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(nm22898) mostow Assignment # 1 - Introductory Physics and 1D Kinematics schultz (WVFAL20
This print-out should have 55 questions.
Multiple-choice questions may continue on
the next column or page nd all choices
before answering.
001 10.0 points
What would equal 5 meters?
4. [T]
5. 1/[T]
6. [L]/[T]2
7. 1/[T]2
1. 50 dm correct
8. [T]/[L]
2. 50 dam
9. [L]
3. 50 mm
10. [L]/[T] correct
4. 50 cm
Explanation:
5. 0.5 km
s = c1 t
s
c1 =
t
[s]
[c1 ] =
= [L]/[T]
[ t]
6. 0.5 hm
Explanation:
(50 dm)
1m
=5m
10 dm
002 10.0 points
What unit of measure would you use for mass?
1. None of these
004 (part 2 of 5) 10.0 points
Find the dimension for the quantity c2 in the
expression s = c2 t2 .
2. liters
1. [T]
3. cubic centimeters
2. 1/[T]
4. grams correct
3. [T]2 /[L]
Explanation:
Liters and cubic centimeters are units of
measure for volume.
4. [L]
003 (part 1 of 5) 10.0 points
s is a distance with unit [L], t is a time with
unit [T] and is an angle in radians.
Find the dimension for the quantity c1 in
the expression s = c1 t.
6. 1/[L]
1. 1/[L]
5. [L]/[T]
7. [L]/[T]2 correct
8. 1/[T]2
9. None of these
2. None of these
10. [T]/[L]
3. [T]2/[L]
Explanation:
mostow (nm22898) Assignment # 1 - Introductory Physics and 1D Kinematics schultz (WVFAL20
1 mL = 1 cc.
s
t2
[s]
[c2 ] = 2 = [L]/[T]2
[ t]
c2 =
010 10.0 points
An electron travels 1.37 m in 5.89 108 s.
How fast is it traveling?
005 (part 3 of 5) 10.0 points
WITHDRAWN
Correct answer: 2.32598 107 m/s.
Explanation:
006 (part 4 of 5) 10.0 points
WITHDRAWN
Let :
s = 1.37 m and
t = 5.89 108 s .
007 (part 5 of 5) 10.0 points
WITHDRAWN
008 10.0 points
How many (whole number of) 90 kg people
can safely occupy an elevator that can hold a
maximum mass of exactly 1 metric ton? A
metric ton is 1.000 103 kg.
Correct answer: 11 people.
v=
s
1.37 m
=
t
5.89 108 s
= 2.32598 107 m/s .
011 10.0 points
Convert an acceleration of 5.7 mi/h/s to m/s2 .
Correct answer: 2.54758 m/s2 .
Explanation:
Explanation:
Let :
1 ton = 1.000 103 kg
R = 90 kg/person .
and
Let :
a = 5.7 mi/h/s .
Using dimensional analysis,
1 person
n = 1.000 10 kg
90 kg
= 11.1111 persons
3
Note that the numerical answer 11.1111 must
be rounded down to 11 people because of
overload considerations.
009 10.0 points
What is equivalent to a cubic centimeter?
1. centiliter
2. milliliter correct
a=
5.7 mi 1.609 km 1000 m
1h
hs
1 mi
1 km 3600 s
= 2.54758 m/s2 .
012 10.0 points
Convert 47.7 mi/h to m/s. 1 mi = 1609 m.
Correct answer: 21.3193 m/s.
Explanation:
1 mi = 1609 m
(47.7 mi/h)
1 h = 3600 s
1609 m 1 h
= 21.3193 m/s .
1 mi 3600 s
3. liter
4. deciliter
Explanation:
013 10.0 points
The average mass of an automobile in the
United States is about 1.409 106 g.
mostow (nm22898) Assignment # 1 - Introductory Physics and 1D Kinematics schultz (WVFAL20
Express this mass in kilograms.
015 (part 1 of 4) 10.0 points
Use signicant gures to nd the sum of the
measurements 759 g, 37.8 g, 0.87 g, and 2.8 g.
