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Scalars and vectors are super important concepts in physics. You have to understand the difference between them if you want to solve physics problems easily, especially in AP Physics exams. In fact, a lot of questions in physics involve understanding how these two types of quantities work.

This guide will break down everything you need to know about scalars and vectors: definitions, formulas, examples, practice questions, and tips to master this topic.

For the official AP Physics 1 syllabus and exam details, you can refer to the College Board’s AP Physics 1 Page.

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What Are Scalars and Vectors?

Scalars

• Definition: Scalars are quantities that only have magnitude (size) and no direction. They are measured by a numerical value alone, making them simpler to work with compared to vectors.
• Examples:
  • Mass (5 kg) – It’s just a number with no direction.
  • Temperature (25°C) – You only measure the amount of heat.
  • Speed (60 m/s) – How fast something is moving, not where it’s heading.
  • Time (3 seconds) – A quantity that flows continuously and has no direction.
  • Distance (10 meters) – How far something has traveled, regardless of direction.

Vectors

• Definition: Vectors are quantities that have both magnitude and direction. Understanding vectors is essential because most physical quantities like force, displacement, and velocity fall under this category.
• Examples:
  • Displacement (10 m, East) – Shows how far and in which direction something moves.
  • Velocity (20 m/s, North) – Speed with a specific direction.
  • Force (5 N, upward) – A push or pull acting in a particular direction.
  • Acceleration (9.8 m/s², downward) – The rate of change of velocity in a certain direction.
  • Momentum (15 kg m/s, right) – The product of mass and velocity, pointing in the same direction as velocity.

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Key Differences Between Scalars and Vectors

Scalars	Vectors
Only magnitude	Magnitude and direction
Added algebraically	Added geometrically
No negative sign	Can have positive or negative direction
Simple calculations	Requires trigonometry when added or subtracted

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Formulas and Operations with Vectors

Vector Addition

• Rule: To add vectors, place them head-to-tail and draw the resultant vector from the tail of the first to the head of the last.
• Formula (For two vectors A and B):
  $$\vec{R} = \vec{A} + \vec{B}$$ 
• Tips: When vectors are in component form, just add the x-components and y-components separately.

Subtracting Vectors

• Rule: Reverse the direction of the vector you want to subtract and then add it to the other vector.
• Formula:
  $$\vec{R} = \vec{A} – \vec{B} = \vec{A} + (-\vec{B})$$ 

Multiplying Vectors

• Dot Product (Scalar Product):
  $$A \cdot B = AB \cos \theta$$ 
• Cross Product (Vector Product):
  $$\vec{A} \times \vec{B} = AB \sin \theta \hat{n}$$ 

Magnitude of a Vector

• Formula:
  $$|\vec{A}| = \sqrt{A_x^2 + A_y^2}$$ 

Unit Vectors

• Definition: Vectors with a magnitude of one that are used to specify direction.
• Example:
  $\hat{i}, \hat{j}, \hat{k}$ 

  are unit vectors along the x, y, and z axes, respectively.

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20 most commonly asked questions on Scalars and Vectors in the AP Physics 1 exam

1. What is the difference between a scalar and a vector quantity?

Solution:

• Scalars have magnitude only (e.g., speed, distance, mass).
• Vectors have both magnitude and direction (e.g., velocity, displacement, force).
• Example:
  • Speed (scalar): 60 km/h
  • Velocity (vector): 60 km/h north

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2. A car travels 60 km north and then 80 km east. What is the resultant displacement?

Solution:

1. Treat the displacement as a right triangle.
2. Use the Pythagorean theorem:
   $$R = \sqrt{(60)^2 + (80)^2} = \sqrt{3600 + 6400} = \sqrt{10000} = 100 \text{ km}$$ 
3. Direction:
   $$\theta = \tan^{-1} \left(\frac{80}{60}\right) = \tan^{-1} (1.33) = 53.1^\circ \text{ east of north}$$ 
   • Resultant displacement: 100 km at 53.1° east of north

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3. Resolve a 50 N force acting at 30° above the horizontal into its components.

