′ can be determined if we know the position xo = x(0) at t = 0:
v=
a
x(0) = xo = (v0 )(0) + (0) + C ′ → C ′ = xo
2
at 2
(eq. 2)
x(t ) = xo + v0t +
2
(2-8)
Example: The position x(t) of a particle
Is given by
x(t) = 8 - 6 t + t2 [m]
v(t) = 0 - 6 + 2 t
a(t) = 2 [m/s2]
[m/s]
v = v0 + at
(eq. 1) ;
If we eliminate the time t
v 2 − v02 = 2a ( x − x0 )
at 2
x = x0 + v0t +
(eq. 2)
2
between equation 1 and equation 2 we get:
(eq. 3)
Below we plot the position x(t ), the velocity v(t ), and the acceleration a versus time t:
at 2
x = x0 + v0t +
2
The x(t) versus t plot is a parabola that
intercepts the vertical axis at x = x0.
v = v0 + at
The v(t) versus t plot is a straight line with
slope = a and intercept = v0.
The acceleration a is a constant.
(2-9)
Important (useful) Mathematical Relationsships
between
x, v and a
for constant acceleration
Variables Related
Equation
Velocity, time, acceleration
v = v0 + at
Initial, final, and average
velocity
vav = ½(v0 + vfinal)
Position, time, velocity
x = x0 + ½(v0 + v)t
Position, time, acceleration
x = x0 + v0t + ½ at2
Velocity, position,
acceleration
v2 = v02 + 2a(x – x0) = v02 + 2aΔx
Free Fall
Close to the surface of the Earth all objects move toward the center of the Earth
with an acceleration whose magnitude is constant and equal to 9.8 m/s2. We
use the symbol g to indicate the acceleration of an object in free fall.
If we take the y-axis to point upward then the
acceleration of an object in free fall a = -g and the
equations for free fall take the form:
a
B
A
(2-10)
y
v = v0 − gt
(eq. 1)
gt 2
x = xo + v0t −
(eq. 2)
2
v 2 − v02 = −2 g ( x − xo )
(eq. 3)
Note: Even though with this choice of axes a < 0, the
velocity can be positive (upward motion from point A to
point B). It is momentarily zero at point B. The velocity
becomes negative on the downward motion from point
B to point A.
Hint: In a kinematics problem, always indicate the axis
as well as the acceleration vector. This simple
precaution helps to avoid algebraic sign errors.
Graphical Integration in Motion Analysis (nonconstant acceleration)
When the acceleration of a moving object is not constant we must use
integration to determine the velocity v(t ) and the position x(t ) of the object.
The integration can be done either using the analytic or the graphical approach:
t
t
t
t
1
1
1
1
dv
a=
→ dv = adt → ∫ dv = ∫ adt → v1 − v0 = ∫ adt → v1 = v0 + ∫ adt
dt
t0
t0
t0
t0
t1
∫ adt
= [ Area under the a versus t curve between t0 and t1 ]
t0
t
t
1
1
dx
v=
→ dx = vdt → ∫ dx = ∫ vdt →
dt
t0
t0
t1
t1
t0
t0
x1 − x0 = ∫ vdt → x1 = x0 + ∫ vdt
t1
∫ vdt
= [ Area under the v versus t curve between t0 and t1 ]
to
(2-11)