Fillable Problem 2C Holt Physics DISPLACEMENT WITH CONSTANT ACCELERATION

Fill Online, Printable, Fillable, Blank Problem 2C Holt Physics DISPLACEMENT WITH CONSTANT ACCELERATION Form

Use Fill to complete blank online OTHERS pdf forms for free. Once completed you can sign your fillable form or send for signing. All forms are printable and downloadable.

Problem 2C Holt Physics DISPLACEMENT WITH CONSTANT ACCELERATION

On average this form takes 3 minutes to complete

The Problem 2C Holt Physics DISPLACEMENT WITH CONSTANT ACCELERATION form is 4 pages long and contains:

  • 0 signatures
  • 0 check-boxes
  • 12 other fields

Country of origin: OTHERS
File type: PDF

BROWSE OTHERS FORMS

Related forms
Fill has a huge library of thousands of forms all set up to be filled in easily and signed.
Fill in your chosen form
Sign the form using our drawing tool
Send to someone else to fill in and sign.
find another form
 
Problem 2C 7
NAME ______________________________________ DATE _______________ CLASS ____________________
Holt Physics
Problem 2C
DISPLACEMENT WITH CONSTANT ACCELERATION
PROBLEM
In England, two men built a tiny motorcycle with a wheel base (the dis-
tance between the centers of the two wheels) of just 108 mm and a wheel’s
measuring 19 mm in diameter. The motorcycle was ridden over a distance
of 1.00 m. Suppose the motorcycle has constant acceleration as it travels
this distance, so that its final speed is 0.800 m/s. How long does it take the
motorcycle to travel the distance of 1.00 m? Assume the motorcycle is ini-
tially at rest.
SOLUTION
Given: v
f
= 0.800 m/s
v
i
= 0 m/s
∆ x = 1.00 m
Unknown: ∆t = ?
Use the equation for displacement with constant acceleration.
∆ x =
1
2
(v
i
+ v
f
)∆t
Rearrange the equation to calculate ∆t.
∆t =
v
2
f
∆
+
x
v
i
∆t ==
0
2
.
.
8
0
0
0
0
s
=
2.50 s
(2)(1.00 m)

0.800
m
s
+ 0
m
s
HRW material copyrighted under notice appearing earlier in this book.
ADDITIONAL PRACTICE
1. In 1993, Ileana Salvador of Italy walked 3.0 km in under 12.0 min. Sup-
pose that during 115 m of her walk Salvador is observed to steadily in-
crease her speed from 4.20 m/s to 5.00 m/s. How long does this increase
in speed take?
2. In a scientific test conducted in Arizona, a special cannon called HARP
(High Altitude Research Project) shot a projectile straight up to an alti-
tude of 180.0 km. If the projectile’s initial speed was 3.00 km/s, how long
did it take the projectile to reach its maximum height?
3. The fastest speeds traveled on land have been achieved by rocket-
powered cars. The current speed record for one of these vehicles is
about 1090 km/h, which is only 160 km/h less than the speed of
sound in air. Suppose a car that is capable of reaching a speed of
Menu
Lesson
Print
Holt Physics Problem Workbook8
NAME ______________________________________ DATE _______________ CLASS ____________________
1.09 × 10
3
km/h is tested on a flat, hard surface that is 25.0 km long. The
car starts at rest and just reaches a speed of 1.09 × 10
3
km/h when it
passes the 20.0 km mark.
a. If the car’s acceleration is constant, how long does it take to make
the 20.0 km drive?
b. How long will it take the car to decelerate if it goes from its maxi-
mum speed to rest during the remaining 5.00 km stretch?
4. In 1990, Dave Campos of the United States rode a special motorcycle
called the Easyrider at an average speed of 518 km/h. Suppose that at
some point Campos steadily decreases his speed from 100.0 percent to
60.0 percent of his average speed during an interval of 2.00 min. What is
the distance traveled during that time interval?
5. A German stuntman named Martin Blume performed a stunt called “the
wall of death.” To perform it, Blume rode his motorcycle for seven
straight hours on the wall of a large vertical cylinder. His average speed
was 45.0 km/h. Suppose that in a time interval of 30.0 s Blume increases
his speed steadily from 30.0 km/h to 42.0 km/h while circling inside the
cylindrical wall. How far does Blume travel in that time interval?
6. An automobile that set the world record for acceleration increased speed
from rest to 96 km/h in 3.07 s. How far had the car traveled by the time
the final speed was achieved?
7. In a car accident involving a sports car, skid marks as long as 290.0 m
were left by the car as it decelerated to a complete stop. The police report
cited the speed of the car before braking as being “in excess of 100 mph”
(161 km/h). Suppose that it took 10.0 seconds for the car to stop. Esti-
mate the speed of the car before the brakes were applied. (REMINDER:
Answer should read, “speed in excess of . . .”)
8. Col. Joe Kittinger of the United States Air Force crossed the Atlantic
Ocean in nearly 86 hours. The distance he traveled was 5.7 × 10
3
km.
Suppose Col. Kittinger is moving with a constant acceleration during
most of his flight and that his final speed is 10.0 percent greater than his
initial speed. Find the initial speed based on this data.
9. The polar bear is an excellent swimmer, and it spends a large part of its
time in the water. Suppose a polar bear wants to swim from an ice floe to
a particular point on shore where it knows that seals gather. The bear
dives into the water and begins swimming with a speed of 2.60 m/s. By
the time the bear arrives at the shore, its speed has decreased to 2.20 m/s.
If the polar bear’s swim takes exactly 9.00 min and it has a constant de-
celeration, what is the distance traveled by the polar bear?
HRW material copyrighted under notice appearing earlier in this book.
Menu
Lesson
Print
Holt Physics Solution ManualII Ch. 2–4
II
Copyright © by Holt, Rinehart and Winston. All rights reserved.
8. v
i
=+245 km/h
a
avg
=−3.0 m/s
2
v
f
= v
i
−(0.200) v
i
Givens Solutions
9. ∆x = 3.00 km
∆t = 217.347 s
a
avg
=−1.72 m/s
2
v
f
= 0 m/s
v
i
= v
avg
=
∆
∆
x
t
==13.8 m/s
t
stop
=
v
f
a
a
−
vg
v
i
=
==8.02 s
−13.8 m/s

