Applications to Motion in a Straight Line Revision Notes for VCE SSCE Mathematical Methods

Applications to Motion in a Straight Line

This topic explores how differentiation helps us understand and analyse the motion of particles moving along a straight line. We'll examine position, velocity, and acceleration, and learn how these quantities relate to each other through calculus.

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Understanding motion in a straight line forms the foundation for more complex physics problems. The concepts you learn here will apply to projectiles, circular motion, and many real-world scenarios.

Position

When studying motion in a straight line, we need a way to describe where a particle is located at any moment. The position of a particle is determined by measuring its distance from a fixed reference point called the origin, denoted by $O$ .

By convention, we consider the direction to the right of the origin as positive, while the direction to the left is negative.

A particle's position at any instant can be specified by a real number $x$ . For example:

If $x = 3$ (using metres), the particle is 3 m to the right of $O$
If $x = -3$ , the particle is 3 m to the left of $O$

Position as a function of time

Often, we can express position as a function of time. We write this as $x(t)$ , which represents the position at time $t$ seconds. This function allows us to track how the particle moves over time.

For instance, if a stone is dropped from the top of a 45-metre cliff, and we measure downward position from the top, the position function might be $x(t) = 5t^2$ for $0 \leq t \leq 3$ , where air resistance is neglected.

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Worked Example: Finding Position at Different Times

A particle moves in a straight line so that its position, $x$ cm, relative to $O$ at time $t$ seconds is given by $x = t^2 - 7t + 6$ , where $t \geq 0$ .

a) Find its initial position.

At $t = 0$ :

x = (0)^2 - 7(0) + 6 = 6

The particle starts 6 cm to the right of $O$ (since the result is positive).

b) Find its position at $t = 4$ .

At $t = 4$ :

x = (4)^2 - 7(4) + 6

x = 16 - 28 + 6

x = -6

At $t = 4$ seconds, the particle is 6 cm to the left of $O$ (since the result is negative).

Distance and displacement

Two important concepts in motion are distance and displacement. While they sound similar, they describe different things.

Displacement is a vector quantity defined as the change in position of a particle. It has both magnitude and direction. Displacement can be positive, negative, or zero.

Distance is a scalar quantity that measures the total length of the path travelled by the particle. Distance only has magnitude (no direction) and is always positive or zero.

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Key Distinction: Distance tells you HOW FAR the particle travelled, while displacement tells you WHERE the particle ended up relative to where it started.

Understanding the difference

Consider a particle that starts at origin $O$ , moves 5 units to the right to point $P$ , then moves 7 units to the left to point $Q$ .

The displacement is the difference between the final and initial positions: $-2 - 0 = -2$ units.

The distance travelled is the total path length: $5 + 7 = 12$ units.

This example shows that displacement tells us where the particle ended up relative to where it started, while distance tells us how far the particle actually travelled.

Velocity

Velocity describes how quickly position changes with time. We distinguish between average velocity over a time interval and instantaneous velocity at a specific moment.

Average velocity

The average velocity over a time interval $[t_1, t_2]$ is the average rate of change of position with respect to time:

\text{average velocity} = \frac{\text{change in position}}{\text{change in time}} = \frac{x_2 - x_1}{t_2 - t_1}

where $x_1$ is the position at time $t_1$ and $x_2$ is the position at time $t_2$ .

Instantaneous velocity

The instantaneous velocity is the rate of change of position at a specific instant. This is found by differentiating the position function with respect to time.

If $x$ is the position at time $t$ , then:

v = \frac{dx}{dt}

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Notice how velocity is the first derivative of position. This relationship is central to understanding motion - differentiation allows us to move from position to velocity, and then from velocity to acceleration.

Interpreting velocity

The sign of velocity tells us the direction of motion:

Positive velocity: the particle is moving to the right
Negative velocity: the particle is moving to the left
Zero velocity: the particle is instantaneously at rest

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Worked Example: Analysing Velocity and Motion

Using the same particle from earlier with position $x = t^2 - 7t + 6$ (in cm) at time $t$ seconds:

a) Find its initial velocity.

First, find the velocity function by differentiating:

v = \frac{dx}{dt} = 2t - 7

At $t = 0$ :

v = 2(0) - 7 = -7

The particle is initially moving to the left at 7 cm/s (negative velocity indicates leftward motion).

b) When does its velocity equal zero, and what is its position at this time?

Set $v = 0$ :

2t - 7 = 0

t = 3.5

At $t = 3.5$ :

x = (3.5)^2 - 7(3.5) + 6

x = 12.25 - 24.5 + 6

x = -6.25

At $t = 3.5$ seconds, the particle is at rest, positioned 6.25 cm to the left of $O$ .

c) What is its average velocity for the first 4 seconds?

We need the positions at $t = 0$ and $t = 4$ :

At $t = 0$ : $x = 6$ cm
At $t = 4$ : $x = -6$ cm (calculated earlier)

Average velocity:

$= \frac{-6 - 6}{4 - 0}$

$= \frac{-12}{4}$

$= -3 \text{ cm/s}$

Speed and average speed

Speed is simply the magnitude (absolute value) of velocity. Unlike velocity, speed has no direction - it only tells us how fast the particle is moving.

