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Unformatted text preview: CHAPTER 2
MOTION IN ONE DIMENSION
• Why objects move
• Displacement and velocity o Average velocity
o Average speed
o Relative velocity
o Instantaneous velocity
o Reading displacementtime graphs
• Acceleration o Average acceleration
o Instantaneous acceleration
o Reading velocitytime graphs
o Free fall
• Equations of motion using calculus (You study) This chapter is about MOTION ... so, why do objects
move?
It is FORCE that causes a change in motion (starting,
stopping, going round corners, etc).
It is the force of the road
on the tires that cause a
car to move!
It is the force of gravity
downward that makes the cat
fall! (Its weight.)
It is the force exerted by the
typists fingers that makes the
keys move!
For the moment we won’t worry about what causes
motion but the rules that tell us how objects move. Displacement and distance traveled
Velocity and average speed 0 x2
finish x1
start d1 The displacement is ( x 2 − x1 ) ( = ∆x ). 0 d1 0 x1
start d2
x2 x3
finish The distance traveled is ( x 3 − x1 ) = (d1 + d 2 ).
d1
x1
start t2 • Average speed = displacement ∆x
=
.
time
∆t distance (d1 + d 2 )
=
.
time
∆t But, in this example, (d1 + d 2 ) > ( x 3 − x1 ), so d2 x3
finish x2 ∆t = ( t 2 − t 1 ) • Average velocity v av = The displacement is ( x 3 − x1 ) . 0 x3
finish t1 The distance traveled is ( x 2 − x1 ) also. x1
start d2 x2 The displacement is ( x 3 − x1 ) .
The distance traveled is (d1 + d 2 ) .
• distance traveled ≥ displacement
• displacement can be positive or negative. average speed > average velocity Dimensions: velocity and speed ⇒
Units: m s , km h , ft s , mi h . [L ]
[ T] Variation of velocity around a circular track for a car
traveling at constant speed.
A The displacement of the
s R car from A is given by θ R DISCUSSION PROBLEM [2.1]:
Estimate the average velocity of the winning car at the
Indianapolis 500 ... (total distance 500 miles in about 2 12
hours). s2 = R 2 + R 2 − 2R ⋅ R cos θ • Car i.e., s = R 2(1 − cos θ) . •
B Note that the displacement is periodic in θ . If the speed is constant, then θ increases
linearly with time, so the magnitude of the velocity
v = s t is periodic in time. ( ) Instantaneous velocity
v ( t) Time
A B A B A B 2.50km
x1 x2
∆t1
∆t 2 (a) From x1 → x 2:
Question 1: A runner runs for 2.5km in a straight line in
= 9.0min and takes 30.0mins to walk back to the starting
point. (a) What is the runner’s velocity for the first
9.0min? (b) What is the average velocity for the time Run
Walk v av = v av = spent walking? (c) What is the average velocity for the
= trip? (e) Plot a position versus time graph for this
problem. (c) v av = ∆t 1 2.5km 2500 m
=
= 4.63m/s
9 min 9 × 60s (b) From x 2 → x1: whole trip? (d) What is the average speed for the whole ( x 2 − x1 ) ( x1 − x 2 )
∆t 2 −2.5km −2500 m
=
= −1.39 m/s.
30min 30 × 60s displacement
=0
total time (d) av. speed = distance 5000 m 5000 m
=
=
total time 39min 2340s = 2.14 m/s. (e) Position versus time graph for this problem; note that
the displacement is the runner’s position relative to the
starting point.
Position (km)
3.0 B
2.0 DISCUSSION PROBLEM [2.2]:
1.0
C A 0
0 10 • The slope of AB is 20
30
Time (min) 40 2.5km
⇒ 4.63m/s, i.e., the average
9 min velocity for the first 9min.
−2.5km
⇒ −1.39 m/s, i.e., the
• The slope of BC is
30 min
average velocity while walking.
Thus, velocity is obtained from the slope of a
displacement vs time plot. You lean out of a window and throw a ball straight down.
It bounces off the sidewalk and returns to your hand 2.0s
later. (a) What is the average velocity of the ball? (b)
What is its average speed? Relative velocity
1 2 v 21
v wg If the velocity of the walkway relative to the ground (i.e.,
the stationary observer) is v wg, and the velocity of is v 21, then the velocity of 2 with
respect to a stationary observer (the relative velocity) is: with respect to 1 2 ( v wg + v 21 )
Note, if 2 is walking in the opposite direction, his velocity relative to the observer is ( v wg − v 21 )
What “happens” in that case if v wg = v 21 ?
