CHAPTER 1: CONVERSIONS AND KINEMATICS CONVERSIONS WORKSHEET SOLUTIONS

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Nov 14, 2013 (3 years and 9 months ago)

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MR. SURRETTE


VAN NUYS HIGH SCHOOL

1

|
P a g e

PHYSICS


CHAPTER 1
: CONVERSIONS AND KINEMATICS

CONVERSIONS WORKSHEET SOLUTIONS


1.

On planet Q, quash is a popular beverage among the natives. The average person on Q consumes 5
guppies of quash per month. There are 11 months per year on Q and an estimated popu
lation of 700
million. Provide an order of magnitude estimate value for the total value of quash consumed per year
(in guppy units).

1A.

(1)
Start
: 11 x 5 x 700,000,000

(2)
Round
: 10
1

x 10
1

x 10
9

(3)
10
11

guppy units


2.

Note the following mathemati
cal expression:

y = x
2
. Which one of the statements is most consistent with this expression?

(a) If y doubles, then x quadruples

(b) y is greater than x

(c) if x doubles, then y doubles

(d) if x doubles, then y quadruples

2A.


(D)
if x doubles, then
y quadruples.

(1) y = x
2

(2) y = 2(
2x
)
2

(3) y = 4x
2

(4)
y = (4)x
2


3.

On planet R, the standard unit of length is the rigby. If a 5.8 foot height astronaut travels to planet R
and is measured to have a height of 121 rigbys, what would be the height i
n rigbys of another astronaut
who measures 6.5 feet in height?

3A.

(1) h = (6.5
foot
)
121 rigbys



5.8
foot



(2)
h = 135.6 rigbys


4.

If a submarine on the surface dives at an angle of 30 degrees with respect to the h
orizontal and
follows a straight line path for a distance of 50 m, how far below the surface will it be (measured in m)?

4A.

(1) y = (sin 30
o
)(50 m)

(2) y = (0.5)(50 m)

(3)
y = 25 m


MR. SURRETTE


VAN NUYS HIGH SCHOOL

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PHYSICS


5.

Suppose it takes you 20 minutes to walk a mile. If you trained s
o that you could walk steadily day
after day, for 5 hours a day, estimate how long it would take you to walk across America (about 3000
miles).

5A.

(1) (1 mile / 20
minutes
) (60
minutes

/ hour) = 3 miles / 1 hour

(2) (1 hour / 3
miles
) (3000
miles

/ t
rip) = 1000 hour / trip

(3) (1000
hour

/ trip) (1
day

/ 5
hours
) (1 month / 30
days
) =
6.67 months


6.

Which point is nearest the x axis?

(a) (3,4) (b) (4,3) (c) (
-

5, 2) (d) (
-

2,
-

5)

6A.

(c )
(
-

5, 2)


7.

A gallon of paint (volume = 3.78 x 10
-
3

m
3
) covers 35 m
2
. What is the thickness of the paint on the
wall (in millimeters)?

7A.

(1)
V = A T

(2) T = V / A

(3) T = (3.78 x 10
-
3

m
3
) / 35 m
2

(4) T = 1.08 x 10
-
4

m

(5)
T = 0.11 mm


8.

Convert 60.0 mph to a speed in km/hr.

8A.


(60
miles

/ 1 ho
ur) (5280
feet

/ 1
mile
) (12
inches

/ 1
foot
) (2.54
cm

/ 1
inch
) (1
m

/ 100

cm
) (1 km /
1000
m
) =
96.6 km/hr


9.

How many seconds are in 3 hours?

9A.

(1) (3
hours
) (60
minutes

/ 1
hour
) (60 s / 1
minute
)

(2)
10,800 s


10.

How far can you travel in 7 hou
rs if you drive 55 miles per hour?

10A.

(1) (7
hours
) (55 miles / 1
hour
)

(2)

385 miles


11.

How many days will it take to earn $160 if your pay is $4.00 per hour? (Assume 8
-
hour work days.)

11A.

(1) (
$
160) (1
hour

/
$
4.00) (1 day / 8
hours
)

(2)
5 days




MR. SURRETTE


VAN NUYS HIGH SCHOOL

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PHYSICS


12.

Calculate the number of laboratory reports Mr. Jones, the chemistry teacher, will have to grade
during the school year if his students each do 15 experiments. Mr. Jones has 4 classes of 24 students
each.

12A.

(1) (4
classes
) (24
students

/ 1
class
)

(15 reports / 1
student
)

(2)
1440 reports


13.

