# Version PREVIEW – Practice 1 – carroll – (11108) 1 This print ```Version PREVIEW – Practice 1 – carroll – (11108)
This print-out should have 10 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
vav =
5 m − (−2 m)
= 1.75 m/s .
4s−0s
003 (part 3 of 5) 10.0 points
Find the average velocity in the time interval
1 s to 4 s.
Serway CP 02 06
001 (part 1 of 5) 10.0 points
The position versus time for a certain object
moving along the x-axis is shown. The objects
initial position is −2 m.
Explanation:
12
Given : xf = 5 m and
xi = 10 m .
b
10
1
position (m)
8
6
b
vav =
b
4
2
b
0
−2
b
−4
−6
1
2
3
4 5
time (s)
6
7
8
004 (part 4 of 5) 10.0 points
Find the average velocity in the time interval
4 s to 8 s.
Explanation:
b
0
5 m − (10 m)
= −1.66667 m/s .
4s−1 s
9
Find the average velocity in the time interval 0 s to 1 s.
Given :
vav =
xf = −5 m
xi = 5 m .
and
−5 m − (5 m)
= −2.5 m/s .
8s−4s
Explanation:
Given : xf = 10 m and
xi = −2 m .
vav =
vav =
xf − x i
∆x
=
∆t
∆t
005 (part 5 of 5) 10.0 points
Find the average velocity in the time interval
0 s to 9 s.
Explanation:
Given : xf = 0 m and
xi = −2 m .
10 m − (−2 m)
= 12 m/s .
1s−0 s
002 (part 2 of 5) 10.0 points
Find the average velocity in the time interval
0 s to 4 s.
Explanation:
Given : xf = 5 m and
xi = −2 m .
vav =
0 m − (−2 m)
= 0.222222 m/s .
9s−0 s
Serway CP 02 15
006 (part 1 of 4) 10.0 points
The position versus time for a certain object
moving along the x-axis is shown. The object’s initial position is −4 m.
Version PREVIEW – Practice 1 – carroll – (11108)
position (m)
6
b
4
b
2
b
b
0
b
−2
b
−4
−6
0
1
2
3
4 5
time (s)
6
7
8
9
Find the instantaneous velocity at 1 s.
Explanation:
The instantaneous velocity is the slope of
the tangent line at that point.
v=
(5 m) − (−4 m)
= 4.5 m/s .
(2 s) − (0 s)
007 (part 2 of 4) 10.0 points
Find the instantaneous velocity at 2.5 s.
010 10.0 points
The Stratosphere Tower in Las Vegas is
1193 ft high. It takes 2.9 min, 28 s to ascend from the ground floor to the top of the
tower using the high-speed elevator.
The acceleration of gravity is 9.81 m/s2 .
Assuming that the elevator starts and ends
at rest, and its acceleration has a constant
magnitude when moving, find the acceleration
of the elevator as a multiple of the acceleration
of gravity.
Explanation:
Let :
h = 1193 ft ,
t1 = 2.9 min + 28 s = 202 s ,
v0 = 0 , and
g = 9.81 m/s2 .
The elevator accelerates uniformly until the
halfway point (where it reaches maximum
speed) and decelerates uniformly until the
top:
Explanation:
v=
(2 m) − (5 m)
= −3 m/s .
(3 s) − (2 s)
008 (part 3 of 4) 10.0 points
Find the instantaneous velocity at 4 s.
Explanation:
v=
(2 m) − (2 m)
= 0 m/s .
(5 s) − (3 s)
009 (part 4 of 4) 10.0 points
Find the instantaneous velocity at 8 s.
Explanation:
v=
(0 m) − (−2 m)
= 1 m/s .
(9 s) − (7 s)
Tipler PSE5 02 106
2
1
1
1
h = v0 ∆t + a (∆t)2 = a (∆t)2
2
2
2
h
h
4h
= 2 = 2
2
(∆t)
t1
1
t1
2
4 (1193 ft)
1m
=
&times;
2
(202 s)
3.281 ft
g
&times;
9.81 m/s2
4 (363.609 m)
=
g
(202 s)2 (9.81 m/s2 )
a=
= 0.00363348 g .
```