1. 1.409 1012 kg
2. 1.409 106 kg
1. Whole number (800 g) correct
3. 1.409 109 kg
2. Hundredths (800.47 g)
4. 1.409 109 kg
3. Tenths (800.5 g)
5. 1.409 1012 kg
4. Three signicant gures (800 g)
6. None of these
5. Tens (800 g)
7. 1.409 100 kg
6. One signicant gure (800 g)
8. 1.409 103 kg
7. None of these
9. 1.409 103 kg correct
Explanation:
1 kg = 1 103 g ,
8. Two signicant gures (800 g)
so
m = 1.409 106 g
1 kg
1 103 g
= 1.409 103 kg .
014
10.0 points
1
Suppose your hair grows at the rate of
inch
23
per day.
Find the rate at which it grows in nanometers per second.
Correct answer: 12.7818 nm/s.
Explanation:
R=
1 day
1h
1 inch
23 day
24 h
3600 s
9 nm
2.54 cm
10
1 in
100 cm
Explanation:
For addition/subtraction:
measurement
least precise
759 g + 37.8 g + 0.87 g + 2.8 g = 800.47 g
The least precise measurement is a whole
number of grams.
016 (part 2 of 4) 10.0 points
3. 3 m
.
Find the quotient
3.566 s
1. One signicant gure (0.9 m/s)
2. Hundredths (0.93 m/s)
3. Two signicant gures (0.93 m/s) correct
4. Tenths (0.9 m/s)
5. Three signicant gures (0.925 m/s)
= 12.7818 nm/s .
6. Thousandths (0.925 m/s)
The distance between atoms in a molecule is
on the order of 0.1 nm, so this suggests how
rapidly atoms are assembled in this protein
synthesis.
7. Four signicant gures (0.9254 m/s)
8. None of these
mostow (nm22898) Assignment # 1 - Introductory Physics and 1D Kinematics schultz (WVFAL20
Explanation:
For multiplication/division: least signicant gures in measured quantities
3. 3 m
= 0.925407 m/s
3.566 s
The least number of signicant gures is 2.
7. Two signicant gures (24 s)
8. Tenths (24.0 s) correct
9. One signicant gure (20 s)
Explanation:
017 (part 3 of 4) 10.0 points
Find the product of 5.07 mm and .
1. Thousandths (15.928 mm)
27.57 s 3.6 s = 23.97 s
The least precise measurement is a tenth of a
second.
2. Two signicant gures (16 mm)
3. Five signicant gures (15.928 mm)
4. Four signicant gures (15.93 mm)
5. Hundredths (15.93 mm)
6. Three signicant gures (15.9 mm) correct
019 (part 1 of 3) 10.0 points
You measure a distance to be 3 miles.
Convert to the proper SI unit using signicant gures.
1. 4.83 km
2. 1.87 km
3. 2 km
7. Whole number (16 mm)
4. 1.9 km
8. None of these
5. 5 km correct
9. Tenths (15.9 mm)
Explanation:
6. 4.8 km
(5.07 mm)( ) = 15.9279 mm
7. 4.827 km
The least number of signicant gures is 3 (
is a constant, not a measured quantity).
8. 1.865 km
018 (part 4 of 4) 10.0 points
Find the dierence of 27.57 s and 3.6 s.
Explanation:
The measurement of 3 miles has only one
signicant gure, so the results should be expressed in only one signicant gure:
1. Whole number (24 s)
2. Hundredths (23.97 s)
3. Four signicant gures (23.97 s)
4. Three signicant gures (24.0 s)
(3 mi)
1.609 km
= 4.827 km ,
1 mi
which should be rounded to 5 km.
020 (part 2 of 3) 10.0 points
WITHDRAWN
5. Tens (20 s)
6. None of these
021 (part 3 of 3) 10.0 points
A marathon is 26.3 miles.
mostow (nm22898) Assignment # 1 - Introductory Physics and 1D Kinematics schultz (WVFAL20
Convert to centimeters using appropriate
signicant gures.
