Motion language & signs | CAIE 9709 Mechanics Notes

Mechanics

Mathematics9709

Choose an origin and a positive direction

The origin is the reference position labelled zero. Position is measured from that origin. A positive position lies on the chosen positive side; a negative position lies on the other side. Neither sign means “good”, “large” or “slow”.

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Distance counts the whole path. Displacement compares only the final position with the initial position, using the chosen positive direction.

Definition

What is meant by displacement?

Model answer: The signed change in position from the initial point to the final point, measured in a stated direction.

Calculation: final position minus initial position.

Key idea

Changing the positive direction changes the signs of displacement, velocity and acceleration, but it does not change the physical journey.

Separate scalars from signed vectors

A scalar has size only. A vector has size and direction; in straight-line motion, its sign records that direction.

Distance is the total path length. It is a scalar and is never negative.
Displacement is the signed change in position. It can be positive, zero or negative.
Speed is the magnitude of velocity. It is never negative.
Velocity is the rate of change of displacement, so its sign gives the direction of motion.

\text{average speed}=\dfrac{\text{total distance}}{\text{total time}}

Symbols

$distance$ = the whole path travelled (m)
$time$ = the total duration of the journey (s)

\text{average velocity}=\dfrac{\text{displacement}}{\text{total time}}

Symbols

$displacement$ = signed final position minus initial position (m)
$time$ = the total duration of the journey (s)

Common mistake

Average speed is not usually the average of the starting and ending speeds. That shortcut works for one constant-acceleration stage only when velocity does not change sign. If the particle reverses, split the journey and use total distance divided by total time.

Compare the signs of velocity and acceleration

Definition

What is meant by acceleration?

Model answer: The rate of change of velocity with respect to time.

If acceleration is constant at $+3\text{ m s}^{-2}$ , velocity increases by $3\text{ m s}^{-1}$ each second. A negative value describes a negative change in velocity, not automatically a decrease in speed.

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Acceleration sign tells you how velocity changes. Compare the signs of v and a before saying whether the particle speeds up or slows down.

If velocity and acceleration have the same sign, speed increases.
If they have opposite signs, speed decreases.
If acceleration is zero, velocity and speed are constant.
A negative acceleration is not automatically a deceleration. A particle moving in the negative direction can speed up while its acceleration is negative.
If velocity is zero at one instant, inspect its sign immediately before and after that instant before claiming a change of direction.

Your turnIndependent check [5 marks]

Positive is east. A particle starts at position 2 m, travels to position 11 m, then finishes at position 5 m after 5 s. Find its distance, displacement, average speed and average velocity. It is later moving west with acceleration $-3\text{ m s}^{-2}$ . State whether its speed is increasing or decreasing.

Show worked answer

\text{distance}=9+6=15\text{ m}

\text{displacement}=5-2=+3\text{ m}

15\div5=3\text{ m s}^{-1}

3\div5=0.6\text{ m s}^{-1}

West is negative, so both velocity and acceleration are negative. The speed is increasing.

Answer

15\text{ m}

+3\text{ m}

3\text{ m s}^{-1}

+0.6\text{ m s}^{-1}

\text{speed increasing}

Choose an origin and a positive direction

Swipe left or right to see the whole diagram →

Distance counts the whole path. Displacement compares only the final position with the initial position, using the chosen positive direction.

Definition

What is meant by displacement?

Model answer: The signed change in position from the initial point to the final point, measured in a stated direction.

Calculation: final position minus initial position.

Key idea

Changing the positive direction changes the signs of displacement, velocity and acceleration, but it does not change the physical journey.

Separate scalars from signed vectors

A scalar has size only. A vector has size and direction; in straight-line motion, its sign records that direction.

Distance is the total path length. It is a scalar and is never negative.
Displacement is the signed change in position. It can be positive, zero or negative.
Speed is the magnitude of velocity. It is never negative.
Velocity is the rate of change of displacement, so its sign gives the direction of motion.

\text{average speed}=\dfrac{\text{total distance}}{\text{total time}}

Symbols

$distance$ = the whole path travelled (m)
$time$ = the total duration of the journey (s)

\text{average velocity}=\dfrac{\text{displacement}}{\text{total time}}

Symbols

$displacement$ = signed final position minus initial position (m)
$time$ = the total duration of the journey (s)

Common mistake

Compare the signs of velocity and acceleration

Definition

What is meant by acceleration?

Model answer: The rate of change of velocity with respect to time.

Swipe left or right to see the whole diagram →

Acceleration sign tells you how velocity changes. Compare the signs of v and a before saying whether the particle speeds up or slows down.

If velocity and acceleration have the same sign, speed increases.
If they have opposite signs, speed decreases.
If acceleration is zero, velocity and speed are constant.
A negative acceleration is not automatically a deceleration. A particle moving in the negative direction can speed up while its acceleration is negative.
If velocity is zero at one instant, inspect its sign immediately before and after that instant before claiming a change of direction.

Your turnIndependent check [5 marks]

Show worked answer

\text{distance}=9+6=15\text{ m}

\text{displacement}=5-2=+3\text{ m}

15\div5=3\text{ m s}^{-1}

3\div5=0.6\text{ m s}^{-1}

West is negative, so both velocity and acceleration are negative. The speed is increasing.

Answer

15\text{ m}

+3\text{ m}

3\text{ m s}^{-1}

+0.6\text{ m s}^{-1}

\text{speed increasing}