Vector Examples

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.

The DIRECTION tells you where the vector is pointing.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

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:Normally vectors will be written in BOLDFACE or with an arrow above the letter:

Vr

= 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.

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?

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/hr100 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

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 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

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 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

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 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

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.

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

We need to use the Pythagorean Theorem to find the magnitude of Vtotal

100 km/hr

600 km/hr

( ) ( )22total 100600V +=

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 V ≈ 608 km/hr100 km/hr

600 km/hr

Vtotal ≈ 608 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 V ≈ 608 km/hr100 km/hr

600 km/hr

Vtotal ≈ 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)

θ

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?

Vtotal ≈ 608 km/hr100 km/hr

600 km/hrθ

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°.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 1.62A 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.

45°

Start Finish

5.80 km

2.00 km

3.50 kmThird

leg

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Problem 1.62A 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.

45°

Start Finish

5.80 km

2.00 km

3.50 kmThird

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.

y

x

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 1.62A 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.

45°

Start Finish

5.80 km

2.00 km

3.50 kmThird

leg

Now we look at each leg of the journey and find the components:

y

x

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 1.62A 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.

45°

Start Finish

5.80 km

2.00 km

3.50 kmThird

leg

y

x

The 2nd leg has components in both x- and y-directions, so we can find the components from our triangle rules:components from our triangle rules:

V2x =

V2y =

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Problem 1.62A 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.

45°

Start Finish

5.80 km

2.00 km

3.50 kmThird

leg

y

x

The 2nd leg has components in both x- and y-directions, so we can find the components from our triangle rules: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 1.62A 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.

45°

Start Finish

5.80 km

2.00 km

3.50 kmThird

leg

y

x

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 1.62A 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.

45°

Start Finish

5.80 km

2.00 km

3.50 kmThird

leg

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:

Vtotal,x =

Vtotal,y =

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Problem 1.62A 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.

45°

Start Finish

5.80 km

2.00 km

3.50 kmThird

leg

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:

Vtotal,x = +5.80 km

Vtotal,y = 0.00 km

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Problem 1.62A 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.

45°

Start Finish

5.80 km

2.00 km

3.50 kmThird

leg

y

x

Finally we can just add everything up and solve for our unknowns.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,xV1y + V2y + V3y = Vtotal,y

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Problem 1.62A 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.

45°

Start Finish

5.80 km

2.00 km

3.50 kmThird

leg

y

x

Finally we can just add everything up and solve for our unknowns.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

2.00 + 2.47 + V3x = 5.80

V3x = +1.33 km

(positive is East)

V1y + V2y + V3y = Vtotal,y

0.00 – 2.47 + V3y = 0.00

V3y = +2.47 km

(positive is North)

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Problem 1.62A 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.

45°

Start Finish

5.80 km

2.00 km

3.50 kmThird

leg

y

x

Now that we have the components, the last step is to combine them to get the 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 1.62A 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.

45°

Start Finish

5.80 km

2.00 km

3.50 kmThird

leg

y

x

Now that we have the components, the last step is to combine them to get the 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 = 2.81 km

( ) ( )2y3,2

x3,3 VVV +=

( ) ( )223 2.471.33V +=

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

For extra fun, find the angle of the vector too!

Problem 1.62A 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.

45°

Start Finish

5.80 km

2.00 km

3.50 kmThird

leg

y

x

Now that we have the components, the last step is to combine them to get the 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 = 2.81 km

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)

( ) ( )2y3,2

x3,3 VVV +=

( ) ( )223 2.471.33V +=

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB