Linear motion

What you'll learn

How to use the constant-acceleration equations for motion in a straight line.
How free-fall motion uses the acceleration of free fall, $g$ .
How motion and collisions can be investigated using light gates, ticker timers, data-loggers and video.
How reaction time, thinking distance, braking distance and stopping distance are linked.

Describing motion in one dimension

Linear motion means motion along a straight line. Because the motion is only along one line, you can choose one direction as positive and the opposite direction as negative.

Definition

Core motion quantities

Displacement, $s$ , is the change in position in a specified direction. Its unit is the metre, m.
Velocity, $v$ , is the rate of change of displacement. Its unit is m s^-1.
Acceleration, $a$ , is the rate of change of velocity. Its unit is m s^-2.
In the constant-acceleration equations, $u$ means initial velocity, $v$ means final velocity, and $t$ means time.

Key Idea

Use signs consistently

Choose a positive direction before substituting into equations. If upward is positive, then an object in free fall near Earth has $-9.81\ \text{m s}^{-2}$ ; if downward is positive, it has $+9.81\ \text{m s}^{-2}$ .

Constant acceleration and the SUVAT equations

Constant acceleration means the acceleration does not change with time. The velocity then changes by equal amounts in equal time intervals.

The equations are often called SUVAT equations because they connect displacement $s$ , initial velocity $u$ , final velocity $v$ , acceleration $a$ and time $t$ .

The velocity-time graph below shows why these equations work: the gradient gives acceleration, and the area under the graph gives displacement.

Velocity-time graph showing constant acceleration, gradient and area for displacement

Key Idea

The four constant-acceleration equations

Use these only for motion in a straight line with constant acceleration:

$v = u + a t$
$\frac{1}{2}(u + v)t$
$\frac{1}{2}at^2$
$v^2 = u^2 + 2as$

Tip

Choosing the right equation

Pick the equation that contains the quantity you want and the quantities you know, but avoids any unknown you do not need. For example, if time is not given, $v^2 = u^2 + 2as$ is often useful.

Common Mistake

Constant acceleration only

These equations are not valid if the acceleration changes significantly, such as for a falling object where air resistance becomes important.

Example

Thrown upwards under gravity

A ball is thrown vertically upwards with initial velocity $s−112.0\ \text{m s}^{-1}$ . Ignore air resistance. Find the time to reach its highest point and the height gained.

Choose upward as positive. At the highest point the final velocity is $v = 0$ , and the acceleration is $-9.81\ \text{m s}^{-2}$ .
Use $v = u + a t$ to find the time:

0 = 12.0 + (- 9.81) t

\frac{12.0}{9.81} = 1.22\ \text{s}

Use $v^2 = u^2 + 2as$ to find the height, because it avoids needing another time calculation:

0^2 = (12.0)^2 + 2(-9.81)s

\frac{(12.0)^2}{2 \times 9.81} = 7.34\ \text{m}

Common Mistake

Forgetting direction

A negative answer is not automatically wrong. It usually means the quantity is in the opposite direction to the positive direction you chose.

Free fall and $g$

Free fall means motion under gravity alone, with no air resistance. Near Earth’s surface, the gravitational field is approximately uniform, so the acceleration is constant.

Definition

Acceleration of free fall

The acceleration of free fall, $g$ , is the acceleration of an object falling freely in a gravitational field. Near Earth’s surface, $\approx 9.81\ \text{m s}^{-2}$ downward.

All objects have the same acceleration of free fall if air resistance is negligible. A feather and a steel ball would fall with the same acceleration in a vacuum.

Investigating motion and collisions

In practical work, the aim is to measure distances and times accurately enough to calculate velocities and accelerations. Electronic timing is useful because it avoids human reaction-time error.

