Transcript U3D1 HW Sol HON 1

Vectors Problem Packet
Homework Solutions
(1-14)
7. A man walks 9 km east and then 15 km west. Find his
distance traveled and his displacement.
Resultant
is Dd
Dd
9 km [E]
Added the vectors
tip-to-tail
15 km [W]
Distance = 9 km + 15 km = 24 km
Dd = 6 km [W]
Vector: direction
and “little hat”
Scalar: No
direction
9. A robin is flying south for the winter at a rate of 40 miles
per hour when it runs into a hurricane blowing due
west at 100 miles per hour. What is the new velocity of
the robin?
Find the resultant,
which is the new
velocity.
R =v
q
40 mi/h
100 mi/h
m
100
Obviously,
s the
tan
q

m2
was
R 2 robin
40 2 40
100
s
km
o
v  107.7
[ S 68 W ]
sped up 100
by the m
h
R  q40
 tan 100 107
68 .7
2
1
2


hurricane!!!
 40 
o
s
10. Tarzan is hanging from one of his vines. His weight (his force) is
210 pounds, directed Southwards (actually, downward would be
more appropriate). A boy sees that Tarzan wants to be pushed on
his vine. The boy pushes with a force of 90 pounds West. The
rope can withstand a 225-lb force. What happens to the “King of
the Apes”? (Find both the magnitude and the direction of the
resultant).
Find the resultant,
which is the force in
the vine.
Since 228.5 lbs is
greater than what
the
rope
can2 take
2
2
lb
R(225
 90
90
210
lbs),
Tarzan
tan q 
lb
will210
fall!
90 

o
2 
2  23
q

tan
R  90 210
210   228.5
1
R=F
q
210 lbs
[Down]
90 lbs [W]
lb
F  228.5 lb [ D 23 W ]
o
11. The world is again being attacked by hostile aliens from outer
space. This looks like a job for Superman! With his super
strength, he is able to fend off the attack. However, he wants to
rid the earth of these dangerous aliens forever. As they fly away,
they must pass close to our sun. They are heading on a bearing
of East with a speed of Worp 8. If Superman can change their
course to a bearing of between [N 61o E] and [N 64o E], they will
crash into the sun. He gives them a velocity of Worp 4 on a
bearing North. What happens? Has Good triumphed again?
o,
4 <worps
Since
61
63.4
<
64
tan q 
the world
is safe!!!!!
8 worps
R
Worp 4 [N]
q
4
q  tan    26.6o
8
1
Worp 8 [E]
o
direction is [ E 26.6 N ]
or ….
o
[ N 63.4 E ]
12. On a television sports review show, a film clip is shown of the great tackle
that ended the 1960 championship football game in Philadelphia. Jim
Taylor of Green Bay was running forward down the field for a touchdown
with a force of 1000 newtons (direction = South). Check Bednardek of
Philadelphia runs across the field “at right angles to Taylor’s path”
(direction = West) and tackles him with a force of 2500 newtons. Find the
resultant of this crash.
Find the resultant,
which is “resulting”
force.
R =F
2500 N
2500
tan q 2
2
R  2500 
1000 2
1000
 2500 
 1000 
q
1000 N
What a hit!!!!!
F  2,693 N [S 68 W ]
2 1
2  68o
q

tan


R  2500  1000  2,692.58 N
o
13. A kite weighs 5 newtons (downward force). A girl
throws the kite straight up in the air with a force of 20
newtons. If the wind is blowing horizontally with a
force of 20 newtons EAST, find the new force acting
upon it.
F
20 N
15 N
q
20 N
F  152  202  25N
 15 
q  tan    36.9o
 20 
20 N
1
5N
F  25N [ E 37 up]
o
14. A clock has a minute hand that is 15 in long. Find:
- the average speed of the minute hand’s tip.
- the inst. speed of the minute hand’s tip as it passes the 3 on the clock.
- the displacement of the minute hand’s tip in moving from 1:30 to 1:45.
speed 
Ddistance
d  15
up
]
 2R2 2[
left
(15 in )45
 94
.2 in
15
Dd
45o
distance
time
15
time  1 hour  60 min  3600 sec
speed 
45 – 45 – 90
Triangle 
94.2 in
 .026 ins
3600 sec
The minute hand’s speed is
constant!!! Thus, the instantaneous
speed equals .026 m/s