6
8. None of these
9. 1 m
1. 42.310 cm
6
2. 410 cm
3. 4.232106 cm
4. 4.2106 cm
5. 0.423103 cm
10. 7 m
Explanation:
Since its nal position is the same as its
initial position, the displacement is 0.
023 (part 2 of 2) 10.0 points
What is the distance traveled?
6. 4.0106 cm
1. 7 m
7. 42.3167106 cm
2. 4 m
8. 4.23106 cm correct
3. 3 m
Explanation:
The measurement of 26.3 miles has three
signicant gures, so the results should be
expressed in three signicant gures:
1.609 km 100 cm 1000 m
(26.3 m)
1 mi
1m
1 km
= 4.23167 106 cm ,
which should be rounded to 4.23 106 cm.
022 (part 1 of 2) 10.0 points
A physics book is moved once around the
perimeter of a table of dimensions 1 m by 3
m.
If the book ends up at its initial position,
what is the magnitude of its displacement?
1. 8 m
4. 1 m
5. 6 m
6. None of these
7. 5 m
8. 2 m
9. 8 m correct
10. 0 m
Explanation:
The distance traveled is the perimeter of
the table:
d = 2 + 2 w = 2 (1 m) + 2 (3 m)
= 8 m.
2. 5 m
6. 2 m
024 10.0 points
An oceanic depth-sounding vessel surveys the
ocean bottom with ultrasonic waves that
travel 1530 m/s in seawater.
How deep is the water directly below the
vessel if the time delay of the echo to the
ocean oor and back is 7 s?
7. 0 m correct
Correct answer: 5355 m.
3. 4 m
4. 3 m
5. 6 m
Explanation:
mostow (nm22898) Assignment # 1 - Introductory Physics and 1D Kinematics schultz (WVFAL20
Let :
v = 1530 m/s and
t = 7 s.
Correct answer: 23.5 m/s.
Explanation:
The sound takes 3.5 s to reach the ocean
oor (and 3.5 s to return), so
Let :
v0 = 14 m/s ,
t = 10 s , and
a = 1.9 m/s2 .
d = v t = (1530 m/s) (3.5 s) = 5355 m .
v = v0 + a t, and
025 (part 1 of 2) 10.0 points
A cyclist maintains a constant velocity of
6.3 m/s headed away from point A. At some
initial time, the cyclist is 252 m from point A.
What will be his displacement from his
starting position after 42 s?
Correct answer: 264.6 m.
v=
v0 + v
v0 + (v0 + a t)
2 v0 + a t
=
=
2
2
2
2 (14 m/s) + (1.9 m/s2 ) (10 s)
2
= 23.5 m/s .
=
Explanation:
Let :
v = 6.3 m/s and
t = 42 s .
The distance is
d = v t = (6.3 m/s) (42 s) = 264.6 m .
026 (part 2 of 2) 10.0 points
What will be his position from point A after
that time?
Correct answer: 516.6 m.
Explanation:
Let :
d0 = 252 m .
The position is
d = d0 + d = 252 m + 264.6 m = 516.6 m .
027 10.0 points
A car is moving at a constant speed of 14 m/s
when the driver presses down on the gas pedal
and accelerates for 10 s with an acceleration
of 1.9 m/s2 .
What is the average speed of the car during
the period?
028 (part 1 of 2) 10.0 points
A person travels by car from one city to another. She drives for 25.2 min at 65.9 km/h,
9.92 min at 91.3 km/h, 36 min at 46.2 km/h,
and spends 19.1 min along the way eating
lunch and buying gas.
Determine the distance between the cities
along this route.
Correct answer: 70.4929 km.
Explanation:
Let :
t1
v1
t2
v2
t3
v3
= 25.2 min ,
= 65.9 km/h ,
= 9.92 min ,
= 91.3 km/h ,
= 36 min , and
= 46.2 km/h .
x = x1 + x2 + x3
= v1 t1 + v2 t2 + v3 t3
= (65.9 km/h)(25.2 min)
+ (91.3 km/h)(9.92 min)
+ (46.2 km/h)(36 min)
= 70.4929 km .
mostow (nm22898) Assignment # 1 - Introductory Physics and 1D Kinematics schultz (WVFAL20
Explanation:
Let :
tother = 19.1 min .
