armington (kma786) – hw0127 – fete – (57165)
1
This printout should have 10 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
A car travels along a straight stretch oF road.
It proceeds For 16
.
2 mi at 56 mi
/
h, then
20
.
8 mi at 44 mi
/
h, and fnally 49
.
8 mi at
34
.
7 mi
/
h.
What is the car’s average velocity during
the entire trip?
Correct answer: 39
.
5053 mi
/
h.
Explanation:
Let :
d
A
= 16
.
2 mi
,
v
A
= 56 mi
/
h
,
d
B
= 20
.
8 mi
,
v
B
= 44 mi
/
h
,
d
C
= 49
.
8 mi
,
and
v
C
= 34
.
7 mi
/
h
.
The total time the car spent on the road is
Δ
t
=
d
A
v
A
+
d
B
v
B
+
d
C
v
C
=
16
.
2 mi
56 mi
/
h
+
20
.
8 mi
44 mi
/
h
+
49
.
8 mi
34
.
7 mi
/
h
= 2
.
19717 h
,
so the average velocity is
v
=
Δ
d
Δ
t
=
d
A
+
d
B
+
d
C
Δ
t
=
16
.
2 mi + 20
.
8 mi + 49
.
8 mi
2
.
19717 h
=
39
.
5053 mi
/
h
.
002
10.0 points
A car makes a 244 km trip at an average
speed oF 40
.
2 km
/
h. A second car starting
1 h later arrives at their mutual destination
at the same time.
What was the average speed oF the second
car For the period that it was in motion?
Correct answer: 48
.
1295 km
/
h.
Explanation:
Let :
d
= 244 km
,
v
= 40
.
2 km
/
h
,
and
Δ
t
= 1 h
.
Average velocity is
v
=
d
t
.
Δ
t
=
t
1

t
2
=
d
v
1

d
v
2
d
v
2
=
d
v
1

Δ
t
v
2
d
=
1
d
v
1

Δ
t
·
v
1
v
1
v
2
=
d v
1
d

v
1
Δ
t
=
(244 km)(40
.
2 km
/
h)
244 km

(40
.
2 km
/
h)(1 h)
=
48
.
1295 km
/
h
.
003 (part 1 of 2) 10.0 points
A runner is jogging at a steady 3
.
1 km
/
hr.
When the runner is 4
.
4 km From the fnish
line, a bird begins ±ying From the runner to
the fnish line at 9
.
3 km
/
hr (3 times as Fast
as the runner). When the bird reaches the
fnish line, it turns around and ±ies back to
the runner.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '07
 Swinney
 mechanics, Velocity, Automobile, Antilock braking system

Click to edit the document details