Transcript University Physics - Erwin Sitompul

Lecture 2
Ch2. STRAIGHT LINE MOTION
University Physics: Mechanics
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
Solution for Homework 1: Truck
You drives a truck along a straight road for 8.4 km at 70 km/h,
at which point the truck runs out of gasoline and stops. Over
the next 30 min, you walk another 2.0 km farther along the
road to a gasoline station.
(a) What is your overall displacement from the beginning of
your drive to your arrival at the station?
x1  0
x2  8.4  2.0  10.4 km
x  x2  x1  10.4 km  0  10.4 km
(b) What is the time interval Δt from the beginning of your
drive to your arrival at the station?
8.4 km
tdrv 
 0.12 h
t  tdrv  twlk
70 km/h
 0.12 h  0.5 h  0.62 h
twlk  30 min  0.5 h
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Solution for Homework 1: Truck
(c) What is your average velocity vavg from the beginning of
your drive to your arrival at the station? Find it both
numerically and graphically.
vavg
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x 10.4 km


t
0.62 h
 16.77 km/h
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Solution for Homework 1: Truck
(d) Suppose that to pump the gasoline, pay for it, and walk
back to the truck takes you another 45 min. What is your
average speed from the beginning of your drive to you
return to the truck with the gas?
total distance  8.4  2  2  12.4 km
total time interval  0.12  0.5  0.75  1.37 hr
savg 
total distance
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t
12.4 km

1.37 hr
 9.05 km/h
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Instantaneous Velocity and Speed
 Instantaneous velocity (or
simply velocity) is the average
velocity over a very short period
of time interval
x
dx
v  lim

t  0 t
dt
 Velocity v at any instant is the
slope of the position-time curve.
 Instantaneous speed (or simply speed) is the
magnitude of velocity, that is, speed is velocity without
any indication of direction
dx
speed 
dt
● What is the
car’s vavg?
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Instantaneous Velocity
instantaneous velocity v = slope of curve at any point
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Example: Elevator Cab
Figure (a) on the left is an x(t) plot for
an elevator cab that is initially
stationary, then moves upward (which
we take to be the positive direction of
x), and then stops. Plot v(t).
 The slope of x(t), which is the v(t), is
zero in the intervals from 0 to 1 s and
from 9 s on  the cab is stationary in
these intervals.
 During the interval bc, the slope is
nonzero constant  the cab moves with
constant velocity.
x
24  4
v
 4 m s
t
83
 Between 1 s and 3 s the cab begins to
move, and between 8 s and 9 s it slows
down  the velocity of the cab varies.
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● Given v(t), can you
determine x(t) exactly?
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Motion with Constant Velocity
dx
v
dt
constant
dx  v dt
 dx   v dt
x  vt  c
Motion with constant
velocity on x-t graph
Erwin Sitompul
Taking at time t0 = 0 the
position is at x0  c = x0
x  x0  vt
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Position, Time, and Velocity
What is the velocity a cyclist for each stage of this trip?
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Acceleration
 Average acceleration is
the ratio of change in
velocity to the time interval.
aavg
v2  v1

t2  t1
v

t
 Instantaneous
acceleration (or simply
acceleration) is the
derivative of the velocity
with respect to time.
v dv
a  lim

t  0 t
dt
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dv d  dx  d 2 x
   2
a
dt dt  dt  dt
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Acceleration
 Compare the a(t) curve with the v(t) curve.
 Each point on the a(t) curve shows the
derivative (slope) of the v(t) curve at the
corresponding time.
 When v is constant, the derivative is zero
 a is also zero.
 When the cab first begin to move, the v(t)
curve has a positive derivation, which
means that a(t) is positive.
 When the cab slows to a stop, the
derivative of v(t) is negative; that is, a(t) is
negative.
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Motion with Constant Acceleration
dv
dt
dv  a dt
a
constant
 dv   a dt
v  at  c
Taking at time t0 = 0 the
velocity equals v0  c = v0
Motions with constant
acceleration on v-t graph
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v  v0  at
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Motion with Constant Acceleration
v  v0  at
dx  v dt
dx  (v0  at ) dt
 dx   (v
0
 at ) dt
1
x  v0t  at 2  c
2
Motions with constant
acceleration on v-t graph
Taking at time t0 = 0 the
position is at x0  c = x0
1 2
x  x0  v0t  at
2
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Motion with Constant Acceleration
1 2
x  x0  v0t  at
2
From these two equations,
the following equations can
be derived:
v  v0  at
v2  v02  2a( x  x0 )
1
x  x0  (v0  v)t
2
1 2
x  x0  vt  at
2
a  constant
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Questions
What is velocity in
intervals A, B, C, D
1 m/s
0 m/s
What is acceleration in
intervals A, B, C
–1.5 m/s
2 m/s2
0 m/s2
–0.5 m/s2
2 m/s
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University Physics: Mechanics
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Example: Porsche
Spotting a police car, you brake a Porsche from a speed of
100 km/h to a speed of 80.0 km/h during a displacement of
88.0 m, at a constant acceleration.
(a) What is that acceleration?
v2  v02  2a( x  x0 )
22.222  27.782  2a(88  0)
22.222  27.782
a
 1.58 m/s2
(2)(88)
1000 m
3600 s
 27.78 m/s
100 km/h  100 
1000 m
80 km/h  80 
3600 s
 22.22 m/s
v  v0  at
v 2  v02  2a ( x  x0 )
x  x0  v0t  12 at 2 x  x0  12 (v0  v )t
x  x0  vt  12 at 2
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University Physics: Mechanics
2/16
Example: Porsche
Spotting a police car, you brake a Porsche from a speed of
100 km/h to a speed of 80.0 km/h during a displacement of
88.0 m, at a constant acceleration.
(b) How much time is required for the given decrease in
speed?
v  v0  at
v  v0
22.22  27.78
t