The total time is
t = t1 + t2 + t3 + tother
= 25.2 min + 9.92 min
+ 36 min + 19.1 min
= 1.50367 h , so
vav =
x
70.4929 km
=
= 46.8807 km/h .
t
1.50367 h
030 10.0 points
A race car is one lap behind the lead race car
when the lead car has 59 laps to go in a race.
If the speed of the lead car is 55.9 m/s, what
must be the average speed of the second car
to catch the lead car just before the end of the
race (i.e., right at the nish line)? Assume 1
lap is 1.34 km.
Correct answer: 56.8475 m/s.
displacement (m)
Correct answer: 46.8807 km/h.
Displacement vs Time
10
8
029 (part 2 of 2) 10.0 points
Determine the average speed for the trip.
6
4
2
0
2
4
6
8
10
02
46
8 10 12 14 16 18 20
time (s)
What is the displacement at 10 s?
1. 2 m
2. 3 m
3. None of these
4. 1 m
5. 1 m
6. 3 m
Explanation:
7. 2 m
Let :
n = 59 laps and
v = 55.9 m/s .
nx
.
The lead car nishes the race in t =
t
In the same amount of time, the trailing car
must have an average speed of
(n + 1) v
(n + 1) x
=
t
n
(60 laps) (55.9 m/s)
=
(59 laps)
8. Unable to determine
9. 0 m
10. 4 m correct
Explanation:
Read the displacement from the graph.
vavg =
= 56.8475 m/s .
032 (part 2 of 2) 10.0 points
What is the velocity at 10 s?
1. 1 m/s
2. 2 m/s
031 (part 1 of 2) 10.0 points
Consider the following graph
3. 3 m/s
mostow (nm22898) Assignment # 1 - Introductory Physics and 1D Kinematics schultz (WVFAL20
the magnitude v = |v| of this slope is the
speed. The curve is steepest (in absolute
magnitude) during the interval III and that is
when the object had the highest speed.
4. 1 m/s
5. None of these
6. 3 m/s
034 (part 2 of 2) 10.0 points
During which interval(s) did the velocity objects remain constant?
7. 4 m/s
8. 2 m/s
1. During none of the three intervals
9. Unable to determine
2. During interval II only
10. 0 m/s correct
3. During interval III only
Explanation:
At 10 s, the slope is 0.
4. During each of the three intervals correct
033 (part 1 of 2) 10.0 points
Consider a moving object whose position x
is plotted as a function of the time t. The
object moved in dierent ways during the
time intervals denoted I, II and III on the
gure.
x
4
Explanation:
For each of the three intervals I, II or III, the
x(t) curve is linear, so its slope (the velocity
v) is constant. Between the intervals, the
velocity changed in an abrupt manner, but it
did remain constant during each interval.
035 (part 1 of 3) 10.0 points
Consider the following graph of motion.
2
80
II
2
III
4
t
6
During these three intervals, when was the
objects speed highest? Do not confuse the
speed with the velocity.
50
r
Ca
40
30
10
0
2. Same speed during each of the three intervals.
rB
a
dC
60
20
1. During interval I
3. Same speed during intervals II and III
C
70
an
rA
Ca
0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6
Time (hr)
How long does it take for the cars to be at
the same position?
4. During interval III correct
1. 0.5 hr correct
5. During interval II
2. 0.3 hr
Explanation:
The velocity v is the slope of the x(t) curve;
A
Ca
rB
I
Distance (km)
6
5. During interval I only
3. 1.4 hr
mostow (nm22898) Assignment # 1 - Introductory Physics and 1D Kinematics schultz (WVFAL20
and car B traveled
4. 0.2 hr
(80 km/hr) (0.4 hr) = 32 km ,
5. 0.6 hr
so after 0.4 hr they are
6. 0.4 hr
(20 + 16 32) km = 4 km
7. 1.5 hr
apart.