This is called a Galilean transformation. Question 2: Two trains, 75km apart and traveling at
15km/h, are approaching each other on parallel tracks.
A bird flies back and forth at 20km/h between the trains
until the trains pass each other. For how long does the
bird fly? How far does the bird fly? However, velocities often
P2 v1 = 15 km/h P2′
P1 v 2 = 15 km/h
75km Average speed of bird ⇒ distance flown s
=.
time
t change ... ! Think of a ball
bouncing on the ground: Displacement (x)
P2 y2 P2′ ∴ s = 20 × t . ∆x But what is t? ... it’s the time before the trains meet.
Imagine you are sitting on the front of the left hand train, y1 P1 ∆t then the right hand train is approaching you at 15km/h + 15km/h = 30 km/h .
So, the time taken until the trains pass is
distance they travel 75
=
= 2.50h.
relative speed
30 ∴ s = 20 × 2.5 = 50 km. t1 t2 Time (t) Average velocity from P1 to P2 is:
∆x x 2 − x1
v1→2 =
=
.
∆t
t 2 − t1
Note that this is less than:
v1→2′ so the average velocity continually changes. Instantaneous velocity “Reading” a displacement  time graph: What is the instantaneous velocity at the point P? Displacement (x)
C
• Displacement (x)
• P2 B D• •E t Instantaneous velocity = slope = dx
dt ∆x A• P ∆t
t1 t2 Time (t) We define the instantaneous velocity at point P ( x1, t1 ) as
the slope of the tangent at the point P.
∆x dx ∴ v ( t ) = Limit ∆t→0
=
,
∆t dt P i.e., the first derivative of x with respect to t at the point tA = 0 v>0 x=0
x
v>0 tB v=0 tC
v<0 tD
tE x v<0 P. Note that normally the instantaneous velocity (the
slope) is a function of time, i.e., v ⇒ v ( t ). The length of the red line shows the magnitude (i.e.,
size) of the velocity x
x
x DISCUSSION PROBLEM [2.3]:
A physics professor is walking to campus when he
realizes he’s forgotten a book and he returns home. His
displacement as a function of time is shown below. At
which point(s) is his velocity: (a) zero? (b) constant and
positive? (c) constant and negative? (d) increasing in Question 3: The position of an object depends on time
according to to the equation x ( t ) = t 2 − 5.0 t + 1.0,
where x is in meters and t in seconds. (a) Find the magnitude? (e) decreasing in magnitude? displacement and average velocity for the time interval
Displacement (x) 3.0s ≤ t ≤ 4.0s. (b) Find the instantaneous velocities at D
C t = 3.0s and t = 4.0s. (c) At what time is the E instantaneous velocity zero?
B
A F
time x( t) x ( t ) = t 2 − 5.0 t + 1.0 In the previous problem, the velocity changed with time, v ( t ) = 2.0 t − 5.0 . t When t1 = 3.0s:
2 x1 = (3.0)2 − 15.0 + 1.0 = −5.0 m. Whenever velocity changes we have ... acceleration. When t 2 = 4.0s: 1 v1 (a) v av = ∆x
, where ∆x = displacement
∆t = x 2 − x1 = −3.0 − ( −5.0) = 2.0 m and ∆t = t 2 − t1 = 1.0s .
2.0 m
∴ v av =
= +2.0 m/s.
1.0s
(b) Instantaneous velocity v =
At t1 = 3.0s: dx
= 2.0 t − 5.0.
dt v1 = 6.0 − 5.0 = 1.0 m/s . At t 2 = 4.0s: v 2 = 8.0 − 5.0 = 3.0 m/s.
dx
(c) When v = 0 ,
= 0, i.e., 2.0 t − 5.0 = 0 .
dt
∴t = 5.0
= 2.5s
2.0 x2 t1 x 2 = (4.0) − 20.0 + 1.0 = −3.0 m. v2 x1 2 t2 x v − v1
Average acceleration ⇒ a = 2
.
t 2 − t1
Dimension: [L ] 1
⇒ [L][ T]−2
[ T][ T] Units: m / s2, ft / s2 , etc.
