Welcome to the World of Moving Forces!

In your previous Mechanics studies, you likely dealt with constant forces—like a steady push or a constant gravity. But in the real world, forces are rarely that simple. Think about riding a bike: the harder the wind blows against you, the more resistance you feel. Or think of a magnet pulling a paperclip; the closer it gets, the stronger the pull becomes. This is linear motion under a variable force.

In this chapter, we will learn how to use calculus to describe exactly how objects move when the force pushing or pulling them is constantly changing. Don't worry if this seems tricky at first; we'll break it down into simple steps!

1. The Foundation: Newton’s Second Law

Everything in this chapter starts with the most famous equation in mechanics: Newton’s Second Law.

\( F = ma \)

Where \( F \) is the resultant force, \( m \) is the mass, and \( a \) is acceleration. In this chapter, \( F \) isn't just a number; it's a function. It might depend on:

  • Time (\( t \)): The force changes as the clock ticks.
  • Displacement (\( x \)): The force changes based on where the object is.
  • Velocity (\( v \)): The force changes based on how fast the object is moving (like air resistance).

Quick Review: Acceleration is the rate of change of velocity. Depending on what information we have, we can write acceleration in two very important ways:

1. \( a = \frac{dv}{dt} \) (Use this when you are dealing with time).

2. \( a = v \frac{dv}{dx} \) (Use this when you are dealing with distance/displacement).

Key Takeaway:

To solve these problems, your first step is always to write \( F = ma \) and then replace \( a \) with either \( \frac{dv}{dt} \) or \( v \frac{dv}{dx} \).


2. Choosing Your Tool: Which Acceleration Formula?

Choosing the right version of acceleration is the "secret sauce" to solving these problems easily. Use this simple guide:

Scenario A: Force depends on Time or Velocity, and you need Time

If the question mentions how long something takes, use:
\( m \frac{dv}{dt} = F \)

Scenario B: Force depends on Displacement or Velocity, and you need Distance

If the question asks how far something moves, use:
\( m v \frac{dv}{dx} = F \)

Analogy: Think of these like two different screwdrivers. You could technically try to use the wrong one, but life is much easier if you match the tool to the screw!

Did you know? Air resistance is a classic variable force. It usually depends on the square of the velocity (\( v^2 \)). This is why fast cars need to be aerodynamic—the faster they go, the "thicker" the air feels!


3. Solving the Equation: Separation of Variables

Once you have set up your equation, you will have a Differential Equation. To solve it, we use a technique called Separation of Variables. This is where your Pure Mathematics (P3) skills come in handy!

Step-by-Step Guide:

1. Set up the equation: Write \( m \frac{dv}{dt} = F \) or \( m v \frac{dv}{dx} = F \).

2. Separate: Move all terms with \( v \) to one side of the equals sign and all terms with \( t \) (or \( x \)) to the other side.

3. Integrate: Add integral signs to both sides.

4. Find the Constant: Use the "initial conditions" (e.g., "at \( t=0, v=2 \)") given in the question to find the constant of integration (\( C \)).

Example: A particle of mass \( m \) is acted on by a force \( kv \).
Setting it up: \( m \frac{dv}{dt} = -kv \) (Force is negative because it's usually resisting motion).
Separating: \( \frac{1}{v} dv = -\frac{k}{m} dt \).
Integrating: \( \ln(v) = -\frac{k}{m} t + C \).

Key Takeaway:

Always remember to put the "integrating partner" (\( dt, dv, \) or \( dx \)) on the top of the fraction, never the bottom!


4. Common Resisting Forces and Terminal Velocity

Many exam questions involve a "resisting force." This is a force that acts in the opposite direction to motion.

Important Tip: When a force opposes motion, remember to use a minus sign in your equation!

Terminal Velocity

Imagine a skydiver falling. Gravity pulls them down (constant force), but air resistance pushes them up (variable force that gets stronger as they go faster). Eventually, the air resistance equals the weight. At this point, the resultant force is zero.

When \( F = 0 \), the acceleration is \( 0 \), and the object reaches its Terminal Velocity.

Quick Review Box:
- If \( a = 0 \), the velocity is constant.
- To find terminal velocity, set the total force expression to zero and solve for \( v \).


5. Avoiding Common Pitfalls

Even the best students can make these mistakes. Watch out for them!

  • The Missing 'v': When using \( a = v \frac{dv}{dx} \), students often forget the \( v \) in front. Remember: \( v \) is for velocity-distance problems.
  • Sign Errors: Always define which direction is positive. If an object is slowing down due to resistance, the force should be negative: \( ma = -ResistingForce \).
  • The Constant of Integration: Don't forget \( +C \)! If you prefer, you can use definite integrals with limits (e.g., from \( u \) to \( v \)) to avoid the constant altogether.
  • Units: Ensure mass is in kg and force is in Newtons. If they give you grams, convert them!

Summary Checklist for Your Exam

Before you finish a problem, ask yourself:

1. Did I use \( \frac{dv}{dt} \) for time or \( v \frac{dv}{dx} \) for displacement?
2. Is my resisting force negative?
3. Did I separate the variables correctly before integrating?
4. Have I used the initial conditions to find \( C \)?
5. Does my answer for Terminal Velocity make sense (is it a constant value)?

Keep practicing! Mechanics is all about seeing the patterns in how things move. You've got this!