8. 0.1 hr
Explanation:
After 0.5 hr, both cars are at the same
positon.
036 (part 2 of 3) 10.0 points
How far apart are they after 0.4 h?
037 (part 3 of 3) 10.0 points
Which car maintained a constant speed for
the entire trip?
1. Car A correct
2. Both cars
1. 7.3 km
3. Unable to determine
2. 10.5 km
4. Car B
3. 11 km
Explanation:
Car A has a graph with a constant slope.
4. 3.0 km
5. 4 km correct
6. 1.5 km
038 10.0 points
The plot shows x(t) for a train moving along
a long, straight track.
x
7. 2 km
8. 6.4 km
9. 9 km
10. 5 km
Explanation:
In 0.5 h, Car A traveled 20 km:
vA =
20 km
= 40 km/hr
0.5 hr
and Car B traveled 40 km:
40 km
= 80 km/hr .
vB =
0.5 hr
In 0.4 hr Car A traveled
(40 km/hr) (0.4 hr) = 16 km
t
Which statement is correct about the motion?
1. The train at rst speeds up, then slows
down.
2. The train in fact moves at constant speed
along a circular path described by the x(t)
curve.
mostow (nm22898) Assignment # 1 - Introductory Physics and 1D Kinematics schultz (WVFAL20
3. The train continually slows down and
comes to rest. correct
v
t
3.
4. The train continually speeds up.
5. The train at rst slows down, then speeds
up.
v
t
4.
6. The train moves at constant velocity.
correct
Explanation:
The slope of x(t) continually decreases to
dx
reach zero. Since vx (t) =
is by denition
dt
the slope of x(t) at each t, the train continually slows down until it stops.
039 (part 1 of 4) 10.0 points
v
v
t
6.
The following 4 questions refer to a toy car
which can move to the right or left along a
horizontal line. The positive direction is to
the right.
Choose the correct velocity-time graph for
each of the following questions. Assume friction is so small that it can be ignored.
v
t
7.
car
g
t
5.
v
v
t
8.
+
O
Which velocity graph shows the car moving
toward the right (away from the origin) at a
steady (constant) velocity?
v
9.
v
1.
2. None of these graphs is correct.
t
t
mostow (nm22898) Assignment # 1 - Introductory Physics and 1D Kinematics schultz (WVFAL20
v
v
t
10.
t
6.
Explanation:
Since the velocity is constant, the graph is
a straight line. Since the car is moving to the
right the velocity is positive.
v
t
7.
v
t
v
t
8.
040 (part 2 of 4) 10.0 points
Which velocity graph shows the car moving
towards the right (away from the origin) reversing direction and then moving to the left?
v
t
9.
v
t
1.
v
t
10.
2. None of these graphs is correct.
v
t
3.
Explanation:
Since the car reverses its direction, the velocity is positive and then negative.
v
t
4.
v
t
v
5.
correct
t
041 (part 3 of 4) 10.0 points
Which velocity graph shows the car moving
toward the left (toward the origin) at a steady
(constant) velocity?
mostow (nm22898) Assignment # 1 - Introductory Physics and 1D Kinematics schultz (WVFAL20
v
v
t
1.
correct
v
v
t
2.
t
9.
t
10.
v
t
3.
Explanation:
Constant velocity is a straight line; toward
the left is negative velocity.
042 (part 4 of 4) 10.0 points
Which velocity graph shows the car increasing
its speed towards the right (away from the
origin) at a steady (constant) rate?
4. None of these graphs is correct.
v
t
5.
v
t
1.
v
t
6.
v
t
2.
v
t
7.
v
t
3.
v
8.
t
correct
v
4.
t
mostow (nm22898) Assignment # 1 - Introductory Physics and 1D Kinematics schultz (WVFAL20
9
v
8
t
5.
v
t
6.
velocity (m/s)
7
6
5
4
3
2
1
7. None of these graphs is correct.
0
1
v
t
8.
