Vector Addition Worksheet - Fill Online, Printable, Fillable ... - Free Printable

Let's solve this Vector Addition Worksheet step by step. We'll find the x and y components of each vector using trigonometry, then add them accordingly.

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🔷 Part I: Find the x and y components of each vector

We use:
- $ x = r \cos(\theta) $
- $ y = r \sin(\theta) $

Angles are measured from the positive x-axis, counterclockwise (standard position).

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#### 1. Vector: 12 m at 30°
- $ r = 12 $ m
- $ \theta = 30^\circ $

$$
x = 12 \cos(30^\circ) = 12 \times \frac{\sqrt{3}}{2} \approx 12 \times 0.866 = 10.39 \text{ m}
$$
$$
y = 12 \sin(30^\circ) = 12 \times 0.5 = 6.00 \text{ m}
$$

✔ x = 10.39 m, y = 6.00 m

---

#### 2. Vector: 25 m/s at 60°
- $ r = 25 $ m/s
- $ \theta = 60^\circ $

$$
x = 25 \cos(60^\circ) = 25 \times 0.5 = 12.5 \text{ m/s}
$$
$$
y = 25 \sin(60^\circ) = 25 \times \frac{\sqrt{3}}{2} \approx 25 \times 0.866 = 21.65 \text{ m/s}
$$

✔ x = 12.5 m/s, y = 21.65 m/s

---

#### 3. Vector: 15 m/s at 150°
- $ r = 15 $ m/s
- $ \theta = 150^\circ $

Note: 150° is in the second quadrant → x negative, y positive.

$$
x = 15 \cos(150^\circ) = 15 \times (-\cos(30^\circ)) = 15 \times (-0.866) \approx -12.99 \text{ m/s}
$$
$$
y = 15 \sin(150^\circ) = 15 \times \sin(30^\circ) = 15 \times 0.5 = 7.5 \text{ m/s}
$$

✔ x = -12.99 m/s, y = 7.5 m/s

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🔷 Add the following vectors

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#### First Pair:
- Vector 1: 12 m/s at 20°
- Vector 2: 12 m/s at 70°

##### Vector 1:
- $ r_1 = 12 $, $ \theta_1 = 20^\circ $
- $ x_1 = 12 \cos(20^\circ) \approx 12 \times 0.9397 = 11.28 \text{ m/s} $
- $ y_1 = 12 \sin(20^\circ) \approx 12 \times 0.3420 = 4.10 \text{ m/s} $

##### Vector 2:
- $ r_2 = 12 $, $ \theta_2 = 70^\circ $
- $ x_2 = 12 \cos(70^\circ) \approx 12 \times 0.3420 = 4.10 \text{ m/s} $
- $ y_2 = 12 \sin(70^\circ) \approx 12 \times 0.9397 = 11.28 \text{ m/s} $

##### Add:
- $ x_{\text{tot}} = x_1 + x_2 = 11.28 + 4.10 = 15.38 \text{ m/s} $
- $ y_{\text{tot}} = y_1 + y_2 = 4.10 + 11.28 = 15.38 \text{ m/s} $

✔ Resultant: $ x_{\text{tot}} = 15.38 $, $ y_{\text{tot}} = 15.38 $

> This means the resultant vector has equal x and y components — it points at 45° with magnitude $ \sqrt{15.38^2 + 15.38^2} \approx 21.77 $ m/s.

---

#### Second Pair:
- Vector 1: 50 m at 135°
- Vector 2: 30 m at 90° (straight down)

##### Vector 1: 50 m at 135°
- $ \theta = 135^\circ $ → second quadrant
- $ x_1 = 50 \cos(135^\circ) = 50 \times (-\cos(45^\circ)) = 50 \times (-0.7071) \approx -35.36 \text{ m} $
- $ y_1 = 50 \sin(135^\circ) = 50 \times \sin(45^\circ) = 50 \times 0.7071 \approx 35.36 \text{ m} $

##### Vector 2: 30 m at 90°
- 90° is straight up → but arrow is pointing down, so it's 270° or -90°
- Wait! The diagram shows a downward arrow labeled 30 m at 90°? That’s inconsistent.

