**Unformatted text preview: **Chapter 2
Motion in One Dimension 2.1 position, Velocity, and Speed
2.2 Instantaneous Velocity and Speed
2.3 Acceleration
2.4 Freely Falling Objects
2.5 Kinematic Equations Derived from
Calculus. Kinematics Kinematics describes motion while ignoring
the agents that caused the motion
For now, we will consider motion in one
dimension Along a straight line We will use the particle model A particle is a point-like object, that has mass but
infinitesimal size Position Position is defined in
terms of a frame of
reference For one dimension the
motion is generally along
the x- or y-axis The object’s position is
its location with respect
to the frame of reference Position-Time Graph The position-time
graph shows the
motion of the
particle (car)
The smooth curve is
a guess as to what
happened between
the data points Displacement Displacement is defined as the change in
position during some time interval Represented as x x = xf - xi
SI units are meters (m), x can be positive or
negative Displacement is different than distance.
Distance is the length of a path followed by a
particle. Vectors and Scalars Vector quantities that need both -magnitude
(size or numerical value) and direction to
completely describe them We will use + and – signs to indicate vector
directions Scalar quantities are completely described
by magnitude only x fi
vaverg
tt Average Velocity The average velocity is the rate at which the
displacement occurs The dimensions are length / time [L/T]
The SI units are m/s
Is also the slope of the line in the position –
time graph Average Speed Speed is a scalar quantity same units as velocity
total distance / total time The average speed is not (necessarily)
the magnitude of the average velocity Instantaneous Velocity Instantaneous velocity is the limit of the
average velocity as the time interval
becomes infinitesimally short, or as the
time interval approaches zero
The instantaneous velocity indicates
what is happening at every point of time x
d
vx litm 0t Instantaneous Velocity The general equation for instantaneous
velocity is The instantaneous velocity can be
positive, negative, or zero Instantaneous Velocity The instantaneous
velocity is the slope
of the line tangent to
the x vs t curve
This would be the
green line
The blue lines show
that as t gets
smaller, they
approach the green
line Instantaneous Speed The instantaneous speed is the
magnitude of the instantaneous velocity
Remember that the average speed is
not the magnitude of the average
velocity v x
f
x
i
x
ax tt Average Acceleration Acceleration is the rate of change of the
velocity Dimensions are L/T2
SI units are m/s² v
d
v
d
x
a
lim tt 2
x
x
x
t
0 Instantaneous Acceleration The instantaneous acceleration is the limit of
the average acceleration as t approaches 0 Instantaneous Acceleration The slope of the
velocity vs. time
graph is the
acceleration
The green line
represents the
instantaneous
acceleration
The blue line is the
average acceleration Acceleration and Velocity When an object’s velocity and acceleration
are in the same direction, the object is
speeding up
When an object’s velocity and acceleration
are in the opposite direction, the object is
slowing down Acceleration and Velocity The car is moving with constant positive
velocity (shown by red arrows maintaining the
same size)
Acceleration equals zero Acceleration and Velocity Velocity and acceleration are in the same direction
Acceleration is uniform (blue arrows maintain the
same length)
Velocity is increasing (red arrows are getting longer)
This shows positive acceleration and positive velocity Acceleration and Velocity Acceleration and velocity are in opposite directions
Acceleration is uniform (blue arrows maintain the
same length)
Velocity is decreasing (red arrows are getting shorter)
Positive velocity and negative acceleration a
v
f
i
tvf
a
i 1D motion with constant
acceleration tf – t i = t x f
i
vavgx
tfi
vavgvavg
1
(2v0
f) 1D motion with constant acceleration In a similar manner we can rewrite equation
for average velocity: and than solve it for xf Rearranging, and assuming vxff
tv0t a
ii
12at2 1D motion with constant acceleration
(1) Using (1) and than substituting into equation for final
position yields
(2)
(2) Equations (1) and (2) are the basic kinematics
equations vx
( 2
x12(v
a
))t
2ffi200ffi 1D motion with constant acceleration
These two equations can be combined to
yield additional equations.
