kinematics in one dimension chapter 2. expectations after this chapter, students will: distinguish...
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Kinematics in One Dimension
Chapter 2
Expectations
After this chapter, students will:
distinguish between distance and displacement distinguish between speed and velocity understand what acceleration is recognize situations in which the kinematic
equations apply
Expectations
After this chapter, students will:
be able to do calculations involving freely-falling objects
Kinematics
Mechanics Kinematics: the descriptive study of motion,
without concern for what causes the motion (space and time only)
Dynamics: the study of what causes and changes the motion of objects (force and inertia)
Displacement and Distance
Displacement: the vector that extends from the starting point to the ending point of an object’s travel.
Distance: the total length of an object’s path (a scalar quantity).
Displacement and Distance
An ant crawls along the edge of a meter stick.
He starts out at x0 = 50 cm
Then he moves along to x1 = 75 cm
Displacement and Distance
Now he changes his mind and goes toward the left,
to x2 = 20 cm …
… and reverses course yet again and goes to
xf = 30 cm, where he collapses in weariness.
Displacement and Distance
What distance did the ant travel? from x0 = 50 cm to x1 = 75 cm: d1 = 25 cm
from x1 = 75 cm to x2 = 20 cm: d2 = 55 cm
from x2 = 20 cm to xf = 30 cm: d3 = 10 cm
total distance = d1 + d2 + d3
= 25 cm + 55 cm + 10 cm = 90 cm
Displacement and Distance
What was the ant’s displacement?
x = xf – x0 = 30 cm – 50 cm = - 20 cm
The magnitude of the displacement was 20 cm; the direction was toward – X (or, as we drew it, leftward).
ant’s initial position
ant’s fi nal position
X
Displacement and Distance
Magnitude of displacement: x = xf – x0
Direction of displacement: from x0 to xf
SI unit: meter (m)
Speed and Velocity
Back to the ant-infested meter stick: but this time, we’ll take a stopwatch with us.
x0 = 50 cm
t0 = 0.0 s t1 = 5.0 s
x1 = 75 cm
Speed and Velocity
Note that speed is a scalar quantity.
Dimensions: length/time SI unit: m/s
x0 = 50 cm
t0 = 0.0 s t1 = 5.0 s
x1 = 75 cm
cm/s 5.0 s 5.0
cm 25
s 0.0 - s 5.0
cm 50 - cm 75 speed average
01
01
tt
xx
Speed and Velocity
Velocity is a vector: its direction is the same as that of the displacement vector.
Magnitude:
x0 = 50 cm
t0 = 0.0 s t1 = 5.0 s
x1 = 75 cm
ntdisplaceme 1
velocity average01
tt
cm/s 5.0 s 0.0 - s 0.5
cm 50 - cm 75
01
01
tt
xx
t
xv
Speed and Velocity
Now, the second part of the journey …
x2 = 20 cm
t2 = 19 s t1 = 5.0 s
x1 = 75 cm
Speed and Velocity
… and the third.
x2 = 20 cm
t2 = 19 s t f = 22 s
xf = 30 cm
Speed and Velocity
Times and positions:
Average speed:
t x
0.0 s 50 cm
5.0 s 75 cm
19 s 20 cm
22 s 30 cm
cm/s 4.1 s 0.0 - s 22
cm 10 cm 55 cm 25 speed ave.
speed ave.0
1
tt
xx
f
ii
Speed and Velocity
Times and positions:
Average velocity:
t x
0.0 s 50 cm
5.0 s 75 cm
19 s 20 cm
22 s 30 cm
left)(or X :direction
)(magnitude cm/s 0.91- s 0.0 - s 22
cm 50 - cm 30
0
0
v
tt
xx v
f
f
Acceleration
We’ve been talking about “average” velocity because velocity may not be constant as motion occurs.
If velocity changes, acceleration occurs.
Acceleration
Acceleration is the time rate of change of velocity.
If an object has a velocity v0 at time t0, and a different velocity v at a later time t, its acceleration is:
dimensions: length / time2 SI units: m/s2
0
0
tt
vva
Acceleration
Note that acceleration may or may not involve a change in speed. If either the magnitude (speed) or the direction (or both) of the velocity vector change, an acceleration has taken place.
In common speech, “acceleration” means that speed is increasing. In physics, acceleration means: speed increases; speed decreases; direction of motion changes; or some combination of these.
Acceleration
Consider an elevator car. At the initial time t0 = 0 s, it is descending with a velocity v0 = -1.2 m/s. It comes to a stop (v1 = 0 m/s) at its floor at time t1 = 1.5 s. The magnitude of its acceleration is
2
01
011 m/s 0.80
s 0 - s 1.5
m/s) (-1.2 - m/s 0
tt
vva v0 = -1.2 m/ s
v1 = 0 m/ sv1 – v0 = +1.2 m/ s
v0 (v1 – v0)
a1 = v1 – v0
t1 – t0
Acceleration
Passengers enter, the door closes, and the elevator starts upward, reaching a constant upward velocity of 1.7 m/s in a time of 3.0 s. The acceleration:
2
02
022 m/s 0.57
s 0 - s 3.0
m/s 0 - m/s .71
tt
vva
v0 = 0 m/ sv2 = +1.7 m/ s
v2 – v0 = +1.7 m/ s
v2 (v2 – v0)
a1 = v1 – v0
t1 – t0
Acceleration
The elevator slows as it approaches the 86th floor. At t3 = 177 s, its velocity (v3) is 1.7 m/s, upward; at t4 = 179.5 s, it stops (v4 = 0 m/s).
When the acceleration vector points in the opposite direction from the initial velocity vector, speed decreases.
