NAME_____________ DATE_________

Pythogarean: **V**^{2}= Vx^{2} + Vy^{2}

Vx=
**V** cos (Ø)

Vy= **V**sin (Ø)

Ø=
tan-1(Vy/Vx)

VECTOR 3 PROBLEMS **ANSWERS!!!**

Draw three
different ways to add the three vectors to get a resultant:

*There are
Six different ways to add these!*

(From
book Chapter 3 pgs 113-117)

23. A submarine dives 110
meters at an angle of 10 degrees below the horizontal. What are the horizontal
and vertical components of the sub’s displacement? See Sample Problem 3B.

* Dx= D*

* *

*Dy= D*

25. A golfer takes two putts
to sink his ball in the hole once he is on the green. The first putt displaces
the ball 6 m east, and the second putt displaces it 5.4 m south. What
displacement would put the ball in the hole on one putt? See Sample Problem 3A

*D**=?, D ^{2}=Dx^{2}+Dy^{2}=
6^{2}+5.4^{2}=36+29.16=65.16, so D=8.072 m*

* *

*Ø = tan ^{-1} (opp/adj) = tan^{-1}(5.4/6)=tan*

* *

*So, D*

27. A roller coaster travels
41.1 meters at an angle of 40 degrees above the horizontal. How far does it
move horizontally and vertically? See Sample Problem 3B

*Dx= D*

* *

*Dy= D*

29. A person walks 10 m east,
300 m south, 150 m at 30 degrees south of west, then 200 m at 60 degrees north
of west. What is the person’s resultant displacement from the starting point?
See Sample Problem 3C.

*But we can change this to
all north/south and east/west by finding the components of the two angled
trips….*

*Dx= D*

*Dy= D*

* *

*Dx= D*

*Dy= D*

* *

*So the trip
with just north/south and east west looks like:*

* *

* *

*Or we could
just add all the north/south and east/west to make one big right triangle….*

*Y= -300 m S +
-75 m S + 173.2 m N = -201.8 South*

*X= +10 m E + -
129.9 m W + -100 m W = -219.9 m West*

*D**=?, D ^{2}=Dx^{2}+Dy^{2}=
219.9^{2}+127.55^{2}=48356 +16269=64625, so D=254.2 m*

* *

*Ø = tan ^{-1} (opp/adj) = tan^{-1}(127.55/219.9)=tan*

* *

*So, D*

* *

* But in the book, the first leg is 100 m,
so Mr. T copied wrong, so *

*The big
triangle should be 129.9 m west, 201.8 m south…, so the hypoteneuse is 239 m at
57.2° South of West*

* *

51. A river flows due east at
1.5 m/s. A boat crosses from the south shore to the north shore by maintaining
a constant velocity of 10 m/s due north relative to the water.

a) What is the
velocity of the boat as viewed by an observer on the shore?

*V**=?, V ^{2}=Vx^{2}+Vy^{2}= 1.5^{2}+10^{2}=2.25
+100=102.25, so V=10.11 m/s*

* *

*Ø = tan ^{-1}
(opp/adj) = tan^{-1}(10/1.5)=tan*

* *

*So, V*

b) If the river is
325 m wide, how far downstream has the boat moved by the time it reaches the
north shore?

*Time is the same in all
directions!*

*Dy=VyT so 325m=10m/s * T
so T = 32.5 seconds*

* *

*Dx=VxT so Dy = 1.5 m/s *
32.5 sec = 48.75 meters*

See Sample Problem
3F

53. A hunter wishes to cross
a river that is 1.5 km wide and that flows with a speed of 5 km/hr. The hunter
uses a small powerboat that moves at a maximum speed of 12 km/hr with respect
to the water. What is the minimum time needed for crossing?

*It doesn’t matter how fast
the river flows!... If he aims straight across relative to the river he will
travel the shortest distance across so T=D/V = 1.5 km/12 km/hr = .125 hrs = 7.5
minutes!*

* *

See Sample Problem 3F

**** Honors:

In problem 53, what
is the hunter’s displacement?

*If t= .125 hrs, then
Dx=VxT=5(.125)=.625km*

* *

*V ^{2}*

*So D=V*

Suppose someone is .2 downstream from the hunter’s original
position and starts 3 minutes after him. What speed and bearing would they have
to have to catch him on the bank?

*.2 km down stream means
they need to go (.625-.2)= .425 km in the x direction*

*If they start 3 minutes
after him that means they have (7.5-3)=4.5 minutes=.075 hrs. Their speed in the
x direction (along the river) would be*

* Vx=Dx/T=.425/.075= 5.666 km/hr.*

* *

*They still need to go
across the 1.5 km river in the y direction, same time of .075
hrs….Vy=Dy/T=1.5/.075=20 km/hr.*

* *

*So V^{2}*

* *