We can eliminate t to obtain Second, we can eliminate the acceleration a
to produce an equation in which acceleration
does not appear: tΔ
v a
fv
i
v
f
2
1
2
x
if2
2
a
x Kinematics with constant acceleration
- Summary Kinematic Equations - summary Kinematic Equations The kinematic equations may be used to
solve any problem involving one-dimensional
motion with a constant acceleration
You may need to use two of the equations to
solve one problem
Many times there is more than one way to
solve a problem Kinematics - Example 1 How long does it take for a train to come to rest if
it decelerates at 2.0m/s2 from an initial velocity of
60 km/h? Kinematics - Example 1 How long does it take for a train to come to
rest if it decelerates at 2.0m/s2 from an initial
velocity of 60 km/h?
km/h Using
to solve for t:
v vwe arearrange
t
f i vf vi
t
a Vf = 0.0 km/h, vi=60 km/h and a= -2.0 m/s2. 0 (60x 1000 / 3600)
t
8.3 s 2.0 A car is approaching a hill at 30.0 m/s when its engine
suddenly fails just at the bottom of the hill. The car moves
with a constant acceleration of –2.00 m/s2 while coasting up
the hill. (a) Write equations for the position along the slope
and for the velocity as functions of time, taking x = 0 at the
bottom of the hill, where vi = 30.0 m/s. (b) Determine the
maximum distance the car rolls up the hill. ti
0 (a) Take
at the bottom of the hill where
xi=0,
=0 vi=30m/s,
=30m/s a=-2m/s2. Use these values in the
general equation 1 2
x f xi vi t at
2 1
x f 0 30.0t m s 2.00 m s 2 t 2
2 x f 30.0t t 2 m (a) Take ti 0 at the bottom of the hill where
xi=0, vi=30m/s, a=-2m/s2. Use these values in the
general equation 2 v f vi at 30.0 m s 2.00 m s t
v f 30.0 2.00t m s The distance of travel, xf, becomes a maximum, xmax
when when v f 0 (turning point in the motion). Use the expressions found in part (a) for v f 30.00 2.00t m / s v f 0 when t=15sec. xmax 30.0t t 2 m 30.0 15.0 15.0 225 m
2 Graphical Look at Motion:
displacement-time curve The slope of the
curve is the velocity
The curved line
indicates the
velocity is changing Therefore, there is
an acceleration Graphical Look at Motion:
velocity-time curve The slope gives the
acceleration
The straight line
indicates a constant
acceleration Graphical Look at Motion:
acceleration-time curve The zero slope
indicates a constant
acceleration Freely Falling Objects A freely falling object is any object
moving freely under the influence of
gravity alone.
It does not depend upon the initial
motion of the object Dropped – released from rest
Thrown downward
Thrown upward Acceleration of Freely Falling Object The acceleration of an object in free fall is
directed downward, regardless of the initial
motion
The magnitude of free fall acceleration is
g = 9.80 m/s2 g decreases with increasing altitude
g varies with latitude
9.80 m/s2 is the average at the Earth’s surface Acceleration of Free Fall We will neglect air resistance
Free fall motion is constantly accelerated
motion in one dimension
Let upward be positive
Use the kinematic equations with
ay = g = -9.80 m/s2 Free Fall Example Initial velocity at A is upward (+)
and acceleration is g (-9.8 m/s2)
At B, the velocity is 0 and the
acceleration is g (-9.8 m/s2)
At C, the velocity has the same
magnitude as at A, but is in the
opposite direction A student throws a set of keys vertically upward to her
sorority sister, who is in a window 4.00 m above. The keys
are caught 1.50 s later by the sister's outstretched hand.
(a) With what initial velocity were the keys thrown? (b)
What was the velocity of the keys just before they were
caught? A student throws a set of keys vertically upward to her
sorority sister, who is in a window 4.00 m above. The keys
are caught 1.50 s later by the sister's outstretched hand.