2
34
344 m/s 0.68-
s 177 - s 179.5
m/s 1.7 - m/s 0
tt
vva
v4 = 0 m/ sv3 = +1.7 m/ s
v4 – v3 = -1.7 m/ s
v3 (v4 – v3)
a4 = v4 – v3
t4 – t3
Kinematic Equations
If acceleration is constant:
From the definition of acceleration:
atvv
ttt
ttavv
ttavvvtt
vva
0
0
00
000
0
:simply interval time thecall weIf
:for Solve
Kinematic Equations
If acceleration is constant, from the definition of average velocity:
tvvx
xvv
t
xv
0
0
2
1
:for Solve 2
Kinematic Equations
If acceleration is constant, we can substitute v from the first kinematic equation into the second:
20
000
00
2
1
22
1
2
1
2
1
attvx
tatvtatvvx
atvvtvvx
Kinematic Equations
If acceleration is constant, we can solve the definition of acceleration for t:
and substitute in the second kinematic equation:
a
vvt
t
vva 00
axvvva
vvx
a
vvvv
a
vvvvtvvx
2 :for Solve 2
22
1
2
1
20
222
02
00000
Kinematic Equations
Why do we insist that acceleration must be constant?
We are implicitly assuming that velocity is a linear function of time.
atvv
mxbybmxy
0
Kinematic Equations: Summary
Valid only when acceleration is constant.
axvv
attvx
tvvx
atvv
2
2
12
1
20
2
20
0
0
Graphical Representations of Motion
Object moving at constant velocity
t
x
t0 t
x0
xx-x0
t-t0
velocity = slope = x – x0
t – t0
Graphical Representations of Motion
Object moving with changing velocity
t
x
t0 t f
x0
xf
xf-x0
t f-t0
average velocity = xf – x0
t f – t0
instantaneous velocity = slope of tangent line at any point
Graphical Representations of Motion
Object moving with constant velocity
t
v
v
tt ft0
x = vt = area under velocity vs. time plot
Graphical Representations of Motion
Object moving with constant acceleration
t
v
tt ft0
v0
vf
slope = acceleration
x = v t = (v0 + v) t
(area under v vs. t plot)
12
not true if v is not a linear function of t (constant acceleration)
Free Fall ≈ Constant Acceleration
Objects at or near Earth’s surface experience a constant acceleration toward the Earth’s center, due to gravitation – if we ignore the resistive effect of the air through which it falls.
The magnitude of this acceleration is 9.807 m/s2.
The symbol of this acceleration is g.
Free Fall ≈ Constant Acceleration
In the presence of air, a falling object has an initial downward acceleration g; the acceleration decreases until the object is falling with a constant velocity called its terminal velocity.
The acceleration due to gravity, g, does not depend on the mass or the size of the object.
Free Fall and the Kinematic Equations
Consider a stone thrown straight upward with an initial velocity v0. It will ascend with its velocity decreasing until it stops momentarily; then it will accelerate back downward.
Let’s calculate the time, ta, of its ascent.
v0
Free Fall and the Kinematic Equations
First, we need to establish a (one-dimensional) coordinate system.
Our positive x-axis will point straight up.
That means the stone’s initial velocity is +v0, and its acceleration is –g.
v0
g
0
+X
Free Fall and the Kinematic Equations
In the first kinematic equation, we know everything except t. We know this because the final velocity, v, is zero (the stone stops momentarily at the top of its trajectory). Another way of looking at this is to say we know v = 0 when t = ta. We can substitute these into the first kinematic equation and solve it for ta.
v0
g
0
+Xatvv 0
Free Fall and the Kinematic Equations
v0
g
0
+X
g
vt
vgt
tgtv
ga
atvatvv
a
a
aa
a
0
0
0
00
:for Solve 0
0
Free Fall and the Kinematic Equations
Now, let’s calculate the height of the ascent.
We know that at the maximum height xmax, the stone stops momentarily; that is, v = 0 when x = xmax.
v0
g
0
+X
Free Fall and the Kinematic Equations
Substituting v = 0, a = -g, and x = xmax into our fourth kinematic equation: v0
g
0
+X
g
vxvgx
gxvaxvv
2 2
20 22
0max
20max
max2
02
02
Free Fall and the Kinematic Equations
As the stone falls back to the height (x = 0) from which it started, we wonder: how long does the descent take?
Will it take the same amount of time that the ascent did?
Let’s find out.
v0
g
0
+X
Free Fall and the Kinematic Equations
For the descent:
v0 = 0, x = -xmax, t = td, and a = -g.
Substitute these into the third kinematic equation:
v0
g
0
+X
g
xtxgt
gttxattvx
dd
dd
maxmax
2
2max
20
2
2
1
2
10-
2
1
Free Fall and the Kinematic Equations
That result doesn’t look much like our result for ta:
But, if we substitute our result for xmax into our equation for td:
ad tg
v
g
v
g
g
v
g
xt
g
vx
02
20
20
max
20
max
22
2
2
g
vta
0
Free Fall and the Kinematic Equations
By now, you probably suspect that the ascent and descent are fully symmetric, and would probably predict that the final velocity of the falling stone is equal in magnitude and opposite in direction to the initial velocity of the ascending stone.
We can check this using the first kinematic equation, substituting our calculated td for t, -g for a, zero for v0, and vf for v.
atvv 0
Free Fall and the Kinematic Equations
The result:
Thus, we have demonstrated that the ascent of an object thrown upward, and its descent to the same height, are indeed fully symmetric.
If the object returns to a different height … “all bets are off.”
0
00 0
vv
g
vgvatvv
f
f