−1.72 m/s
2
0 m/s − 13.8 m/s

−1.72 m/s
2
3.00 × 10
3
m

217.347 s
10. ∆x =+5.00 × 10
2
m
∆t = 35.76 s
v
i
= 0 m/s
∆t = 4.00 s
v
max
= v
avg
+ (0.100) v
avg
v
f
= v
max
= (1.100)v
avg
= (1.100)
∆
∆
x
t
= (1.100)
5.0
3
0
5
×
.7
1
6
0
s
2
m
=+15.4 m/s
a
avg
=
∆
∆
t
v
=
v
f
∆
−
t
v
i
==+3.85 m/s
2
15.4 m/s − 0 m/s

4.00 s
1. ∆x = 115 m
v
i
= 4.20 m/s
v
f
= 5.00 m/s
∆t =
v
i
2∆
+
x
v
f
=
4.20
(
m
2)
/
(
s
1
+
15
5.
m
00
)
m/s
=
(2
9
)
.
(
2
1
0
1
m
5
/
m
s
)
= 25.0 s
Additional Practice 2C
2. ∆x = 180.0 km
v
i
= 3.00 km/s
v
f
= 0 km/s
∆t =
v
i
2∆
+
x
v
f
==
3
3
.
6
0
0
0
.0
km
km
/s
= 1.2 × 10
2
s
(2)(180.0 km)

3.00 km/s + 0 km/s
3. v
i
= 0 km/h
v
f
= 1.09 × 10
3
km/h
∆x = 20.0 km
∆x = 5.00 km
v
i
= 1.09 × 10
3
km/h
v
f
= 0 km/h
a. ∆t =
v
i
2
+
∆x
v
f
=
∆t ==
b. ∆t =
v
i
2
+
∆x
v
f
=
∆t ==33.0 s
10.0 × 10
3
m

(1.09 × 10
3
km/h)
36
1
0
h
0s

1
1
00
k
0
m
m
(2)(5.00 × 10
3
m)

(1.09 × 10
3
km/h + 0 km/h)
36
1
0
h
0s

1
1
00
k
0
m
m
132 s
40.0 × 10
3
m

(1.09 × 10
3
km/h)
36
1
0
h
0s

1
1
00
k
0
m
m
(2)(20.0 × 10
3
m)

(1.09 × 10
3
km/h + 0 km/h)
36
1
0
h
0s

1
1
00
k
0
m
m
v
i
=
245
k
h
m

36
1
0
h
0s
 
1
1
0
k
3
m
m
=+68.1 m/s
v
f
= (1.000 − 0.200) v
i
= (0.800)(68.1 m/s) =+54.5 m/s
∆t =
v
f
a
a
−
vg
v
i
=
54.5 m
−3
/s
.0
−
m
6
/
8
s
.
2
1m/s
= = 4.5 s
−13.6 m/s

