# 5.16: Resultant of Two Displacements

**At Grade**Created by: CK-12

**Practice**Resultant of Two Displacements

You are outside playing frisbee on a warm afternoon with your friends. You are trying to throw the frisbee to your friend, but unfortunately, the wind is blowing and keeps pushing your frisbee away from your friend and off course. If you are throwing the frisbee at 20 miles per hour at your friend who is due North of you, and the wind is blowing toward the East at 5 miles per hour, what is the actual trajectory of the frisbee?

Keep reading, and by the end of this Concept, you'll understand how to apply vector addition to exactly this sort of problem.

### Watch This

Adding Vectors Part 1 (Resultant Velocity)

### Guidance

We can use vectors to find direction, velocity, and force of moving objects. In this section we will look at a few applications where we will use resultants of vectors to find speed, direction, and other quantities. A displacement is a distance considered as a vector. If one is 10 ft away from a point, then any point at a radius of 10 ft from that point satisfies the condition. If one is 28 degrees to the east of north, then only one point satisfies this.

#### Example A

A cruise ship is traveling south at 22 mph. A wind is also blowing the ship eastward at 7 mph. What speed is the ship traveling at and in what direction is it moving?

**Solution:** In order to find the direction and the speed the boat is traveling, we must find the resultant of the two vectors representing 22 mph south and 7 mph east. Since these two vectors form a right angle, we can use the Pythagorean Theorem and trigonometric ratios to find the magnitude and direction of the resultant vector.

First, we will find the speed.

\begin{align*}22^2 + 7^2 & = x^2 \\ 533 & = x^2 \\ 23.1 & = x\end{align*}

The ship is traveling at a speed of 23.1mph.

To find the direction, we will use tangent, since we know the opposite and adjacent sides of our triangle.

\begin{align*}\tan \theta & = \frac{7}{22} \\ \tan^{-1} \frac{7}{22} & = 17.7^\circ\end{align*}

The ship’s direction is \begin{align*}S 17.7^\circ E\end{align*}.

#### Example B

A hot air balloon is rising at a rate of 13 ft/sec, while a wind is blowing at a rate of 22 ft/sec. Find the speed at which the balloon is traveling as well as its angle of elevation.

**Solution:** First, we will find the speed at which our balloon is rising. Since we have a right triangle, we can use the Pythagorean Theorem to find calculate the magnitude of the resultant.

\begin{align*}x^2 & = 13^2 + 22^2 \\ x^2 & = 653 \\ x & = 25.6\ ft / \sec\end{align*}

The balloon is traveling at rate of 25.6 feet per second.

To find the angle of elevation of the balloon, we need to find the angle it makes with the horizontal. We will find the \begin{align*}\angle A\end{align*} in the triangle and then we will subtract it from \begin{align*}90^\circ\end{align*}.

\begin{align*}\tan A & = \frac{22}{13} \\ A & = \tan^{-1} \frac{22}{13} \\ A & = 59.4^\circ\end{align*}

Angle with the horizontal \begin{align*}= 90 - 59.4 = 30.6^\circ\end{align*}.

The balloon has an angle of elevation of \begin{align*}30.6^\circ\end{align*}.

#### Example C

Continuing on with the previous example, find:

a. How far from the lift off point is the balloon in 2 hours? Assume constant rise and constant wind speed. (this is *total displacement*)

b. How far must the support crew travel on the ground to get under the balloon? (*horizontal displacement*)

c. If the balloon stops rising after 2 hours and floats for another 2 hours, how far from the initial point is it at the end of the 4 hours? How far away does the crew have to go to be under the balloon when it lands?

**Solution:**

a. After two hours, the balloon will be 184,320 feet from the lift off point (25.6 ft/sec multiplied by 7200 seconds in two hours).

b. After two hours, the horizontal displacement will be 158,400 feet (22ft/sec multiplied by 7200 seconds in two hours).

c. After two hours, the balloon will have risen 93,600 feet. After an additional two hours of floating (horizontally only) in the 22ft/sec wind, the balloon will have traveled 316,800 feet horizontally (22ft/second times 14,400 seconds in four hours).

We must recalculate our resultant vector using Pythagorean Theorem.

\begin{align*}x = \sqrt{93600^2 + 316800^2} = 330338\ ft.\end{align*}

The balloon is 330,338 feet from its initial point. The crew will have to travel 316,800 feet or 90 miles (horizontal displacement) to be under the balloon when it lands.

### Vocabulary

**Displacement:** A ** displacement** is a distance considered as a vector.

### Guided Practice

1. Does \begin{align*}|\vec{a} + \vec{b}| = |\vec{a}| + |\vec{b}|\end{align*}? Explain your answer.

2. A plane is traveling north at a speed of 225 mph while an easterly wind is blowing the plane west at 18 mph. What is the direction and the speed of the plane?

3. Two workers are pulling on ropes attached to a tree stump. One worker is pulling the stump east with 330 Newtons of forces while the second working is pulling the stump north with 410 Newtons of force. Find the magnitude and direction of the resultant force on the tree stump.

**Solutions:**

1. When two vectors are summed, the magnitude of the resulting vector is almost always different than the sum of the magnitudes of the two initial vectors. The only times that \begin{align*}|\vec{a} + \vec{b}| = |\vec{a}| + |\vec{b}|\end{align*} would be true is when 1) the magnitude of at least one of the two vectors to be added is zero, or 2) both of the vectors to be added have the same direction.

2. Speed (magnitude): \begin{align*}\sqrt{18^2 + 225^2} = 225.7\end{align*} and its direction is \begin{align*}\tan \theta = \frac{18}{225} = N 4.6^\circ W\end{align*}.

3. The magnitude is \begin{align*}\sqrt{330^2 + 410^2} = 526.3\end{align*} Newtons and the direction is \begin{align*}\tan^{-1} \left (\frac{410}{330} \right ) = E 51.2^\circ N\end{align*}.

### Concept Problem Solution

When you draw each of the vectors, one for your throw of the frisbee and one for the wind, they look like this:

And combining them, as we saw in this Concept, results in this:

The diagonal line in above diagram is the result of both of the vector displacements, and is therefore the actual path of the frisbee.

### Practice

A ship is traveling north at 30 mph. A constant 5 mph wind is coming from the east (blowing west).

- What is the ship's actual speed?
- In what direction is the ship moving?
- How far does the ship travel in 2 hours?
- If the ship travels for 2 hours and then just floats for another two hours, how far has the trip traveled?
- If the ship travels for 2 hours and then just floats for another two hours, how far from its starting point is the ship?

A balloon is rising at a rate of 20 cm/s, while a wind is blowing at a constant rate of 15 cm/s.

- What is the balloon's actual speed?
- What is the balloon's angle of elevation?
- At what rate would the wind have to be blowing for the balloon to be traveling at 50 cm/s?
- After 3 hours, how far must the support crew travel on the ground to get under the balloon?

A plane is traveling west at a speed of 500 mph while a northerly wind is blowing the plane south at 30 mph.

- What is the actual speed of the plane?
- What is the actual direction of the plane?
- How far will the plane travel in 4 hours?
- How long will it take for the plane to travel exactly 3000 miles?

A plane is takes off at a rate of 230 mph with an angle of 30° to the ground.

- What is the rate at which the plane is receding from the ground?
- What is its ground speed?

### Notes/Highlights Having trouble? Report an issue.

Color | Highlighted Text | Notes | |
---|---|---|---|

Please Sign In to create your own Highlights / Notes | |||

Show More |

### Image Attributions

In this Concept you will learn to interpret the resultant of vector addition as a physical situation involving the combination of two displacements.