But the angle is drawn from the positive x-axis, and it says 90° — that would be upward.

However, the arrow is pointing down, so likely it's meant to be 270° or -90°, but the label says 90°.

Wait: if it's labeled as 90°, that’s upward. But the arrow is pointing down.

Let’s recheck: the diagram shows a downward arrow with "90°" marked from the positive x-axis?

No — 90° from positive x-axis is up. But the arrow is down, so it must be 270° or -90°.

But the worksheet says 90°, and the arrow is down — contradiction.

Wait: sometimes angles are measured from the vertical or from the direction of motion.

But standard convention is from positive x-axis, counterclockwise.

So if the arrow is pointing down, it should be 270° or -90°.

But the label says 90° — that can’t be right.

Wait: look again — the arrow is pointing down, and the angle is drawn from the positive x-axis, going clockwise to the vector? Or counterclockwise?

In physics, angles are usually counterclockwise from positive x-axis.

So if the vector is downward, its direction is 270° or -90°.

But the worksheet says 90° — that’s upward.

This is a problem.

Wait: maybe the angle is measured from the positive y-axis?

But no — the diagrams show angles from the x-axis.

Wait — let's read carefully: the arrow is pointing down, and the angle is labeled 90° — but that doesn't make sense.

Actually, if the vector is pointing straight down, and the angle is measured from the positive x-axis, then:

- Downward = 270° or -90°
- But the label says 90° — which is upward.

So unless the angle is measured from the positive y-axis, it's incorrect.

But in all previous cases, angles are from x-axis.

Possibility: the angle is measured from the positive y-axis, but that’s unusual.

Alternatively, the label is wrong, or the arrow is misdrawn.

Wait — look at the diagram: the arrow is pointing down, and the angle is drawn from the positive x-axis, but it's labeled 90° — that’s impossible.

Unless... the angle is measured clockwise from the positive x-axis?

But that’s not standard.

Wait — perhaps the angle is measured from the negative y-axis?

No — better to assume it's a mistake.

Wait — another possibility: the vector is 30 m, and the angle is 90° — meaning it's vertical, but the direction is down, so we take:

→ $ \theta = 270^\circ $ or $ -90^\circ $

But the label says 90°, which is confusing.

Wait — actually, the angle is shown as 90°, but it's measured from the positive x-axis to the vector, and the vector is down, so the angle should be 270°, not 90°.

So likely, the label is wrong — it should be 270°, or the arrow should be up.

But since the arrow is clearly down, and the label says 90°, that suggests either:

- It's a typo, or
- The angle is measured from the positive y-axis, but that’s rare.

Wait — let’s check the diagram: it shows a downward arrow, and the angle is drawn from the positive x-axis, and it says 90° — but that would mean the vector is up, not down.

So probably, the angle is 270°, and the label is wrong.

But the worksheet says 90°.

Wait — perhaps the angle is measured from the positive y-axis, and it's 90° from there?

But that would be horizontal, not vertical.

I think there's a mistake in labeling.

But let’s suppose the vector is 30 m downward — regardless of the angle label.

Then:

- Direction: 270° or -90°
- So $ x_2 = 30 \cos(270^\circ) = 30 \times 0 = 0 $
- $ y_2 = 30 \sin(270^\circ) = 30 \times (-1) = -30 \text{ m} $

So even if the angle is labeled 90°, the direction is down, so we interpret it as 270°.

Alternatively, if the angle is 90°, then it's up, so $ y = +30 $, but the arrow is down — contradiction.