Mr. Jones plans to do an experiment with his classes which requires 2 test tubes per student. How
many dozens of test tubes will Mr. Jones need to supply to the students in his 4 classes of 24 students
each
?

13A.


(1) (4
classes
) (24
students
/1
class
) (2
test tubes
/1
student
) (1 dozen test tubes / 12

test tubes
)

(2)
16 dozen test tubes



MR. SURRETTE


VAN NUYS HIGH SCHOOL

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PHYSICS


CHAPTER 1
: CONVERSIONS AND KINEMATICS

KINEMATICS WORKSHEET SOLUTIONS


1.

A basketball rolls 10 m across a gym floor,
hits a wall, and rebounds backwards 10 m all in a time of
4.0 seconds. A sketch of x versus time t is shown.




1a.

What is the peak value of
x

in the figure above?

A.
10 meters

because the wall is 10 meters away.


1b.

What is the ball's instantaneous

velocity at t = 2 s?

A.

v at 2 seconds = 0

because it is momentarily stopped against the wall.


1c.

A sketch of the ball’s velocity versus time is shown below. What are v
o

and v?




A.

INITIAL VELOCITY:

(1) v
AVG

= [(x



x
o
) / (t



t
o
)]

(2) v
AVG

= [(1
0m
-

0m) / (2s
-

0s)]

(3)
v
o

=
5 m/s

FINAL VELOCITY:

(4) Same magnitude as initial velocity, but traveling in the opposite direction.

(5) v =
-

v
o

(6)
v =
-

5 m/s


MR. SURRETTE


VAN NUYS HIGH SCHOOL

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PHYSICS


2.

A European sports car dealer claims that his product will accelerate at a constant r
ate from rest to a
speed of 120 km/hr in 7 s. What is the acceleration?

2A.

(1) v = (120

km
/1
hour
)(1000 m/
km
)(1
hour
/3600 s)

(2)
v = 33.3 m/s

(3)
v = v
o

+ at

(4) v =
0

+ at

(5) v = at

(6) a = v / t

(7) a = (
33.3 m/s
) / (7 s)

(8)
a = 4.76 m/s
2


3
.
A car traveling at 30 m/s slows to a stop in 100 m.

3a.

What is the time required to stop?

A.


GIVEN: v
o

= 30 m/s, v = 0 m/s, d = 100 m

FIND: t

(1)
d = ½ (v + v
o
)t

(2) 2 [d = ½ (v + v
o
) t]

(3) 2d = (v + v
o
)t

(4) t = (2d) / (v + v
o
)

(5) t = [2(
100m) / (0 m/s + 30 m/s)]

(6)
t = 6.67 s


3b.

What is the acceleration?

A.

(1) a = [(v
-

v
o
) / (t


t
o
)]

(2) a = [(0 m/s
-

30 m/s) / 6.67 s] (Note: The 6.67 seconds is from
3a.
)

(3)

a =
-

4.50 m/s
2


4.
A ball is thrown vertically upward at a speed t
hat causes it to return to its starting point 15.0 seconds
later.

4a.

What is the ball's displacement in this interval?

A.

0 meters
. The final position coincides with the initial position.


4b.

Determine the relative direction of the initial (v
o
) and t
he final (v) velocities.

A.

Final velocity is the negative of initial velocity:
v
o

=
-

v


4c.

Find the initial velocity.

A.

Realize that the ball stops momentarily in the air at maximum height:

(1)
v = v
o

+ at

(2) v = v
o

-

g
t

(3)
0

= v
o

-

gt

(4) v
o

= gt

(5) v
o

= (9.8 m/s
2
)(7.5 sec)

(6)
v
o

= 73.5 m/s

MR. SURRETTE


VAN NUYS HIGH SCHOOL

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PHYSICS


5.

A rock, released at rest from the top of a tower, hits the ground after falling for 2.00s. What is the
height of the tower?

5A.

(1)
d = v
o
t + ½ at
2

(2)

d =
v
o
t



½ gt
2

(3) d =
-

½ gt
2

(4)

d

=
-

½ (9.8 m/s
2
)(2 s)
2

(5)

d =
-

19.6 m

(6)

h = 19.6 m


6.

A rock is thrown downward from the top of a tower with an initial speed of 12 m/s. If the rock hits
the ground after 3.25 s, what is the speed of the rock as it hits the ground?

6A.

(1)
v = v
o

+ at

(2)

v = v
o

-

g
t

(3) v = (
-

12 m/s)


(9.8 m/s
2
)(3.25 s)

(4)

v

=
-

12.0 m/s


31.9 m/s

(5)
v =
-

43.9 m/s


7.