2
1
2
3
4567
time (s)
What is the position at 2 seconds?
8
9
Correct answer: 19 m.
v
t
9.
Explanation:
The initial position given in the problem is
10 m.
9
v
8
t
10.
7
6
5
Explanation:
Since the cars speed is increasing at a constant rate, the slope of the graph is a constant.
v
velocity (m/s)
4
3
2
1
0
1
t
043 (part 1 of 3) 10.0 points
Consider the plot below describing motion
along a straight line with an initial position of
x0 = 10 m.
1
2
3
456
time (s)
7
8
9
2
The position at 2 seconds is 10 meters plus
the area of the triangle (shaded in the above
plot)
1
x = 10 m + (2 s 0 s)
2
mostow (nm22898) Assignment # 1 - Introductory Physics and 1D Kinematics schultz (WVFAL20
(9 m/s 0 m/s)
= 19 m ;
however, it can also be calculated:
1
( tf ti ) 2
2
= (10 m) + (0 m/s) (2 s 0 s)
1
+ (4.5 m/s2 ) (2 s 0 s)2
2
= 19 m .
x = xi + vi (tf ti ) +
however, it can also be calculated
1
( tf ti ) 2
2
= 49 m + (0 m/s) (8 s 6 s)
1
+ (0.666667 m/s2 ) (8 s 6 s)2
2
= 47.6667 m .
x = xi + vi (tf ti ) +
044 (part 2 of 3) 10.0 points
What is the position at 6 seconds?
046 10.0 points
When Maggie applies the brakes of her car,
the car slows uniformly from 14.4 m/s to 0
m/s in 2.13 s.
How far ahead of a stop sign must she apply
her brakes in order to stop at the sign?
Correct answer: 49 m.
Correct answer: 15.336 m.
Explanation:
The position is 19 m plus the area of the
trapezoid from 2 s to 6 s
Explanation:
Let :
1
(6 s 2 s)
2
(6 m/s + 9 m/s)
x = 19 m +
= 49 m ;
however, it can also be calculated:
1
( tf ti ) 2
2
= (19 m) + (9 m/s) (6 s 2 s)
1
+ (0.75 m/s2 ) (6 s 2 s)2
2
= 49 m .
x = xi + vi (tf ti ) +
045 (part 3 of 3) 10.0 points
What is the position at 8 seconds?
Correct answer: 47.6667 m.
Explanation:
The position is 49 m minus the area of the
triangle from 6 s to 8 s
1
(8 s 6 s)
2
(1.33333 m/s 0 m/s)
x = (49 m) +
= 47.6667 m ;
vi = 14.4 m/s ,
vf = 0 m/s , and
t = 2.13 s .
The equation simplies to
vi + vf
1
t = vi t
2
2
1
= (14.4 m/s) (2.13 s)
2
= 15.336 m .
x =
047 (part 1 of 2) 10.0 points
A speeder passes a parked police car at a
constant speed of 25.2 m/s. At that instant,
the police car starts from rest with a uniform
acceleration of 2.16 m/s2 .
How much time passes before the speeder
is overtaken by the police car?
Correct answer: 23.3333 s.
Explanation:
Let :
v = 25.2 m/s ,
a = 2.16 m/s2 ,
as = 0 , and
v0p = 0 .
mostow (nm22898) Assignment # 1 - Introductory Physics and 1D Kinematics schultz (WVFAL20
1
x = vi t + a t2
2
Under constant acceleration,
2
2
vf = vi + 2 a x = 0
xs = xp
1
v t = a t2
2
0 = a t2 2 v t
0 = (a t 2 v ) t
2 (25.2 m/s)
2v
=
= 23.3333 s .
t=
a
2.16 m/s2
048 (part 2 of 2) 10.0 points
How far does the speeder travel before being
overtaken by the police car?
Correct answer: 588 m.