 3.52 s
a
1.58
v  v0  at
v 2  v02  2a ( x  x0 )
x  x0  v0t  12 at 2 x  x0  12 (v0  v )t
x  x0  vt  12 at 2
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Example: Porsche
Car accelerating and decelerating
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a > 0, a = 0, a < 0
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Illustration: Overtaking Maneuver
 Velocity vs. Time
● Can you determine the
exact value of vblue car?
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 Position vs. Time
 Both cars move with
constant velocity
 Red car starts moving 4
seconds after the blue car
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2/20
Trivia: How to Stop?
You are driving a car with ever changing velocity but constant
speed of 2 m/s. On your right is a steep cliff with the height of
50 cm. Directly in front of you there is a horse and behind you
an elephant, both of which travel at your own speed. On your
left there is a fire truck blocking you. How do you stop your
car?
Solution:
Simple. Just ask the merry-go-round operator to stop!
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University Physics: Mechanics
2/21
Questions
Which time periods indicates
that an object moves at
constant speed?
What are
(a) Initial direction of travel? –
(b) Final direction of travel? +
(c) Does the particle stop
momentarily? Yes
(d) Is the acceleration positive
or negative? +
(e) Is the acceleration
constant or varying?
Constant
E, where a = 0
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University Physics: Mechanics
2/22
Questions
How far does the runner travel in 16 s?
x  x0  12 (v0  v)t
x  (v0  v)t
1
2
x4  12 (4  4)4
 16
x2  12  (8  8)  8
 64
x1  12  (0  8)  2
8
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x3  12 (8  4)2
 12
xtotal  x1  x2  x3  x4
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2/23
Example: Race
A caravan moves with a constant velocity of 60 km/h along a
straight road when it passes a roadster which is at rest. Exactly
when the caravan passes the roadster, the roadster starts to
move with an acceleration of 4 m/s2.
x0  0, t0  0
vcaravan  60km h  16.67 m s
(a) How much time does the roadster need to catch up the
caravan?
xcaravan  xroadster
x0,caravan  vcaravan t  x0,roadster  v0,roadster t  12 aroadstert 2
0  16.67t  0  0t  12 4t 2
2t 2  16.67t  0
t (t  8.33)  0
t1  0, t2  8.33 s
Thus, the roadster will catch up the caravan
after 8.33 s.
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x  x0  vt
v  v0  at
x  x0  v0t  12 at 2
v 2  v02  2a ( x  x0 )
x  x0  12 (v0  v )t
x  x0  vt  12 at 2
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2/24
Example: Race
A caravan moves with a constant velocity of 60 km/h along a
straight road when it passes a roadster which is at rest. Exactly
when the caravan passes the roadster, the roadster starts to
move with an acceleration of 4 m/s2.
x0  0
Distance traveled
(b) How far does the roadster already move when it catches
up the caravan?
xroadster  x0,roadster  v0,roadstert  12 aroadstert 2
 0  (0)(8.33)  12 (4)(8.33)2
xcaravan
x  x0  vt
v  v0  at
 138.9 m
 x0,caravan  vcaravant
 0  (16.67)(8.33)
 138.9 m
x  x0  v0t  12 at 2
v 2  v02  2a ( x  x0 )
Both the vehicles travel 138.9 m before they
pass each other again.
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x  x0  12 (v0  v )t
x  x0  vt  12 at 2
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2/25
Example: Particle’s Movement
The position of a particle is given by x = 4t2 – 2t + 10, where x is
the distance from origin in meters and t the time in seconds.
(a) Find the displacement of the particle for the time interval
from t = 1 s to t = 2 s.
x1  4(1)2  2(1)  10  12
x2  4(2)2  2(2)  10  22
x  x2  x1  22 12  10 m
(b) Find also the. average velocity for the above given time
interval.
x2  x1 22  12

 10 m s
vavg 
2 1
t2  t1
(c) Find the instantaneous velocity of the particle at t = 0.5 s.
dx
v
 8t  2
dt
At t = 0.5 s, v  8(0.5)  2  2 m s
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University Physics: Mechanics
2/26
Homework 2: Aprilia vs. Kawasaki
An Aprilia and a Kawasaki are separated by 200 m when they
start to move towards each other at t = 0.
200 m
The Aprilia moves with initial velocity 5 m/s and acceleration
4 m/s2. The Kawasaki runs with initial velocity 10 m/s and
acceleration 6 m/s2.
(a) Determine the point where the two motorcycles meet
each other.
(b) Determine the velocity of Aprilia and Kawasaki by the
time they meet each other.
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University Physics: Mechanics
2/27
Homework 2A: Running Exercise
You come late to a running exercise
and your friends already run 200 m
with constant speed of 4 m/s.
The athletic trainer orders you to
catch up your friends within 1 minute.
(a) If you run with constant speed, determine the minimum
speed you have to take so that you can fulfill the trainer’s
order.
(b) If you run with minimum speed, determine the point where
you catch up your friends.
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University Physics: Mechanics
2/28
Homework 2B: Runway Length
1. An electron moving along the x axis has a position given by
x = 16te–t m, where t is in seconds. How far is the electron
from the origin when it momentarily stops?
2. An airplane lands at a speed of 160 mi/h and decelerates at
the rate of 10 mi/h/s. If the plane travels at a constant speed
of 160 mi/h for 1.0 s after landing before applying the
brakes, what is the total displacement of the aircraft between
touchdown on the runway and coming to rest?
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University Physics: Mechanics
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