Physics 6A Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB A VECTOR describes anything that has both a MAGNITUDE and a DIRECTION The MAGNITUDE describes the size of the vector ID: 175518
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Slide1
Vector Examples
Physics 6A
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide2
A
VECTOR
describes anything that has both a
MAGNITUDE and a DIRECTIONThe MAGNITUDE describes the size of the vector.The DIRECTION tells you where the vector is pointing.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide3
Here’s a typical example:
An airplane is flying East at a velocity of 600 kilometers per hour.
Here the magnitude (speed) is 600 km/hr and the direction is East.A diagram of this vector might look like this:
V= 600 km/hr
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide4
Here’s a typical example:
An airplane is flying East at a velocity of 600 kilometers per hour.
Here the magnitude (speed) is 600 km/hr and the direction is East.A diagram of this vector might look like this:
V= 600 km/hr
Normally vectors will be written in
BOLDFACE
or with an arrow above the letter:
= 600 km/
hr
East or
V
=600 km/
hr
East
Notice that the diagram above has a non-bold V – this means magnitude.
So V=600 km/
hr
describes the magnitude
of .You might also see this as || = 600 km/hr.The absolute value bars indicate magnitude.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide5
Let’s try a quick example with our airplane.
Suppose that while this plane is flying 600 km/hr East, it then encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY of the airplane?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide6
Let’s try a quick example with our airplane.
Suppose that while this plane is flying 600 km/hr East, it then encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY of the airplane?
The plane gets blown off course, of course. We need to find the new VELOCITY.
Like any vector, it will have a MAGNITUDE (speed) and a DIRECTIONHere is a diagram:
100 km/hr
600 km/hr
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide7
Let’s try a quick example with our airplane.
Suppose that while this plane is flying 600 km/hr East, it then encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY of the airplane?
The plane gets blown off course, of course. We need to find the new VELOCITY.
Like any vector, it will have a MAGNITUDE (speed) and a DIRECTIONHere is a diagram:
So how do we find the new SPEED of the plane?
100 km/hr
600 km/hr
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide8
Let’s try a quick example with our airplane.
Suppose that while this plane is flying 600 km/hr East, it then encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY of the airplane?
The plane gets blown off course, of course. We need to find the new VELOCITY.
Like any vector, it will have a MAGNITUDE (speed) and a DIRECTIONHere is a diagram:
So how do we find the new SPEED of the plane?Answer: Add the vectors together
Is the new speed just 600 km/hr +100 km/hr = 700 km/hr?
100 km/hr
600 km/hr
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide9
Let’s try a quick example with our airplane.
Suppose that while this plane is flying 600 km/hr East, it then encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY of the airplane?
The plane gets blown off course, of course. We need to find the new VELOCITY.
Like any vector, it will have a MAGNITUDE (speed) and a DIRECTIONHere is a diagram:
So how do we find the new SPEED of the plane?Answer: Add the vectors together
Is the new speed just 600 km/hr +100 km/hr = 700 km/hr?
NO
Note: if the wind blows East, then we
can
simply add the numbers together.
100 km/hr
600 km/hr
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide10
Let’s try a quick example with our airplane.
Suppose that while this plane is flying 600 km/hr East, it then encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY of the airplane?
The plane gets blown off course, of course. We need to find the new VELOCITY.
Like any vector, it will have a MAGNITUDE (speed) and a DIRECTIONHere is a diagram:
So how do we find the new SPEED of the plane?
The new speed will be the magnitude of the SUM of the 2 vectors. In this case, the sum will look like the hypotenuse of a right triangle, labeled
total
above.
100 km/hr
600 km/hr
total
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide11
Let’s try a quick example with our airplane.
Suppose that while this plane is flying 600 km/hr East, it then encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY of the airplane?
The plane gets blown off course, of course. We need to find the new VELOCITY.
Like any vector, it will have a MAGNITUDE (speed) and a DIRECTIONHere is a diagram:
We need to use the Pythagorean Theorem to find the magnitude of
total
100 km/hr
600 km/hr
V
total
≈ 608 km/hr
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
total
Slide12
Let’s try a quick example with our airplane.
Suppose that while this plane is flying 600 km/hr East, it then encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY of the airplane?
The plane gets blown off course, of course. We need to find the new VELOCITY.
Like any vector, it will have a MAGNITUDE (speed) and a DIRECTIONHere is a diagram:
100 km/hr
600 km/hr
V
total
≈ 608 km/hr
Now we have the magnitude, but what is the direction?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
total
Slide13
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide14
Let’s try a quick example with our airplane.