Average speed for a time interval $[t_1, t_2]$ is calculated as:

\text{average speed} = \frac{\text{distance travelled}}{t_2 - t_1}

Units of measurement

Common units for velocity and speed include:

1 metre per second = 1 m/s = 1 m $\cdot$ s $^{-1}$
1 centimetre per second = 1 cm/s = 1 cm $\cdot$ s $^{-1}$
1 kilometre per hour = 1 km/h = 1 km $\cdot$ h $^{-1}$

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Important Conversions:

To convert from km/h to m/s:

1 \text{ km/h} = \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{5}{18} \text{ m/s}

To convert from m/s to km/h:

1 \text{ m/s} = \frac{18}{5} \text{ km/h}

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Worked Example: Calculating Average Speed

d) Determine the average speed for the first 4 seconds.

From our previous work, we know the particle:

Starts at $x = 6$ cm at $t = 0$
Reaches $x = -6.25$ cm at $t = 3.5$ (where it stops)
Ends at $x = -6$ cm at $t = 4$

The particle changed direction at $t = 3.5$ , so we must calculate distances for each part of the journey:

Distance from $t = 0$ to $t = 3.5$ : $|(-6.25) - 6| = 12.25$ cm

Distance from $t = 3.5$ to $t = 4$ : $|(-6) - (-6.25)| = 0.25$ cm

Total distance travelled: $12.25 + 0.25 = 12.5$ cm

Average speed:

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Critical Point: Speed is the magnitude of velocity, but average speed is NOT equal to the magnitude of average velocity.

In the example above, the average velocity was $-3$ cm/s, which has magnitude 3 cm/s, but the average speed is 3.125 cm/s. This is because the particle changed direction during the journey.

Acceleration

Acceleration measures how quickly velocity changes with respect to time.

Average acceleration

The average acceleration over a time interval $[t_1, t_2]$ is:

\text{average acceleration} = \frac{v_2 - v_1}{t_2 - t_1}

where $v_2$ is the velocity at time $t_2$ and $v_1$ is the velocity at time $t_1$ .

Instantaneous acceleration

The instantaneous acceleration at time $t$ is found by differentiating the velocity function:

a = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{dx}{dt}\right) = \frac{d^2x}{dt^2}

This means acceleration is the second derivative of position with respect to time.

Interpreting acceleration

Acceleration can be positive, negative, or zero:

Zero acceleration: the particle moves at constant velocity
Positive acceleration: when combined with positive velocity, the particle speeds up; when combined with negative velocity, it slows down
Negative acceleration: when combined with positive velocity, the particle slows down; when combined with negative velocity, it speeds up

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Key Principle for Speeding Up or Slowing Down:

If acceleration and velocity have the same sign, the particle is speeding up.

If they have opposite signs, the particle is slowing down.

Note that a particle can be instantaneously at rest (velocity = 0) while still having non-zero acceleration.

The most common units for acceleration are cm/s $^2$ and m/s $^2$ .

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Worked Example: Analysing Position, Velocity, and Acceleration

A particle moves in a straight line so that its position, $x$ cm, relative to $O$ at time $t$ seconds is given by $x = t^3 - 6t^2 + 5$ , where $t \geq 0$ .

a) Find its initial position, velocity and acceleration, and describe its motion.

Position: $x = t^3 - 6t^2 + 5$

Velocity: $v = \frac{dx}{dt} = 3t^2 - 12t$

Acceleration: $a = \frac{dv}{dt} = 6t - 12$

At $t = 0$ :

x = 0 - 0 + 5 = 5 \text{ cm}

v = 0 - 0 = 0 \text{ cm/s}

a = 0 - 12 = -12 \text{ cm/s}^2

Initially, the particle is instantaneously at rest at a position 5 cm to the right of $O$ , with an acceleration of $-12$ cm/s $^2$ .

b) Find the times when it is instantaneously at rest and determine its position and acceleration at those times.

The particle is at rest when $v = 0$ :

3t^2 - 12t = 0

3t(t - 4) = 0

t = 0 \text{ or } t = 4

We already analysed $t = 0$ above. At $t = 4$ :

x = (4)^3 - 6(4)^2 + 5

x = 64 - 96 + 5

x = -27 \text{ cm}

a = 6(4) - 12

a = 24 - 12

a = 12 \text{ cm/s}^2

After 4 seconds, the particle is again at rest, now positioned 27 cm to the left of $O$ , with acceleration of 12 cm/s $^2$ .

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Worked Example: Finding Acceleration

A car starts from rest and moves a distance $s$ metres in $t$ seconds, where $s = \frac{1}{6}t^3 + \frac{1}{4}t^2$ .

What is the initial acceleration and the acceleration when $t = 2$ ?

The position is given by:

s = \frac{1}{6}t^3 + \frac{1}{4}t^2

Velocity:

v = \frac{ds}{dt} = \frac{1}{2}t^2 + \frac{1}{2}t

Acceleration:

a = \frac{dv}{dt} = t + \frac{1}{2}

At $t = 0$ :

a = 0 + \frac{1}{2} = 0.5 \text{ m/s}^2

At $t = 2$ :

a = 2 + \frac{1}{2} = 2.5 \text{ m/s}^2

The initial acceleration is 0.5 m/s $^2$ and the acceleration at $t = 2$ is 2.5 m/s $^2$ .

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Key Points to Remember:

Position $x$ describes where a particle is relative to origin $O$ . Right is positive, left is negative.
Displacement is the change in position (vector), while distance is the total path travelled (scalar).
Velocity $v = \frac{dx}{dt}$ is the rate of change of position. Positive velocity means moving right, negative means moving left.
Speed is the magnitude of velocity. Average speed is NOT the magnitude of average velocity.
Acceleration $a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$ is the rate of change of velocity. When acceleration and velocity have the same sign, the particle speeds up; when they have opposite signs, it slows down.

Applications to Motion in a Straight Line (VCE SSCE Mathematical Methods): Revision Notes