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Pre-Calculus Vectors

Vector applications (velocity, force, navigation)

20 practice questions 0 video lessons Theory + worked examples

Vector Applications

California Pre-Calculus • Standard N-VM.3 • Vectors

Vector Applications is a topic in Vectors in the California Common Core State Standards. It is aligned to Standard N-VM.3, which requires students to solve problems involving velocity and other quantities represented by vectors.

Vector applications combine velocities and forces — ground speed is airspeed plus wind, and a net force is the vector sum, zero at equilibrium.

California Pre-Calculus › Vectors › Vector Applications  —  Standard N-VM.3

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Theory

Vectors model any quantity with size and direction — velocity, force, and displacement. Real problems add these vectors to find a single resultant:

  • Navigation: ground velocity = airspeed (or water speed) \(+\) wind (or current).
  • Forces: the net force is the vector sum; if it is \(\langle 0,0\rangle\), the object is in equilibrium.
Set it up with components. Resolve each vector into horizontal and vertical parts, add, then read off magnitude and direction.
Airplane velocity with wind The plane's ground velocity is the vector sum of its airspeed and the wind velocity. airspeed wind ground
Ground velocity is airspeed \(+\) wind.
Forces on an object Several force vectors act on a point; their vector sum is the net force. lift weight F₁ F₂
The net force is the vector sum of all forces.

Resultant magnitude and direction:

\[R=\sum v_i,\qquad \|R\|=\sqrt{R_x^2+R_y^2},\qquad \theta=\arctan\dfrac{R_y}{R_x}\]
the resultant is the sum of the vectors; its magnitude and direction come from its components
Equilibrium means the resultant is the zero vector: all forces cancel.

How to solve a vector application

  1. Resolve each quantity into components.
  2. Add the components to get the resultant.
  3. Find the resultant's magnitude and direction.
  4. Interpret in context (speed, heading, net force).
Example 1 — Ground speed with wind
A plane flies east at \(300\ \text{mph}\); a wind blows north at \(40\ \text{mph}\). Find the ground speed.
Solution

Add the perpendicular velocities and take the magnitude.

\(R\)\(=\)\(\langle 300,40\rangle\)
\(\|R\|\)\(=\)\(\sqrt{300^2+40^2}\approx 302.7\ \text{mph}\)
ground speed is about 302.7 miles per hour
Example 2 — Direction of travel
For that plane, find the direction north of east.
Solution

The angle above east is \(\arctan\dfrac{40}{300}\).

\(\theta\)\(=\)\(\arctan\dfrac{40}{300}\approx 7.6^\circ\)

About \(7.6^\circ\) north of east.

about 7.6 degrees north of east
Example 3 — Resultant force
Two forces \(\langle 5,0\rangle\) lb and \(\langle 0,12\rangle\) lb act on an object. Find the net force's magnitude.
Solution

Add the forces, then take the magnitude.

\(R\)\(=\)\(\langle 5,12\rangle\)
\(\|R\|\)\(=\)\(\sqrt{25+144}=13\ \text{lb}\)
net force is 13 pounds
Example 4 — Equilibrium
Forces \(\langle 4,7\rangle\) and \(\langle -4,-7\rangle\) act on a point. What is the net force?
Solution

They are opposites, so they cancel.

\(R\)\(=\)\(\langle 4-4,\ 7-7\rangle\)
\(=\)\(\langle 0,0\rangle\)

The object is in equilibrium.

net force is zero, so the object is in equilibrium

Common pitfalls

Add vectors, not just speeds. A crosswind changes both speed and direction.
Resolve into components first. Don't add magnitudes at different angles directly.
Equilibrium is a zero resultant, not merely small forces.

Frequently asked questions

How do vectors model velocity with wind?

The ground velocity is the vector sum of the craft's own velocity and the wind (or current) velocity.

How do you find a resultant force?

Resolve each force into components, add them, and take the magnitude and direction of the sum.

What does equilibrium mean for vectors?

The resultant is the zero vector — all the forces cancel and there is no net force.

Why can't you just add speeds?

Because direction matters. Two \(300\)-unit vectors at different angles do not give a \(600\)-unit resultant; you must add components.

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