A ball is released and it rolls down a hill with constant acceleration. At the end of one second, it has
traveled 2 m. How far has it

traveled after six seconds?

7A.

GIVEN: x
1

= 2 m

FIND: x
6

(1)

x
1

=
v
o
t

+ ½ at
2

(2) x
1

= ½ at
2

(3)

2
[x
1

= ½ at
2
]

(4)

2x
1

= at
2

(5)

a = (2x
1

/ t
2
)

(6)

a = 2 (2 m) / (1 s)
2

(7)
a = 4 m/s
2

(8)

x
6

=
v
o
t

+ ½ at
2

(9)

x
6

= ½ (
4 m/s
2
)(6 s)
2

(10)

x
6

= 7
2 m


8.

A locomotive slows from 28 m/s to zero in 12 s. What distance does it travel?

8A.

(1)

d = ½ (v + v
o
)t

(2) d = ½ (
0 m/s

+ 28 m/s)(12 s)

(3)

d = ½ (28 m/s)(12 s)

(4)

d = 168 m



MR. SURRETTE


VAN NUYS HIGH SCHOOL

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PHYSICS


CHAPTER 1
: CONVERSIONS AND KINEMATICS

QUIZ SOLUTIONS


1.

Solve

this question as an “order of magnitude” problem as solved in class. On planet Y, yiggy is a
popular beverage among the natives. The average person on Y consumes 8.9 yaks of yiggy per month.
There are 9 months per year on Y and an estimated population
of 1 billion. Provide an order
-
of
-
magnitude estimate value for the total volume of yiggy consumed per year (in yak units).

1A.

(1)
Start
: 8.9 x 9 x 1,000,000,000

(2)
Round
: 10
1

x 10
1

x 10
9

(3)
10
11

yiggy units


2.

If a bird on the ground flies at an

angle of 20 degrees with respect to the horizontal and follows a
straight line path for a distance of 80 m, how far above the ground will it be (measured in m)?

2A.

(1) y = (sin 20
o
)(80 m)

(2) y = (0.342)(80 m)

(3)
y = 27.4 m


3.
A ball is thrown vert
ically upward at a speed that causes it to return to its starting point 6.0 seconds
later. Find the initial velocity.

3A.

Realize that the ball stops momentarily in the air at maximum height:

(1)
v = v
o

+ at

(2) v = v
o

-

g
t

(3)
0

= v
o

-

gt

(4) v
o

=
gt

(5) v
o

= (9.8 m/s
2
)(3.0 sec)

(6)
v
o

= 29.4 m/s


4.

A rock, released at rest from the top of a tower, hits the ground after falling for 3.00s. What is the
height of the tower?

4A.

(1)
d = v
o
t + ½ at
2

(2)

d =
v
o
t



½ gt
2

(3) d =
-

½ gt
2

(4)

d

=
-

½ (9.8 m/s
2
)(3 s)
2

(5)

d =
-

44.1 m

(6)

h = 44.1 m


MR. SURRETTE


VAN NUYS HIGH SCHOOL

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PHYSICS


5.

A ball is released and it rolls down a hill with constant acceleration. At the end of one second, it has
traveled 2 m. How far has it traveled after four seconds?

5A.

GIVEN: x
1

= 2 m

FIND
: x
4

(1)

x
1

=
v
o
t

+ ½ at
2

(2) x
1

= ½ at
2

(3)

2
[x
1

= ½ at
2
]

(4)

2x
1

= at
2

(5)

a = (2x
1

/ t
2
)

(6)

a = 2 (2 m) / (1 s)
2

(7)
a = 4 m/s
2

(8)

x
4

=
v
o
t

+ ½ at
2

(9)

x
4

= ½ (
4 m/s
2
)(4 s)
2

(10)

x
4

= 32.0 m


6.

A locomotive slows from 25 m/s to zero in 11

s. What distance does it travel?

6A.

(1)

d = ½ (v + v
o
)t

(2) d = ½ (
0 m/s

+ 25 m/s)(11 s)

(3)

d = ½ (25 m/s)(11 s)

(4)

d = 137.5 m


7.

A rock is thrown downward from the top of a tower with an initial speed of 8 m/s. If the rock hits the
ground af
ter 2.50 s, what is the speed of the rock as it hits the ground?

7A.

(1)
v = v
o

+ at

(2)

v = v
o

-

g
t

(3) v = (
-

8 m/s)


(9.8 m/s
2
)(2.50 s)

(4)
v =
-

32.5 m/s