Instantaneous acceleration a = Limit ∆t→0
v but v =
∆v P t ∆t ∴a = ∆v dv
=,
∆t dt dx
.
dt d2x
.
dt 2 Instantaneous acceleration a = C v ( t) B • • •D dv
.
dt In the previous example, the acceleration was a function
of time, i.e., a ⇒ a ( t ) . However, we will only consider •
•
Velocity  time graph t F A E motion at constant (or uniform) acceleration:
dv
⇒ constant
i.e., a =
dt
Example: • a>0
tA = 0 Free fall v ( t) v=0 Slope = dv dt = a v a>0 tB tC at v>0 vo a=0 vo v>0
a<0 tD a<0
v=0
a<0
tF v<0 t From earlier: v>0 tE Time 0 a= dv v − v o
=
dt
t ∴ v = v o + at Average velocity: v av = 1
( v + v o ) *.
2 * Only true at constant acceleration. Motion at constant acceleration
initial
vo final
v xo x to = 0 t x We have from before:
1
∆x = x − x o = v av t = ( v o + v ) t ... ... (1)
2 But v = v o + at . Question 4: A car, starting from rest at x = 50 m,
accelerates at a constant rate of 8 m/s2 . (a) What is its 1
∴ x − x o = (2 v o + at ) t ,
2
1
i.e., x − x o = v o t + at 2
2 Also t = velocity after 10s? (b) How far has it traveled in those 0 ≤ t ≤ 10s? v − vo
.
a Substitute in (1), then x − x o = 1
(v − vo ) .
(vo + v)
2
a After rearrangement we get v 2 = v o2 + 2a ( x − x o ) 10s? (c) What is its average velocity over the interval initial
vo = 0 final
v=? xo = 50m Galileo Galilei (15641642): x=? to = 0 x xo : v o = 0 : t = 0 t = 10s Using the equations of motion: x:v:t (a) v = v o + at = 0 + 8 × 10 = 80m/s.
1
(b) x − x o = v o t + at 2 .
2 1
∴ x = x o + v o t + at 2
2
1
= 50 + × 8 × (10)2 = 450 m.
2 •
In “free fall”, i.e., in the absence of air resistance, all
objects fall with a constant acceleration (downward).
a ⇒ g = 9.81m/s2 (32.2 ft/s2 ) 1
( v o + v ) = 40m/s.
2 v = 9.81m/s
v = 19.6 m/s after 3s (c) v av = after 1s
after 2s Starting from rest: v = 29.4 m/s The velocity of the object increases by 9.81m/s each
second during free fall, i.e., ** Only true at constant acceleration ** v = v o + at = gt . =0
1
1
Also y − y o = v o t + at 2 = gt 2 .
2
2 and they all objects fall the same distance in the same
time:
40 CAREFUL: If the yaxis is chosen so that y increases 1
( y − yo ) = gt2 = 4.91t2
2 The distance fallen
by an object in
freefall, starting
from rest. upward, then as the object falls, y < y o, i.e., g = −9.81m/s2. On the other hand, if you choose y to 20 increase downward, then y > y o, and g = +9.81m/s2 .
time (s) Note that the velocity, v, and displacement ( y − y o ) do 0 1 2 3 not depend on the mass of the object! So, in the absence
of air resistance, all objects fall at the same rate:
An apple and feather 30 released at the same v = gt = 9.81t time, in the absence of 20
The velocity of an object in
freefall, starting from rest. fall). 10
time (s)
0 air resistance (free 1 2 3 Question 5: A rocket is fired vertically with a constant
I05_10.mov upward acceleration of 20.0 m/s2 . After 25.0s, the engine Apollo 15 astronaut David Scott drops a geological is shutoff and the rocket continues rising. The rocket hammer (1.32kg) and a falcon feather (0.03kg) at the eventually stops rising and then falls back to the ground. same time on the Moon (July/August 1971). Find (a) the highest point the rocket reaches, (b) the total
time the rocket is in the air, and (c) the speed of the
rocket just before it hits the ground. Neglect air
resistance, 1.895 × 10 4 m : v 2 = 0 : t 2 y2 : v 2 = 0 : t 2 Take upward
direction as
positive. 6250m : 500m/s : t1 = 25.0s y1 : v 1 : t1 y = 0 : v 3 : t3 yo = 0 : v o = 0 : t o = 0 y o = 0 : a = +20.0 m/s2 : t1 = 25.0s : v o = 0 : y1 = ?