Solution:

1. Horizontal Component:
   $$F_x = F \cos(\theta) = 50 \cos(30^\circ) = 50 \times 0.866 = 43.3 \text{ N}$$ 
2. Vertical Component:
   $$F_y = F \sin(\theta) = 50 \sin(30^\circ) = 50 \times 0.5 = 25 \text{ N}$$ 
   • Components:
     $F_x = 43.3 \text{ N}, F_y = 25 \text{ N}$ 

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4. Find the resultant of two vectors: 5 m at 0° and 12 m at 90°.

Solution:

1. Use the Pythagorean theorem:
   $$R = \sqrt{(5)^2 + (12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ m}$$ 
2. Direction:
   $$\theta = \tan^{-1} \left(\frac{12}{5}\right) = 67.4^\circ \text{ above the x-axis}$$ 
   • Resultant vector: 13 m at 67.4°

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5. A boat moves 4 km north and 3 km east. What is the displacement?

Solution:

1. Use the Pythagorean theorem:
   $$R = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ km}$$ 
2. Direction:
   $$\theta = \tan^{-1} \left(\frac{3}{4}\right) = 36.9^\circ \text{ east of north}$$ 
   • Displacement: 5 km at 36.9° east of north

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6. If a vector has a magnitude of 10 units and makes an angle of 60° with the x-axis, find its components.

Solution:

1. Horizontal component:
   $$x = 10 \cos(60^\circ) = 10 \times 0.5 = 5$$ 
2. Vertical component:
   $$y = 10 \sin(60^\circ) = 10 \times 0.866 = 8.66$$ 
   • Components:
     $x = 5$ 

     , $y = 8.66$ 

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7. Calculate the dot product of vectors

$\mathbf{A} = 3\mathbf{i} + 4\mathbf{j}$

and

$\mathbf{B} = 5\mathbf{i} + 12\mathbf{j}$

.

Solution:

 

$$\mathbf{A} \cdot \mathbf{B} = (3 \times 5) + (4 \times 12) = 15 + 48 = 63$$

 

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8. Calculate the cross product of vectors

$\mathbf{A} = 2\mathbf{i} + 3\mathbf{j}$

and

$\mathbf{B} = 4\mathbf{i} – \mathbf{j}$

.

Solution:

 

$$\mathbf{A} \times \mathbf{B} = (2)(-1) – (3)(4) = -2 – 12 = -14 \mathbf{k}$$

 

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9. Find the angle between two vectors:

$\mathbf{A} = 3\mathbf{i} + 4\mathbf{j}$

and

$\mathbf{B} = 5\mathbf{i} + 12\mathbf{j}$

.

Solution:

1. Dot product:
   $$\mathbf{A} \cdot \mathbf{B} = (3)(5) + (4)(12) = 15 + 48 = 63$$ 
2. Magnitudes:
   $$\mathrm{\mid }\mathbf{A}\mathrm{\mid }=\sqrt{3^2+4^2}=\sqrt{9+16}=5$$ 

   $$|\mathbf{A}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$$ 

   $$|\mathbf{B}| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = 13$$ 

3. Angle:
   $$\cos(\theta) = \frac{63}{(5)(13)} = \frac{63}{65} = 0.969$$ 

   $$\theta = \cos^{-1}(0.969) = 14.5^\circ$$ 

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10. A plane flies 300 km east and then 400 km north. What is its total displacement?

Solution:

 

$$R = \sqrt{300^2 + 400^2} = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \text{ km}$$

 

$$\theta = \tan^{-1} \left(\frac{400}{300}\right) = 53.1^\circ \text{ north of east}$$

 

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$$F_y = 20 \sin(45^\circ) = 20 \times 0.707 = 14.14 \text{ N}$$

• Components: $F_x = 14.14$ N (Horizontal), $F_y = 14.14$ N (Vertical).

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13. Two vectors are $\mathbf{A} = 7 \mathbf{i} – 4 \mathbf{j}$ and $\mathbf{B} = 5 \mathbf{i} + 12 \mathbf{j}$. Find their sum and magnitude.

Solution:

1. Addition of Vectors:

$$\mathbf{R} = \mathbf{A} + \mathbf{B} = (7 + 5) \mathbf{i} + (-4 + 12) \mathbf{j} = 12 \mathbf{i} + 8 \mathbf{j}$$

2. Magnitude of Resultant:

$$|\mathbf{R}| = \sqrt{(12)^2 + (8)^2} = \sqrt{144 + 64} = \sqrt{208} = 14.42$$

• Resultant Vector: $12 \mathbf{i} + 8 \mathbf{j}$, Magnitude = 14.42.

