Vector Examples
Physics 6A
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A VECTOR describes anything that has both a MAGNITUDE and a DIRECTION
The MAGNITUDE describes the size of the vector.
The DIRECTION tells you where the vector is pointing.
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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:
r
V = 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 V.
You might also see this as |V| = 600 km/hr.
The absolute value bars indicate magnitude.
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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 UCSB
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 DIRECTION
Here is a diagram:
100 km/hr
600 km/hr
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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 DIRECTION
Here is a diagram:
100 km/hr
600 km/hr
So how do we find the new SPEED of the plane?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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 DIRECTION
Here is a diagram:
100 km/hr
600 km/hr
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?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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 DIRECTION
Here is a diagram:
100 km/hr
600 km/hr
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.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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 DIRECTION
Here is a diagram:
100 km/hr
600 km/hr
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 Vtotal above.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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 DIRECTION
Here is a diagram:
100 km/hr
600 km/hr
We need to use the Pythagorean Theorem to find the magnitude of Vtotal
2
2
Vtotal = (600) + (100)
Vtotal ≈ 608 km/hr
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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 DIRECTION
Here is a diagram:
100 km/hr
Vtotal ≈ 608 km/hr
600 km/hr
Now we have the magnitude, but what is the direction?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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 DIRECTION
Here is a diagram:
100 km/hr
Vtotal ≈ 608 km/hr
θ
600 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)
θ ≈ 9.5°
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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?
100 km/hr
θ
600 km/hr
Vtotal ≈ 608 km/hr
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.
Next we will work out a (slightly harder) problem from the textbook.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 1.62
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.
5.80 km
Start
2.00 km
Finish
45°
Third
3.50 km
leg
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 1.62
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.
5.80 km
Start
2.00 km
Finish
y
45°
Third
3.50 km
x
leg
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.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 1.62
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.
5.80 km
Start
2.00 km
Finish
y
45°
Third
3.50 km
x
leg
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 1.62
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.
5.80 km
Start
2.00 km
Finish
y
45°
Third
3.50 km
x
leg
The 2nd leg has components in both x- and y-directions, so we can find the
components from our triangle rules:
V2x =
V2y =
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 1.62
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.
5.80 km
Start
2.00 km
Finish
y
45°
Third
3.50 km
x
leg
The 2nd leg has components in both x- and y-directions, so we can find the
components from our triangle rules:
V2x = +3.50*cos(45°) ≈ 2.47 km
(positive for East)
V2y = -3.50*sin(45°) ≈ -2.47 km
(negative for South)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 1.62
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.
5.80 km
Start
2.00 km
Finish
y
45°
Third
3.50 km
x
leg
The 3rd leg has unknown components, but we can write them as unknowns:
V3x = ?
V3y = ?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 1.62
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.
5.80 km
Start
2.00 km
Finish
y
45°
Third
3.50 km
x
leg
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:
Vtotal,x =
Vtotal,y =
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 1.62
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.
5.80 km
Start
2.00 km
Finish
y
45°
Third
3.50 km
x
leg
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:
Vtotal,x = +5.80 km
Vtotal,y = 0.00 km
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 1.62
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.
5.80 km
Start
2.00 km
Finish
y
45°
Third
3.50 km
x
leg
Finally we can just add everything up and solve for our unknowns.
We will have 2 equations – one for each direction:
V1x + V2x + V3x = Vtotal,x
V1y + V2y + V3y = Vtotal,y
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 1.62
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.
5.80 km
Start
2.00 km
Finish
y
45°
Third
3.50 km
x
leg
Finally we can just add everything up and solve for our unknowns.
We will have 2 equations – one for each direction:
V1x + V2x + V3x = Vtotal,x
V1y + V2y + V3y = Vtotal,y
2.00 + 2.47 + V3x = 5.80
0.00 – 2.47 + V3y = 0.00
V3x = +1.33 km
V3y = +2.47 km
(positive is East)
(positive is North)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 1.62
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.
5.80 km
Start
2.00 km
Finish
y
45°
Third
3.50 km
x
leg
Now that we have the components, the last step is to combine them to get the
magnitude and direction of the vector for the 3rd leg.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 1.62
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.
5.80 km
Start
2.00 km
Finish
y
45°
Third
3.50 km
x
leg
Now that we have the components, the last step is to combine them to get the
magnitude and direction of the vector for the 3rd leg.
Use the Pythagorean Theorem
to find the magnitude:
V3 =
(V3,x )2 + (V3,y )2
V3 =
(1.33 )2 + (2.47 )2
For extra fun, find the
angle of the vector too!
V3 = 2.81 km
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Problem 1.62
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.
5.80 km
Start
2.00 km
Finish
y
45°
Third
3.50 km
x
leg
Now that we have the components, the last step is to combine them to get the
magnitude and direction of the vector for the 3rd leg.
Use the Pythagorean Theorem
to find the magnitude:
V3 =
(V3,x ) + (V3,y )
V3 =
(1.33 )2 + (2.47 )2
2
V3 = 2.81 km
2
Use tan(θ) = Vy/Vx 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