(a) With what initial velocity were the keys thrown? (b)
What was the velocity of the keys just before they were
caught?
(a) 1 2
y f yi vi t at
2 4.00 1.50 vi 4.90 1.50 vi 10.0 m s upward 2 A student throws a set of keys vertically upward to her
sorority sister, who is in a window 4.00 m above. The keys
are caught 1.50 s later by the sister's outstretched hand.
(a) With what initial velocity were the keys thrown? (b)
What was the velocity of the keys just before they were
caught?
(b) v f vi at 10.0 9.80 1.50 4.68 m s
v f 4.68 m s downward A ball is dropped from rest from a height h above the
ground. Another ball is thrown vertically upwards from the
ground at the instant the first ball is released. Determine the
speed of the second ball if the two balls are to meet at a
height h/2 above the ground. A ball is dropped from rest from a height h above the
ground. Another ball is thrown vertically upwards from the
ground at the instant the first ball is released. Determine the
speed of the second ball if the two balls are to meet at a
height h/2 above the ground. 1 2
1-st ball: y1 h gt
2
2-nd
ball: 1 2
y2 vi t gt
2 h
y1 2 h
1 2 h gt
2
2 h
h 1 h vi g 2
g 2 g t vi gh h
g A freely falling object requires 1.50 s to travel the last 30.0 m
before it hits the ground. From what height above the ground
did it fall? A freely falling object requires 1.50 s to travel the last 30.0 m
before it hits the ground. From what height above the ground
did it fall?
Consider the last 30 m of fall. We find its speed 30 m
above the ground:
1 y f yi v yi t 2 ay t 2 1
2
2
0 30 m vyi 1.5 s 9.8 m s 1.5 s 2
30 m 11.0 m
vyi 12.6 m s.
1.5 s A freely falling object requires 1.50 s to travel the last 30.0 m
before it hits the ground. From what height above the ground
did it fall?
Now consider the portion of its fall above the 30 m point. We
assume it starts from rest
v 2 v 2 2a y y
yf 12.6 m s 2 0 2 9.8 m s y y Its original height was then: yi y f i 2 160 m 2 s 2
19.6 m s 2 8.16 m. 30 m 8.16 m 38.2 m Motion Equations from Calculus Displacement
equals the area
under the velocity –
time curve lim tn 0 v
n tf xn tn v x (t )dt
ti The limit of the sum
is a definite integral Kinematic Equations – General
Calculus Form dv x
ax dt
t v xf v xi a x dt
0 dx
vx dt
t x f xi v x dt
0 vxxfixfxi
avixt
12xt2 Kinematic Equations – Calculus Form
with Constant Acceleration The integration form of vf – vi gives The integration form of xf – xi gives The height of a helicopter above the ground is given by
h = 3.00t3, where h is in meters and t is in seconds. After
2.00 s, the helicopter releases a small mailbag. How long
after its release does the mailbag reach the ground? The height of a helicopter above the ground is given by
h = 3.00t3, where h is in meters and t is in seconds. After
2.00 s, the helicopter releases a small mailbag. How long
after its release does the mailbag reach the ground? y 3.00t 3 t 2.00 s dy
vy 9.00t 2 36.0 m s dt y 3.00 2.00 24.0 m
3 The height of a helicopter above the ground is given by
h = 3.00t3, where h is in meters and t is in seconds. After
2.00 s, the helicopter releases a small mailbag. How long
after its release does the mailbag reach the ground?