Explanation:
The distance may be obtained with either
car:
d = v t = (25.2 m/s) (23.3333 s)
= 588 m
or
vi 2
2 x
(266 km/h)2 1000 m
=
2 (1150 m)
1 km
a=
1h
3600 s
2
2
= 2.37373 m/s2 .
050 10.0 points
An engineer in a locomotive sees a car stuck
on the track at a railroad crossing in front of
the train. When the engineer rst sees the
car, the locomotive is 350 m from the crossing
and its speed is 23 m/s.
If the engineers reaction time is 0.54 s,
what should be the magnitude of the minimum deceleration to avoid an accident?
Correct answer: 0.783518 m/s2 .
Explanation:
121
2.16 m/s2 (23.3333 s)2
at =
2
2
= 588 m .
d=
049 10.0 points
The French National Railroad holds the
worlds speed record for passenger trains in
regular service. A TGV (tres grand vitesse,
or very great speed) train traveling at a speed
of 266 km/h requires 1.15 km to come to an
emergency stop.
Find the braking acceleration for this train,
assuming constant acceleration.
Correct answer: 2.37373 m/s2 .
Explanation:
Let :
d = 350 m ,
v = vi = 23 m/s ,
t = 0.54 s .
and
While the engineer reacts, the train moves
d = v t = (23 m/s) (0.54 s) = 12.42 m
forward, so it now has to decelerate to rest
within a displacement of
x = dd = 350 m12.42 m = 337.58 m ,
2
2
vf = vi + 2 a x = 0
2
vi
(23 m/s)2
=
2 x
2 (337.58 m)
= 0.783518 m/s2 ,
a=
Let :
vi = 266 km/h ,
vf = 0 m/s , and
x = 1.15 km = 1150 m .
which has a magnitude of 0.783518 m/s2 .
and
mostow (nm22898) Assignment # 1 - Introductory Physics and 1D Kinematics schultz (WVFAL20
Applying kinematics,
051 (part 1 of 2) 10.0 points
A plane landing on a small tropical island has
just 69 m of runway on which to stop.
If its initial speed is 41 m/s, what is the
maximum acceleration of the plane during
landing, assuming it to be constant?
Explanation:
2
v=
s = 69 m and
v0 = 41 m/s .
2
v0
+ 2as = 0
2
v0
(41 m/s)2
a=
=
2s
2(69 m)
= 12.1812 m/s2 .
052 (part 2 of 2) 10.0 points
How long does it take for the plane to stop
with this acceleration?
Correct answer: 3.36585 s.
Explanation:
v = v0 + a t = 0
41 m/s
v0
=
t=
a
12.1812 m/s2
= 3.36585 s .
053 10.0 points
A car traveling initially at 9.9 m/s accelerates
at the rate of 0.675 m/s2 for 2.79 s.
What is its velocity at the end of the acceleration?
Correct answer: 11.7832 m/s.
Explanation:
Let :
v0 = 9.9 m/s ,
a = 0.675 m/s2 ,
t = 2.79 s .
= 9.9 m/s + (0.675 m/s2 )(2.79 s)
= 11.7832 m/s .
054 (part 1 of 2) 10.0 points
An electron has an initial speed of
1.97 105 m/s.
If it undergoes an acceleration of
2.9 1014 m/s2 , how long will it take to reach
a speed of 6.8 105 m/s?
Correct answer: 12.1812 m/s2 .
Let :
vf = v0 + at
and
Correct answer: 1.66552 109 s.
Explanation:
Let : v0 = 1.97 105 m/s ,
a = 2.9 1014 m/s2 ,
v = 6.8 105 m/s .
and
v = v0 + a t
v v0
t=
a
6.8 105 m/s 1.97 105 m/s
=
2.9 1014 m/s2
= 1.66552 109 s .
055 (part 2 of 2) 10.0 points
How far has it traveled in this time?
Correct answer: 0.000730329 m.
Explanation:
12
at
2
= (1.97 105 m/s) (1.66552 109 s)
1
+ (2.9 1014 m/s2 )
2
(1.66552 109 s)2
x = x0 + v0 t +
= 0.000730329 m .

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