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14. A car travels 80 km west and then 60 km south. Find the magnitude and direction of the resultant displacement.

Solution:

1. Magnitude:

$$R = \sqrt{(80)^2 + (60)^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100 \text{ km}$$

2. Direction:

$$\theta = \tan^{-1} \left( \frac{60}{80} \right) = \tan^{-1}(0.75) = 36.87^\circ \text{ south of west}$$

• Resultant Displacement: 100 km at $36.87^\circ$ south of west.

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15. Find the magnitude and direction of the vector $\mathbf{C} = -8\mathbf{i} + 6\mathbf{j}$.

Solution:

1. Magnitude:

$$|\mathbf{C}| = \sqrt{(-8)^2 + (6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$$

2. Direction:

$$\theta = \tan^{-1} \left( \frac{6}{8} \right) = \tan^{-1}(0.75) = 36.87^\circ \text{ above the negative x-axis}$$

• Vector $\mathbf{C}$: Magnitude = 10, Direction = $36.87^\circ$ above the negative x-axis.

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16. Calculate the work done by a force $\mathbf{F} = 10 \mathbf{i} + 5 \mathbf{j}$ acting over a displacement $\mathbf{d} = 4 \mathbf{i} + 3 \mathbf{j}$.

Solution:

$$W = \mathbf{F} \cdot \mathbf{d} = (10 \times 4) + (5 \times 3) = 40 + 15 = 55 \text{ J}$$

• Work Done: 55 J.

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17. A vector has a magnitude of 20 units and is directed 30° south of east. Find its components.

Solution:

1. Horizontal Component:

$$x = 20 \cos(30^\circ) = 20 \times 0.866 = 17.32$$

2. Vertical Component (negative because it is southward):

$$y = 20 \sin(30^\circ) = 20 \times 0.5 = 10$$

• Components: $17.32$ (east), $-10$ (south).

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18. Determine if two vectors $\mathbf{A} = 3\mathbf{i} + 4\mathbf{j}$ and $\mathbf{B} = -4\mathbf{i} + 3\mathbf{j}$ are perpendicular.

Solution:

1. Dot Product Calculation:

$$\mathbf{A} \cdot \mathbf{B} = (3)(-4) + (4)(3) = -12 + 12 = 0$$

• Since the dot product is zero, the vectors are perpendicular.

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19. A vector $\mathbf{A}$ has components $(6, -8)$. Find the unit vector in the direction of $\mathbf{A}$.

Solution:

1. Magnitude of $\mathbf{A}$:

$$|\mathbf{A}| = \sqrt{(6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$

2. Unit Vector:

$$\hat{A} = \left( \frac{6}{10}, \frac{-8}{10} \right) = (0.6, -0.8)$$

• Unit Vector: $(0.6, -0.8)$.

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20. Two vectors are given: $\mathbf{A} = 4\mathbf{i} + 3\mathbf{j}$ and $\mathbf{B} = 6\mathbf{i} – 2\mathbf{j}$. Find the angle between them.

Solution:

1. Dot Product:

$$\mathbf{A} \cdot \mathbf{B} = (4)(6) + (3)(-2) = 24 – 6 = 18$$

2. Magnitudes:

$$|\mathbf{A}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$ $$|\mathbf{B}| = \sqrt{6^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 6.32$$

3. Angle Calculation:

$$\cos \theta = \frac{18}{(5)(6.32)} = \frac{18}{31.6} = 0.57$$ $$\theta = \cos^{-1}(0.57) = 55.2^\circ$$

• Angle between $\mathbf{A}$ and $\mathbf{B}$: $55.2^\circ$

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Understanding Scalars and Vectors is the first step to getting a solid grasp of AP Physics 1. Use the resources above to build your understanding and practice effectively. Whether you go with self-study or get some extra guidance, you’re on the right track to doing great!

• For more detailed practice problems and guidance, check out our AP Physics 1 Course.

• Need extra help? You can Schedule a Free Class with our experts to clear your doubts.

If you are looking to expand your physics knowledge, check out our blog on kinematic equations, covering topics like displacement, velocity and acceleration.