The equation of motion of the mailbag is: 1 2
1
yb
yb ybi vi t gt 24.0 36.0t 9.80 t 2
2
2
2
t 7.96 s 0 24.0 36.0t 4.90t 0 Automotive engineers refer to the time rate of change of
acceleration as the "jerk." If an object moves in one
dimension such that its jerk J is constant, (a) determine
expressions for its acceleration ax(t), velocity vx(t), and
position x(t), given that its initial acceleration, speed, and
position are axi , vxi, and xi , respectively. (b) Show that
2
a x2 a xi 2 j (v x v xi ) Automotive engineers refer to the time rate of change of acceleration
as the "jerk." If an object moves in one dimension such that its jerk J
is constant, (a) determine expressions for its acceleration ax(t), velocity
vx(t), and position x(t), given that its initial acceleration, speed, and
position are axi , vxi, and xi , respectively. (b) Show that 2
a x2 a xi 2 j (v x vxi ) (a) da
J constant
dt a J dt Jt c1
a Jt ai da Jdt
a ai when t 0 c1 ai Automotive engineers refer to the time rate of change of acceleration as the "jerk." If an
object moves in one dimension such that its jerk J is constant, (a) determine
expressions for its acceleration ax(t), velocity vx(t), and position x(t), given that its initial
acceleration, speed, and position are axi , vxi, and xi , respectively. (b) Show that 2
a x2 a xi 2 j (v x v xi ) (a) dv
a
dt
dv adt
v v vi adt when t 1 2
Jt ai dt Jt ai t c2
2
1 2
0 c2 vi
v
Jt ai t vi 2 Automotive engineers refer to the time rate of change of acceleration as the
"jerk." If an object moves in one dimension such that its jerk J is constant,
(a) determine expressions for its acceleration ax(t), velocity vx(t), and
position x(t), given that its initial acceleration, speed, and position are axi , vxi,
and xi , respectively. (b) Show that 2
a x2 a xi 2 j (v x vxi ) dx
(a)
dt
dx vdt
v x 1 2
Jt a
t v
i
i dt 2 1 ai t 2 vi t c3
2 vdt 1
Jt 3
6
x xi when
x t 0 c3 x i x 1 3 1
Jt ai t 2 vi t xi
6
2 (b) a 2 Jt ai 2 J 2t 2 ai2 2 Jai t a 2 ai2 J 2 t 2 2 Jai t
a 2 ai2 1 2J Jt 2 ai t 2 Recall the expression for vv: 1 Jt 2 ai t vi
2
1 2 v vi 2 Jt ai t 2 a 2
ai 2 J v vi The acceleration of a marble in a certain fluid is
proportional to the speed of the marble squared, and is
given (in SI units) by a = –3.00 v2 for v > 0. If the marble
enters this fluid with a speed of 1.50 m/s, how long will it
take before the marble's speed is reduced to half of its
initial value? The acceleration of a marble in a certain fluid is proportional to the
speed of the marble squared, and is given (in SI units) by a = –3.00 v2
for v > 0. If the marble enters this fluid with a speed of 1.50 m/s, how
long will it take before the marble's speed is reduced to half of its initial
value? dv
a 3.00v 2
dt
dv 3.00v 2
dt vi 1.50 m s v v vi v 2 dv 3.00 t dt t 0 1 1
1 1 3.00t or 3.00t .
v vi
v vi The acceleration of a marble in a certain fluid is proportional to the
speed of the marble squared, and is given (in SI units) by a = –3.00 v2
for v > 0. If the marble enters this fluid with a speed of 1.50 m/s, how
long will it take before the marble's speed is reduced to half of its initial
value? dv
a 3.00v 2
dt vi
v
2 vi 1.50 m s 1
t 0.222 s
3.00vi A test rocket is fired vertically upward from a well. A
catapult gives it initial velocity 80.0 m/s at ground level. Its
engines then fire and it accelerates upward at 4.00 m/s2
until it reaches an altitude of 1 000 m. At that point its
engines fail and the rocket goes into free fall, with an
acceleration of –9.80 m/s2. (a) How long is the rocket in
motion above the ground? (b) What is its maximum
altitude? (c) What is its velocity just before it collides with
the Earth? A test rocket is fired vertically
upward from a well. A catapult
gives it initial velocity 80.0
m/s at ground level. Its
engines then fire and it
accelerates upward at 4.00
m/s2 until it reaches an altitude
of 1 000 m. At that point its
engines fail and the rocket goes
into free fall, with an
acceleration of –9.80 m/s2. (a)
How long is the rocket in
motion above the ground? (b)
What is its maximum altitude?
