#### Transcript Chapter 2

```Chapter 2
Motion Along a Straight Line
Motion Along a Straight Line
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In this chapter we will study kinematics, i.e., how objects move along
a straight line.
The following parameters will be defined:
Displacement
Average velocity
Average speed
Instantaneous velocity
Average and instantaneous acceleration
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For constant acceleration we will develop the equations that give us
the velocity and position at any time.
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we will study the motion under the influence of gravity close to the
surface of the Earth.
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Finally, we will study a graphical integration method that can be used
to analyze the motion when the acceleration is not constant.
Ch 2-1 Motion Along a Straight Line
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Motion of an object along a straight line
Object is point mass
Motion along a horizontal or vertical or
inclined (line with finite slope) line
Motion: Change in position
No change in position, body at rest
Ch 2-3 Position and Displacement
 Axis are used to define
position of an object
 Position of an object defined
with respect to origin of an
axis
 Position x of an object is its
distance from the origin at
any time t
 Displacement x, a vector,
is change in position.
 x = xf-xi
 When an object changes its
position, it has a velocity
Check Point 2-1
 Here are three pairs of
initial and final positions,
respectively along x axis.
Which pair give a negative
displacement?
 a) -3 m, + 5 m
 b) -3 m, -7 m
 c) 7 m, -3 m
 Ans:
 x=xf-xi
a) x=xf-xi=5-(-3)=+8
b) x=xf-xi=-7-(-3)=-4
c) x=xf-xi=-3-(+7)=-10
 Ans: b and c
Ch 2-4 Average Velocity, Average Speed
 Average Velocity vavg= x/ t
 vavg = (xf-xi) /(tf-ti)=Dispalcemnt/time
Average speed Savg: a scalar
Savg = total distance/ total time
Instantaneous Velocity v:
v= lim (x/ t)
t0
Position-time graph used to define
object position at any time t and
to calculate its velocity
 v is slope of the line on positiontime graph
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Check Point 2-3
 The following equations give the
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position x(t) of a particle in
four situations ( in each
equation , x is in meters, t is in
seconds, and t>0):
1) x = 3t-2
2) x=-4t2-2
3) x= 2/t2
4) x=-2
a) In which situation velocity v
of the particle constant?
b) In which v is in the negative
direction?
 Ans: v=dx/dt
1) v=+3 m/s
2) v=-8t m/s
3) v = -4/t3
4) v=0
 a) 1 and 4
 c) 2 and 3
Ch 2-6 Acceleration
 When an object changes its
velocity, it undergoes an
acceleration
 Average acceleration aavg
aavg = v/ t
= (vf-vi) /(tf-ti)
 Instantaneous acceleration a:
a= lim (v/ t)
t0
= dv/dt=d2x/dt2
 Velocity-time graph used to
define object velocity at any
time t and calculate its
acceleration
 a is slope of the line on
velocity-time graph
Ch 2-6 Acceleration
If the sign of the velocity and
acceleration of a particle are
the same, the speed of particle
increases.
If the sign are opposite, the
speed decreases.
Check Point 2-4
 A wombat moves along an x
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axis. What is the sign of its
acceleration if it is moving:
a) in the positive direction with
increasing speed,
b) in the positive direction with
decreasing speed
c) in the negative direction with
increasing speed,
d) in the negative direction with
decreasing speed?
Ans:
a)
b)
c)
d)
plus
minus
minus
plus
Ch 2-7 Constant Acceleration
 Motion with constant
acceleration has :
 Variable Slope of positiontime graph
 Constant Slope of velocity time graph
 Zero Slope of acceleration time graph
 For constant acceleration
a =aavg= (vf-vi)/(tf-ti)
vavg= (vf+vi)/2
Equations for Motion with Constant Acceleration
v=v0+at
x-x0=v0t+(at2)/2
v2=v02+2a(x-x0)
x-x0=t(v+v0)/2
x-x0 =vt-(at2)/2
Check Point 2-5
 The following equations give the
position x(t) of a particle in
four situations:
1) x=3t-4
2) x=-5t3+4t2+6
3) x=2/t2-4/t
4) x=5t2-3
 To which of these situations do
the equations of Table 2-1
apply?
Ans: Table 2-1 deals with
constant acceleration case
hence calculate
acceleration for each
equation:
1) a = d2x/dt2=0
2) a = d2x/dt2=-30t+8
3) a = d2x/dt2 = 12/t4-8/t2
4) a = d2x/dt2 = 10
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Ans: 1 and 4 ( constant
acceleration case)
Ch 2-9 Free Fall Acceleration
 Free fall acceleration ‘g’
dde to gravity
 g directed downward
towards Earth’s center
along negative y-axis
 with a = -g = -9.8
m/s2
 equations of motion
with constant
acceleration are valid
Check Point 2-6
a) What is the sign of
the ball’s
displacement for the
ascent, from the
release point to the
highest point?
B) What is it for the
descent , from the
highest point back
to to the release
point
c) What is the ball’s
acceleration at its
highest point?
a) (a) plus
(upward
displacement
on y axis);
(b) minus
(downward
displacement
on y axis);
(c) a = −g =
−9.8m/s2
Ch 2-10 Graphical Integration in Motion Analysis
 Integration of v-t graph
to obtain displacement x
x =x-x0=vt =v dt
but
v dt= area between
velocity curve and time
axis from t0 to t
 Integration of a-t graph
to obtain velocity v
Similarly v =v-v0= a dt
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a dt = area between
acceleration curve and
time axis from t0 to t
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