1
1
(a) y1 − y o = v o t1 + at12 = × 20.0 × (25.0)2
2
2
= 6250 m. and v1 = v o + at1 = 20.0 × 25.0 = 500 m/s.
Now, between y1 and y 2, a = − g = −9.81m/s2 (free fall). v 22 = v12 + 2a ( y 2 − y1 ),
i.e., ( y 2 − y1 ) = v 2 − v12 2 2 2a = −500
2 × ( −9.81) (b) If the time between y1 and y 2 is t 2 − t1 = δt then
v − v1 −500
v2 = v1 + aδt , i.e., δt = 2
=
= 51.0s .
a
−9.81
So, the total “climb” time ⇒ t 2 = t1 + δt
= 25.0 + 51.0 = 76.0s .
The time to fall is t 3 − t 2 = ∆t , then
1
( y o − y 2 ) = v2 ∆t + a( ∆t )2
2 2( y o − y 2 )
2 × (0 − 1.90 × 104 )
=
= 62.2s .
i.e., ∆t =
a
−9.81
∴ Total time ⇒ t 3 = t 2 + ∆t = 76.0 + 62.2 = 138.2s . = 12700 m.
The maximum height is ( y 2 − y o )
= ( y 1 − y o ) + ( y 2 − y1 )
= 6250 + 12700 m = 1.895 × 10 4 m y3 = 0 : v 3 : t 3 yo = 0 : v o = 0 : t o = 0 (19.0 km). (c) Let final velocity be v 3, then v3 = v2 + a∆t
= 0 − 9.81 × 62.2 = −610.2 m/s .
? Measurement of reaction time ... 1
From earlier: d = gt 2.
2
∴t = d
4.91 Measurement of reaction time ... d 0.50 Reaction distance = speed × reaction time
x (m) 0.40
0.30 mi/h
30 0.10
00 reaction time (s)
0.1 0.2 0.3 0.4 ** Also “holds” for socalled hangtime
See useful notes on website
Relate reaction time to distance travelled driving a car Reaction distance = speed × reaction time . 17.9
22.4 60 0.20 13.4 50 d
4.91 m/s 40 t= 26.8 70 31.3 Question 6: It is often claimed that basketball superstars
have hang times of at least 1 second. Is that reasonable? If the hang time is 1 second, it means that the “rise” time
and “fall” time are each 12 s. But a fall time of 12 s
means the distance fallen is
1
1
d = gt 2 = × 9.81 × (0.5)2 = 1.23m,
2
2
which, in turn, is a very difficult “height” to achieve.
Measurements made on Michael Jordan give him a hang
time of 0.92s. So, a hang time close to 1s is possible but
unlikely.
For more details go to
http://www.bizesor.com/brenkus/
and click on the FSN sport link. Equations of motion using calculus
Earlier we derived the equations of motion graphically.
We can also use calculus. We start with the definition of
dv
= a , where v is the velocity and t is
acceleration, i.e.,
dt
time. If we assume that a is constant, we can integrate
this expression with respect to t to get v: v = ∫ dv = ∫ adt = at + C1,
where C1 is an integration contant. To get C1 we consider
the initial conditions, i.e., what is v when t = 0? Let us To get C2 we use the initial conditions, i.e., what is x
when t = 0? Let us assume that x = x o when t = 0. Then
C2 = x o . So, we get
1
1
x = x o + v o t + at 2 ⇒ ( x − x o ) = v o t + at 2 ,
2
2 which is the same expression we obtained earlier.
If we know how the velocity of an object varies with
time, we can also use calculus to determine the
displacement over a period of time. For example, put v = v o when t = 0, then C1 = v o, so
v = v o + at , v ( t) which is the same as an equation we derived earlier.
dx
Also, we know that v = . So,
dt
x = ∫ dx = ∫ vdt = ∫ ( v o + at )dt 1
= v o t + at 2 + C2 ,
2 where C2 is another constant. t t1 t2 given the velocitytime plot above, what is the
displacement of the object over the time period t1 → t 2? We split the time period t1 → t 2 into a large number of
equal time intervals, ∆t .
v ( t) If ∆t is very small, then vi the displacement from v1 time t1 to t1 + ∆t is
∆x1 ≈ v1∆t . t
t1 ∆t t2 So, the total Question 7: An object, starting from rest at x o = 1.00 m,
experiences a nonconstant acceleration given by ( x ( t 2 ) − x ( t1 ) ≈ v1∆t + v 2 ∆t K ≈ ∑i ( v i ∆t ) .
t2 x ( t 2 ) − x ( t1 ) = limit ( ∑i v i ∆t ) = ∫ vdt ,
t1 i.e., the displacement is the area under the velocitytime
curve from t1 to t 2. So, if we know the equation of the
velocitytime curve we can determine the displacement
over any time period.