So the only logical conclusion is: the vector is 30 m downward, so:

✔ $ x_2 = 0 $, $ y_2 = -30 $ m

Even though it says 90°, the arrow is down, so we’ll go with downward, i.e., $ \theta = 270^\circ $

So:

- $ x_2 = 30 \cos(270^\circ) = 0 $
- $ y_2 = 30 \sin(270^\circ) = -30 $ m

Now back to vector 1: 50 m at 135°

We already calculated:
- $ x_1 = -35.36 $ m
- $ y_1 = 35.36 $ m

Now add:

- $ x_{\text{tot}} = x_1 + x_2 = -35.36 + 0 = -35.36 $ m
- $ y_{\text{tot}} = y_1 + y_2 = 35.36 + (-30) = 5.36 $ m

✔ x_tot = -35.36 m, y_tot = 5.36 m

---

✔ Final Answers:

---

Part I: Components

| Vector | Magnitude | Angle | x-component | y-component |
|--------|-----------|-------|-------------|-------------|
| 1 | 12 m | 30° | 10.39 m | 6.00 m |
| 2 | 25 m/s | 60° | 12.5 m/s | 21.65 m/s |
| 3 | 15 m/s | 150° | -12.99 m/s | 7.5 m/s |

---

Vector Addition

#### First Pair:
- Vector 1: 12 m/s at 20°
- $ x_1 = 11.28 $, $ y_1 = 4.10 $
- Vector 2: 12 m/s at 70°
- $ x_2 = 4.10 $, $ y_2 = 11.28 $
- Sum:
- $ x_{\text{tot}} = 15.38 $ m/s
- $ y_{\text{tot}} = 15.38 $ m/s

#### Second Pair:
- Vector 1: 50 m at 135°
- $ x_1 = -35.36 $ m, $ y_1 = 35.36 $ m
- Vector 2: 30 m downward (assumed 270°)
- $ x_2 = 0 $, $ y_2 = -30 $ m
- Sum:
- $ x_{\text{tot}} = -35.36 $ m
- $ y_{\text{tot}} = 5.36 $ m

---

📝 Summary of Answers to Fill In:

---

Top row (components):

1. 12 m at 30°
→ $ x = 10.39 $ m, $ y = 6.00 $ m

2. 25 m/s at 60°
→ $ x = 12.5 $ m/s, $ y = 21.65 $ m/s

3. 15 m/s at 150°
→ $ x = -12.99 $ m/s, $ y = 7.5 $ m/s

---

First addition:

- $ x_1 = 11.28 $, $ x_2 = 4.10 $ → $ x_{\text{tot}} = 15.38 $
- $ y_1 = 4.10 $, $ y_2 = 11.28 $ → $ y_{\text{tot}} = 15.38 $

---

Second addition:

- $ x_1 = -35.36 $, $ x_2 = 0 $ → $ x_{\text{tot}} = -35.36 $
- $ y_1 = 35.36 $, $ y_2 = -30 $ → $ y_{\text{tot}} = 5.36 $

---

⚠️ Note:

The 90° label on the downward vector is likely a mistake. If it were truly 90°, the vector would point up, but the arrow is down. So we assumed it’s 270° or downward.

If the label is correct and the arrow is wrong, then $ y = +30 $, but that contradicts the diagram.

So based on the arrow, we used downward.

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✔ Final Answer Box:

```
Part I: Components

1. 12 m at 30°:
x = 10.39 m
y = 6.00 m

2. 25 m/s at 60°:
x = 12.5 m/s
y = 21.65 m/s

3. 15 m/s at 150°:
x = -12.99 m/s
y = 7.5 m/s

Addition:

First pair:
x₁ = 11.28 + x₂ = 4.10 = x_tot = 15.38
y₁ = 4.10 + y₂ = 11.28 = y_tot = 15.38

Second pair:
x₁ = -35.36 + x₂ = 0 = x_tot = -35.36
y₁ = 35.36 + y₂ = -30 = y_tot = 5.36
```

Let me know if you'd like the resultant magnitude and direction for the sums too!