(c) What is its velocity just
before it collides with the
Earth? Let point 0 be at ground level and
point 1 be at the end of the engine
burn. Let point 2 be the highest
point the rocket reaches and point 3
be just before impact. The data in
the table are found for each phase
of the rocket’s
motion.
2
(0 to 1): v 80.0 2 2 4.00 1 000 f so v f 120 m s 120 80.0 4.00 t t 10.0 s A test rocket is fired vertically
upward from a well. A catapult
gives it initial velocity 80.0
m/s at ground level. Its
engines then fire and it
accelerates upward at 4.00
m/s2 until it reaches an altitude
of 1 000 m. At that point its
engines fail and the rocket goes
into free fall, with an
acceleration of –9.80 m/s2. (a)
How long is the rocket in
motion above the ground? (b)
What is its maximum altitude?
(c) What is its velocity just
before it collides with the
Earth? (1 to 2): 0 120 2 9.80 x f xi
2 x f xi 735 m 0 120 9.80t t 12.2 s This is the time of maximum height of
the rocket. A test rocket is fired
vertically upward from a
well. A catapult gives it
initial velocity 80.0 m/s at
ground level. Its engines
then fire and it accelerates
upward at 4.00 m/s2 until it
reaches an altitude of 1 000
m. At that point its engines
fail and the rocket goes into
free fall, with an acceleration
of –9.80 m/s2. (a) How long
is the rocket in motion above
the ground? (b) What is its
maximum altitude? (c) What
is its velocity just before it
collides with the Earth? v 2f 0 2 9.80 1 735 (2 to 3): v f 184m / s v f 184 9.80 t
(a):
(b):
(c): t 18.8 s ttotal 10 12.2 18.8 41.0 s x f xi total 1.73 km vfinal 184 m s t
0 Launch x v a 0.0 0 80 +4.00 #1 End Thrust 10.0 1 000 120 +4.00 #2 Rise Upwards 22.2 1 735 0 –9.80 #3 Fall to Earth 41.0 0 –184 –9.80 An inquisitive physics student and mountain climber
climbs a 50.0-m cliff that overhangs a calm pool of water.
He throws two stones vertically downward, 1.00 s apart,
and observes that they cause a single splash. The first
stone has an initial speed of 2.00 m/s. (a) How long after
release of the first stone do the two stones hit the water?
(b) What initial velocity must the second stone have if
they are to hit simultaneously? (c) What is the speed of
each stone at the instant the two hit the water? inquisitive
physics
student
and
mountain
climber climbs a 50.0-m
cliff that overhangs a calm
pool of water. He throws two
stones vertically downward,
1.00 s apart, and observes
that they cause a single
splash. The first stone has
an initial speed of 2.00 m/s.
(a) How long after release of
the first stone do the two
stones hit the water? (b)
What initial velocity must
the second stone have if
they
are
to
hit
simultaneously? (c) What is
the speed of each stone at
the instant the two hit the
water?
An (a) y f vi1t 1 2
1
at 50.0 2.00t 9.80 t 2
2
2 2 4.90t 2.00t 50.0 0
t 2.00 2.00 2 4 4.90 50.0
2 4.90 Only the positive root is physically
meaningful: t 3.00 s
after the first stone is thrown. (b) 1 2
y f vi 2 t at
2 t 3.00 1.00 2.00 s 1
2
50.0 vi 2 2.00 9.80 2.00 2
vi 2 15.3 m s
downward
(c) v1 f vi1 at 2.00 9.80 3.00 31.4 m s v2 f vi 2 at 15.3 9.80 2.00 34.8 m s downward
downward ...

View
Full Document

- Spring '19
- Ha
- Acceleration, Velocity