Mathematics (9660) | Motion in a straight line with variable acceleration

M1.2: Motion in a Straight Line with Variable Acceleration

Hello! Welcome to the exciting world where motion isn't always neat and predictable. So far in Mechanics (M1.1), you've dealt with motion under constant acceleration, using the familiar SUVAT equations. But let's be honest—in the real world, things rarely move with perfectly constant acceleration!

Imagine a rocket launch or a car braking hard. The acceleration changes constantly! This is variable acceleration, and to handle it, we must step up our mathematical game using the power of Calculus (Differentiation and Integration) from your Pure Mathematics (P1) knowledge.

Don't worry if this seems tricky at first. We are simply defining the fundamental relationships between position, speed, and how they change over time.

The Core Concepts: Defining Position, Velocity, and Acceleration

In this chapter, we are dealing with a particle moving along a straight line. We define its position using three key variables, all measured as functions of time, \(t\):

• Displacement (\(s\)): The particle’s position relative to a fixed origin point, \(O\). Measured in metres (m).
• Velocity (\(v\)): The rate of change of displacement. Measured in metres per second (\(\text{m s}^{-1}\)).
• Acceleration (\(a\)): The rate of change of velocity. Measured in metres per second squared (\(\text{m s}^{-2}\)).

Did you know? Since we are using calculus, \(s, v\), and \(a\) are usually expressed as functions of time, for example: \(s(t) = 2t^3 - 5t\).

Section 1: The Calculus Connection (The Mechanic's Toolkit)

The entire topic hinges on two simple, yet vital, ideas: differentiation and integration are inverses of each other.

1.1 Moving Downstream: From Displacement to Acceleration (Differentiation)

When you know the position of a particle as a function of time, you can differentiate it to find its velocity, and differentiate again to find its acceleration.

Analogy: Think of this as going downstream. It's generally the easier direction to travel, just like differentiation is often easier than integration.

Key Formulae (Differentiation):

Velocity is the derivative of displacement:
\[v = \frac{ds}{dt}\]

Acceleration is the derivative of velocity:
\[a = \frac{dv}{dt}\]

Acceleration is the second derivative of displacement:
\[a = \frac{d^2s}{dt^2}\]

Step-by-Step Example (Differentiation):

1. If the displacement is given by \(s = t^3 - 4t^2 + 5t\).
2. Find the velocity by differentiating \(s\) with respect to \(t\):
   \[v = \frac{ds}{dt} = 3t^2 - 8t + 5\]
3. Find the acceleration by differentiating \(v\) with respect to \(t\):
   \[a = \frac{dv}{dt} = 6t - 8\]

1.2 Moving Upstream: From Acceleration to Displacement (Integration)

When you know the acceleration of a particle as a function of time, you must integrate it to find velocity, and integrate velocity to find displacement.

Analogy: This is like going upstream. It's often harder because you need extra information to complete the journey!

Key Formulae (Integration):

Velocity is the integral of acceleration:
\[v = \int a \, dt\]

Displacement is the integral of velocity:
\[s = \int v \, dt\]

Section 2: The Critical Role of the Constant of Integration (C)

This is where most students make mistakes! When you perform indefinite integration, you must always add the constant of integration, \(C\). This constant represents the starting value or initial condition of the particle.

Since we integrate twice to go from acceleration to displacement, you will have two constants (\(C_1\) and \(C_2\)) to find.

The Trick: You must use the information given about the particle's motion, usually at time \(t=0\), to find these constants.

Step-by-Step Example (Integration):

A particle starts from the origin (\(s=0\)) with an initial velocity of \(5 \text{ m s}^{-1}\) at \(t=0\). The acceleration is given by \(a = 6t - 2\).

1. Find the Velocity Function (\(v\)):
   \[v = \int a \, dt = \int (6t - 2) \, dt\] \[v = 3t^2 - 2t + C_1\]
2. Use Initial Conditions to Find \(C_1\):
   We know that when \(t=0\), \(v=5\). Substitute these values:
   \(5 = 3(0)^2 - 2(0) + C_1\)
   \(\implies C_1 = 5\)
   So, the velocity function is \(v = 3t^2 - 2t + 5\).
3. Find the Displacement Function (\(s\)):
   \[s = \int v \, dt = \int (3t^2 - 2t + 5) \, dt\] \[s = t^3 - t^2 + 5t + C_2\]
4. Use Initial Conditions to Find \(C_2\):
   We know that when \(t=0\), the particle starts at the origin, so \(s=0\).
   \(0 = (0)^3 - (0)^2 + 5(0) + C_2\)
   \(\implies C_2 = 0\)
   So, the displacement function is \(s = t^3 - t^2 + 5t\).

Common Mistakes to Avoid!

• Forgetting the \(+ C\): If you forget the constant of integration, your calculated position or velocity will be wrong by a fixed offset. Always remember to add \(C\) (or \(C_1, C_2\)) immediately after integrating.
• Confusing Initial Conditions: Make sure you know what the initial conditions mean:
  

  "Starts from the origin" \(\implies s=0\) when \(t=0\).

  "Starts from rest" \(\implies v=0\) when \(t=0\).

• Misidentifying the Function Type: Remember that AS Level M1 problems are restricted to functions that you can integrate using P1 rules (typically polynomials or powers like \(t^{1/2}\)). Always rewrite expressions like \(\frac{1}{t^2}\) as \(t^{-2}\) before integrating/differentiating.

Section 3: Displacement vs. Distance and the Importance of Direction

Remember the distinction between displacement (\(s\)) and distance travelled.

• Displacement is the vector distance from the origin \(O\). It can be positive or negative.
• Distance travelled is a scalar quantity (always positive).

When acceleration is variable, the particle might change direction. If it changes direction, you must calculate the distance travelled in segments.

A particle changes direction when its velocity (\(v\)) is zero.

Step-by-Step: Finding Total Distance Travelled

1. Find the velocity function, \(v(t)\).
2. Set \(v(t) = 0\) and solve for \(t\). These are the times when the particle stops and potentially reverses direction.
3. Calculate the displacement, \(s\), at the starting time, the reversal times, and the end time.
4. Calculate the distance travelled between each segment (always positive). Sum these absolute distances to find the total distance travelled.

Example: If a particle starts at \(s=0\), moves to \(s=5\), then reverses to \(s=2\) at \(t=5\).
Total distance travelled up to \(t=5\) is: (Distance 1) + (Distance 2) = \((5 - 0) + |2 - 5| = 5 + 3 = 8\).