− 3.0 m/s
2
Menu
Lesson
Print
Section Two — Problem Workbook Solutions II Ch. 2–5
II
Copyright © by Holt, Rinehart and Winston. All rights reserved.
4. v
i
= v
avg
= 518 km/h
v
f
= (0.600) v
avg
∆t = 2.00 min
v
avg
=
518
k
h
m

60
1
m
h
in

1
1
0
k
3
m
m
= 8.63 × 10
3
m/min
∆x =
1
2
(v
i
+ v
f
)∆t =
1
2
[v
avg
+ (0.600) v
avg
]∆t =
1
2
(1.600)(8.63 × 10
3
m/min)(2.00 min)
∆x = 13.8 × 10
3
m = 13.8 km
Givens Solutions
5. ∆t = 30.0 s
v
i
= 30.0 km/h
v
f
= 42.0 km/h
∆x =
1
2
(v
i
+ v
f
)∆t =
1
2
(30.0 km/h + 42.0 km/h)
36
1
0
h
0s
(30.0 s)
∆x =
1
2
72.0
k
h
m

36
1
0
h
0s
(30.0 s)
∆x = 3.00 × 10
−1
km = 3.00 × 10
2
m
1. v
i
= 186 km/h
v
f
= 0 km/h = 0 m/s
a =−1.5 m/s
2
∆t =
v
f
−
a
v
i
==
−
−
5
1
1
.5
.7
m
m
/s
/
2
s
= 34 s
0 m/s − (186 km/h)
36
1
0
h
0s

1
1
0
k
3
m
m

−1.5 m/s
2
Additional Practice 2D
6. v
f
= 96 km/h
v
i
= 0 km/h
∆t = 3.07 s
∆x =
1
2
(v
i
+ v
f
)∆t =
1
2
(0 km/h + 96 km/h)
36
1
0
h
0s

1
1
0
k
3
m
m
(3.07 s)
∆x =
1
2
96 × 10
3
m
h
(8.53 +×10
−4
h) = 41 m
7. ∆x = 290.0 m
∆t = 10.0 s
v
f
= 0 km/h = 0 m/s
v
i
=
2
∆
∆
t
x
− v
f
=
(2)(
1
2
0
9
.
0
0
.0
s
m)
− 0 m/s =
(Speed was in excess of 209 km/h.)
58.0 m/s = 209 km/h
8. ∆x = 5.7 × 10
3
km
∆t = 86 h
v
f
= v
i
+ (0.10) v
i
v
f
+ v
i
=
2
∆
∆
t
x
v
i
(1.00 + 0.10) + v
i
=
2
∆
∆
t
x
v
i
=
(2)
(
(
2
5
.
.
1
7
0
×
)(
1
8
0
6
3
h
k
)
m)
= 63 km/h
9. v
i
= 2.60 m/s
v
f
= 2.20 m/s
∆t = 9.00 min
∆x =
1
2
(v
i
+ v
f
)∆t =
1
2
(2.60 m/s + 2.20 m/s)(9.00 min)
m
60
in
s
=
1
2
(4.80 m/s)(5.40 × 10
2
s)
∆x = 1.30 × 10
3
m = 1.30 km
2. v
i
=−15.0 m/s
v
f
= 0 m/s
a =+2.5 m/s
2
v
i
= 0 m/s
v
f
=+15.0 m/s
a =+2.5 m/s
For stopping:
∆t
1
=
v
f
a
− v
i
=
0 m/s
2
−
.5
(−
m
1
/
5
s
.
2
0 m/s)
=
1
2
5
.5
.0
m
m
/s
/
2
s
= 6.0 s
For moving forward:
∆t
2
=
v
f
a
− v
i
==
1
2
5
.5
.0
m
m
/s
/
2
s
= 6.0 s
∆t
tot
=∆t
1
+∆t
2
= 6.0 s + 6.0 s = 12.0 s
15.0 m/s − 0.0 m/s

2.5 m/s
2
Menu
Lesson
Print
Download
Save PDF to desktop
Email
Email completed PDF
Send for Signing
Email for others to sign
Print
Print document
Choose a Field
Click on a field and
select the person that should complete it
Want to add a signature?
You can add a signature field
and assign it to someone else
to sign
Choose a Field
Click on a field and
select the person that should complete it

Problem 2C Holt Physics DISPLACEMENT WITH CONSTANT ACCELERATION

This document is locked as it has been sent for signing.

You have successfully completed this document.

Other parties need to complete fields in the document. You will recieve an email notification when the document has been completed by all parties.

This document has been signed by all parties.

Completed 16 October 2020

Message

View Audit Log
Print With Audit Log
Duplicate Document