Suppose that while this plane is flying 600 km/hr East, it then encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY of the airplane?
The plane gets blown off course, of course. We need to find the new VELOCITY.
Like any vector, it will have a MAGNITUDE (speed) and a DIRECTIONHere is a diagram:
100 km/hr
600 km/hr
V
total
≈ 608 km/hr
Now we have the magnitude, but what is the direction?
We use our right-triangle rules for this:
tan(
θ
) = opposite/adjacent
= 100/600
θ
= tan
-1(1/6)
θθ ≈ 9.5°
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
total
Slide15
Suppose that while this plane is flying 600 km/hr East, it then encounters a wind blowing North at 100 km/hr. How does this affect the VELOCITY of the airplane?
V
total
≈ 608 km/hr
100 km/hr
600 km/hr
θ
θ
≈ 9.5°
Finally! We have our answer!
The plane will speed up to 608 km/hr, but be pushed off-course by an angle of 9.5°.
Note: the angle may be described as 9.5° North of East.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
total
Slide16
Vector Notation
V
total
≈ 608 km/hr
100 km/hr
600 km/hr
θ
θ
≈ 9.5°
You may see a vector represented in a few different ways:
Next we will work out a (slightly harder) problem from a textbook.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
V
x
=600 km/
hr
;
V
y
=100 km/
hr
(600 km/hr,100 km/
hr
)
These all refer to the same vector.
total
Slide17
A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSBSlide18
The first step is to set up a coordinate system. The most convenient thing to do is just call East the x-direction and North the y-direction, as shown.
y
x
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.Slide19
Now we look at each leg of the journey and find the components:
First leg: This one is easy. 2.00 km in the x-direction; 0.00 km in the y-direction
We could label everything to keep it all organized: If the first leg is represented by the vector
V1, we can write the components separately.V1x
= +2.00 km (note: positive indicates East – negative would be West)V1y = 0.00 km
y
x
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.Slide20
y
x
The 2
nd
leg has components in both x- and y-directions, so we can find the components from our triangle rules:
V
2x
=
V
2y
=
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.Slide21
y
x
The 2
nd
leg has components in both x- and y-directions, so we can find the components from our triangle rules:
V
2x
= +3.50*cos(45
°) ≈ 2.47 km (positive for East)
V
2y
= -3.50*sin(45
°) ≈ -2.47 km (negative for South)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.Slide22
y
x
The 3
rd
leg has unknown components, but we can write them as unknowns:
V
3x
= ?
V
3y
= ?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.Slide23
y
x
The next step is the only one where you have to think about the problem.
Since we know that the total journey from start to finish is 5.80 km East, we know the components:
V
total,x
=
V
total,y
=
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.Slide24
y
x
The next step is the only one where you have to think about the problem.
Since we know that the total journey from start to finish is 5.80 km East, we know the components:
V
total,x
= +5.80 km
V
total,y
= 0.00 km
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.Slide25
y
x
Finally we can just add everything up and solve for our unknowns.
We will have 2 equations – one for each direction:
V
1x
+ V
2x
+ V
3x
= V
total,x
V
1y
+ V
2y
+ V
3y
= V
total,yPrepared by Vince ZacconeFor Campus Learning Assistance Services at UCSBA sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.Slide26
y
x
Finally we can just add everything up and solve for our unknowns.
We will have 2 equations – one for each direction:
V
1x
+ V
2x
+ V
3x
= V
total,x
2.00 + 2.47 + V
3x
= 5.80
V
3x
= +1.33 km
(positive is East)
V1y + V2y + V3y = Vtotal,y0.00 – 2.47 + V3y = 0.00V3y = +2.47 km(positive is North)Prepared by Vince ZacconeFor Campus Learning Assistance Services at UCSB
A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.Slide27
y
x
Now that we have the components, the last step is to combine them to get the magnitude and direction of the vector for the 3
rd
leg.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.Slide28
y
x
Now that we have the components, the last step is to combine them to get the magnitude and direction of the vector for the 3
rd
leg.
Use the Pythagorean Theorem to find the magnitude:
V
3
= 2.81 km
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
For extra fun, find the angle of the vector too!
A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.Slide29
y
x
Now that we have the components, the last step is to combine them to get the magnitude and direction of the vector for the 3
rd
leg.
Use the Pythagorean Theorem to find the magnitude:
V
3
= 2.81 km
Use tan(
θ
) = V
y
/V
x
to find the direction
tan(
θ
) = 2.47/1.33
Θ = tan-1(1.86)
Θ = 61.7° North of East(this angle is measured above the x-axis)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of her starting point. Find the magnitude of the third leg of the journey.