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Relative Velocity
Introduction
| Relative Velocity |
Each velocity is labeled by two subscripts: the first
refers to the object, the second to the reference frame in which it has a
velocity.
For any two objects or reference frames, A and B, the velocity of A
relative to B has the same magnitude, but opposite direction, as the
velocity of B relative to A. vBA = -vAB
The resultant velocity on the left side has the subscripts left when
inner and outer subscripts on the right cancel. vAB = vAC
+ vCB
If the above equation is correct then vCB = vAB
- vAC
All other rules for vectors apply. |
Table of Contents
Relative Velocity Problems
| 1. |
Plane A flies from Fayetteville with a velocity of
900. km/h at 240.o. Plane B flies from Raleigh with a velocity of
500. km/h at 160.o.
What is the velocity of A with respect to B?
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| 2. |
A boat is at the bottom shore with the current going from right to left.
The boat’s speed in still water is
vBW = 40.0 km/h. If the boat is
to travel directly across a river whose current has speed vWS = 11.0
km/h, at what angle must the boat head?
|
| 3. |
The desired velocity of the boat compared to
the shore is 50.0 km/h straight across the river. The current is
traveling from left to right at 15.0 km/h. What is the necessary
velocity of the boat compared to the water?
|
| 4. |
A plane whose airspeed is 350. km/h heads due
North compared to the air. A 125 km/h
southeast wind (wind coming from the south east) is blowing. What is the
velocity of the plane with respect to the ground. |
Table of Contents
Relative Velocity Answers
| 1. |
Plane A flies from Fayetteville with a velocity of
900. km/h at 240.o. Plane B flies from Raleigh with a velocity of
500. km/h at 160.o.
What is the velocity of A with respect to B?
|
|
A
= Plane A
B = Plane B
G = Ground
vAB = vAG
+ vGB
|
The resultant velocity on the left side has the subscripts left when
inner and outer subscripts on the right cancel.
vAB = vAC
+ vCB
|
| vAB
= vAG - vBG
vAB = vAG
+ ( - vBG )
vBG = 500. km/h at 160.o
thus
- vBG = 500. km/h at
340.o |
For any two objects or reference frames, A and B, the velocity of A
relative to B has the same magnitude, but opposite direction, as the
velocity of B relative to A.
vBA = -vAB
Either Add or substract 180.o
keeping the answer between 0o and 360.o.
|
 |
North
= 500. km/h x Sin 70.o
=
470. km/h
West = 500. km/h x Cos 70.o
= 171 km/h
South = 900. km/h x Sin 30.o
= 450.
km/h
West = 900. km/h x Cos 30.o
= 779 km/h
|
Break
the two vectors into components. |
| North-South
- North positive and South negative
East-West - West positive and East
negative
North-South = + 470.
km/h - 450. km/h = + 20. km/h (North)
East-West = + 171 km/h
+ 779 km/h = + 950. km/h (West)
|
For
each dimension, assign a positive and negative.
Add the vectors in the same dimension. |
 |
resultant
= (( 20. km/h)2 + (950. km/h)2) 1/2
resultant = 950. km/h
20. km/h
angle = Cos -1 ( ---------------)
950. km/h
angle = 89o
heading = 360.o -
89o = 271o
Resultant = 950. km/h at 271o
|
Draw
the two components and solve for the resultant vector |
|
|
|
| 2. |
A boat is at the bottom shore with the current going from right to left.
The boat’s speed in still water is
vBW = 40.0 km/h. If the boat is
to travel directly across a river whose current has speed vWS = 11.0
km/h, at what angle must the boat head?
|
|
B
= Boat
W = Water
S = Shore
vBS = vBW
+ vWS |
The resultant velocity on the left side has the subscripts left when
inner and outer subscripts on the right cancel.
vAB = vAC
+ vCB
|
|
vBS
= ((40.0 km/h)2 + (11.0 km/h)2) 1/2
vBS = 41.5 km/h
11.0 km/h
angle = Sin -1 ( --------------- )
41.5 km/h
angle = 15.4o
|
Since
vBW only gives a speed without a direction;
calculate vBS if vBW is straight across the
river. |
| The
boat will need to head to the right at 15.4o to go
straight
across the river. |
The
boat should head the opposite direction from the calculated
direction. |
|
|
|
| 3. |
The desired velocity of the boat compared to
the shore is 50.0 km/h straight across the river. The current is
traveling from left to right at 15.0 km/h. What is the necessary
velocity of the boat compared to the water?
|
|
B
= Boat
W = Water
S = Shore
vBS = vBW
+ vWS
|
The resultant velocity on the left side has the subscripts left when
inner and outer subscripts on the right cancel.
vAB = vAC
+ vCB
|
| vBW
= vBS - vWS
vBW = vBS
+ vSW
vSW = 15.0 km/h
from right to left |
For any two objects or reference frames, A and B, the velocity of A
relative to B has the same magnitude, but opposite direction, as the
velocity of B relative to A.
vBA = -vAB
Either Add or substract 180.o
keeping the answer between 0o and 360.o.
|
 |
vBW
= ((50.0 km/h)2 + (15.0 km/h)2) 1/2
vBW = 52.2 km/h
15.0 km/h
Angle = Sin -1 ( ---------------- )
50.0 km/h
Angle = 17.5o
Boat must heat to the left at an angle
of 17.5o and a speed of 52.2 km/h |
Draw
the addition of
vBS and vSW
and
calculate the value using trig. |
|
|
|
| 4. |
A plane whose airspeed is 350. km/h heads due
North compared to the air. A 125 km/h
southeast wind (wind coming from the south east) is blowing. What is the
velocity of the plane with respect to the ground.
|
|
Plane
= P
Air = A
Ground = G
vPG = vPA
+ vAG
|
The resultant velocity on the left side has the subscripts left when
inner and outer subscripts on the right cancel.
vAB = vAC
+ vCB
|
 |
Since
vPA is due North, it is all ready
a component.
If the air is coming from the Southeast,
it is heading Northwest. This is a heading of 315o.
West = 125 km/h x Cos 45o
= 88.4 km/h
North = 125 km/h x Sin 45o
=
88.4 km/h
|
Break
the two vectors into components. |
North-South
North is positive and South is negative
East-West West is positive and
East is negative
North-South = 350. km/h +
88.4 km/h = + 438 km/h (North)
East-West = + 88.4 km/h (West)
|
For
each dimension, assign a positive and negative.
Add the vectors in the same dimension.
|
 |
vPG
= ((88.4 km/h)2 + (438 km/h)2) 1/2
vPG = 447 km/h
438 km/h
angle = Sin -1 ( --------------- )
447 km/h
angle = 78.5o
vPG = 447 km/h at 348.5o
|
Draw
the two components and solve for the resultant vector |
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Table of Contents
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