Note, the above expression is true even if the
acceleration is changing, i.e., when our earlier
equations of motion cannot be applied. At any time t, what is (a) the instantaneous velocity, and
(b) the position of the object? (c) What is the The sum becomes exact if ∆t → 0 . Then
∆t→0 ) a ( t ) = 1.50 + 0.20 t m/s2 . displacement from t1 → t 2 is displacement of the object over the time period from
2.0 → 4.0s? Because the acceleration is not constant we cannot apply
the simple equations of motion. (a) If a ( t ) = 1.50 + 0.20 t , then dv
= 1.50 + 0.20 t .
dt ∴ v = ∫ dv = ∫ (1.50 + 0.20 t )dt 1
= 1.50 t + 0.20 t 2 + C1,
2 Using the result in (b)
t2 )[ (
t i.e., v ( t ) = 1.50 t + 0.10 t 2 , 2
2
3
∫ 1.50 t + 0.10 t dt = 0.75t + 0.033t + C2
1 since the initial velocity of object is zero. 4
2 = [(0.75 × 16 + 0.033 × 64) − (0.75 × 4 + 0.033 × 8)] (b) If v = dx
, the position is given by
dt ( = 10.85m. Check using (b): ) x ( t = 2) = 1.00 + 0.75 × 22 + 0.033 × 23 = 4.26 m. 1
1
= 1.50 t 2 + 0.10 t 3 + C2,
3 2 x ( t = 4) = 1.00 + 0.75 × 42 + 0.033 × 43 = 15.11m x = ∫ dx = ∫ vdt = ∫ 1.50 t + 0.10 t 2 dt 2 ∴ ∆x = 15.112 − 4.264 = 10.85m.
3 i.e., x ( t ) = 1.00 + 0.75t + 0.033t ,
since the initial position of the object is x o = 1.00 m. We can also determine the area under the curve by
“counting rectangles” on the velocitytime graph ... (c) By definition, the displacement is
t2 t2 t1 t1 ( ) ∆x t1 → t 2 = ∫ vdt = ∫ 1.50 t + 0.10 t 2 dt . Question 8: A mass on the end of a spring is oscillating
v ( m/s) upanddown and its position relative to the tabletop is y = y o − A cos(ωt ). v ( t) = 1.50 t + 0.10 t 2 Find the general expressions for (a) the instantaneous
velocity, and (b) the instantaneous acceleration of the
mass. If y o = 0.50 m, A = 0.30 m and ω = 5.00s−1, what
are (c) the maximum velocity, and (d) the maximum
acceleration of the mass?
t(s) Area of each rectangle = 1.0 m/s × 0.50s = 0.50 m. In the
range 2.0s ≤ t ≤ 4.0s, there are 18 complete rectangles
plus parts that make approximately 4 more. Therefore,
there are ~ 22 rectangles, so the area under the curve is ~ 22 × 0.50 m ≈ 11m. yo A y Actual area by integration is 10.85m
intitial conditions (b) the instantaneous acceleration is a ( t ) =
y( t) = Aω 2 cos(ωt ). 1.0m yo dv d 2 y
=
dt dt 2 A a ( m/s 2 )
y t(s) 0 1.0 2.0 3.0 intitial conditions 4.0 10
5 t(s) 0 (a) the instantaneous velocity is v ( t ) = dy
dt −5
−10 = + Aω sin(ωt ) .
v ( m/s) (d) Maximum acceleration when ωt = 0 , 1.5 t(s) ∴ a max = Aω 2 = 7.50m/s2 . 0
−1.5 (c) Maximum velocity when ωt = π 2,
∴ v max = Aω = 1.50m/s. This type of oscillatory motion is called simple harmonic
motion (SHM). You will see a number of examples of
SHM in various physics courses. ...
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This note was uploaded on 07/13/2011 for the course PHY 2048 taught by Professor Guzman during the Spring '08 term at FAU.
 Spring '08
 Guzman
 Physics, Acceleration

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