Common methods include:

Ticker timers: dots are made at regular time intervals on a tape attached to a moving trolley. Increasing spacing means increasing speed.
Light gates: an interrupt card of length $L$ blocks a beam for time $Δt\Delta t$ , so the speed is $\frac{L}{\Delta t}$ .
Air tracks or low-friction runways: these reduce friction for trolley or glider experiments.
Data-loggers and video analysis: position can be recorded repeatedly, then velocities and accelerations can be calculated from the data.
Collision experiments: measure velocities just before and just after the collision, usually with light gates or video, while keeping the motion along one straight line.

Example

Finding acceleration using light gates

A trolley carries a card of length $m0.050\ \text{m}$ . It blocks the first light gate for $s0.100\ \text{s}$ and the second for $s0.0625\ \text{s}$ . The time between the two speed measurements is $s0.80\ \text{s}$ . Find the acceleration.

Calculate the speed at the first light gate:

\frac{0.050}{0.100} = 0.50\ \text{m s}^{-1}

Calculate the speed at the second light gate:

\frac{0.050}{0.0625} = 0.80\ \text{m s}^{-1}

Use $v = u + a t$ :

\frac{v-u}{t} = \frac{0.80 - 0.50}{0.80} = 0.375\ \text{m s}^{-2}

So the acceleration is about $s−20.38\ \text{m s}^{-2}$ .

Determining $g$ in the laboratory

Two standard arrangements are the trapdoor/electromagnet method and the light-gate method. Both are designed to remove timing by human reaction.

Laboratory arrangements for determining acceleration of free fall using an electromagnet and trapdoor, and using light gates

Electromagnet and trapdoor

A steel ball is held by an electromagnet. When the current is switched off, the timer starts and the ball falls. When it hits the trapdoor switch, the timer stops.

If the ball is released from rest and downward is positive:

\frac{1}{2}gt^2

A good method is to repeat the experiment for several heights and plot $s$ against $t^2$ . The gradient is $g2\frac{g}{2}$ , so $\times \text{gradient}$ .

Light gates and timer

An object falls through two light gates separated by a known distance $s$ . If its speed at the first gate is $u$ and at the second gate is $v$ , then:

v^2 = u^2 + 2gs

So:

\frac{v^2-u^2}{2s}

Example

Determining g with two light gates

A falling card has speed $s−11.00\ \text{m s}^{-1}$ at the first light gate and $s−11.72\ \text{m s}^{-1}$ at the second. The gates are separated by $m0.100\ \text{m}$ . Calculate $g$ .

Take downward as positive, so the acceleration is $g$ and the displacement between gates is positive.
Rearrange $v^2 = u^2 + 2gs$ :

\frac{v^2-u^2}{2s}

Substitute the data:

\frac{(1.72)^2 - (1.00)^2}{2(0.100)} = 9.6\ \text{m s}^{-2}

This is close to the accepted value, allowing for experimental uncertainty.

Common Mistake

Relying on one drop

One timing measurement gives a weak value for $g$ . Repeats, several heights, and a graph help reduce the effect of random uncertainty and make anomalies easier to spot.

Reaction time and stopping distances

When a driver sees a hazard, the car does not stop instantly. The total stopping distance is split into thinking distance and braking distance.

Car stopping distance split into thinking distance and braking distance

Definition

Stopping-distance terms

Reaction time is the time between seeing a hazard and starting to brake.
Thinking distance is the distance travelled during the reaction time.
Braking distance is the distance travelled while the brakes slow the vehicle to rest.
Stopping distance is thinking distance plus braking distance.

During the thinking distance, the car is still moving at its original speed:

s_{\text{thinking}} = vt_{\text{reaction}}

If the braking acceleration is constant and the vehicle slows from speed $u$ to rest, then:

0 = u^2 + 2as

Since braking acceleration is opposite to the motion, $a$ is negative if forward is positive.

Key Idea

Speed matters twice

For a fixed reaction time, thinking distance is proportional to speed. For constant braking deceleration, braking distance is proportional to speed squared, so higher speed has a very large effect.

Factors that increase thinking distance include tiredness, alcohol, drugs, distractions and poor visibility. Factors that increase braking distance include higher speed, worn tyres, poor brakes, wet or icy roads, and reduced friction.

Example

Calculating stopping distance

A car travels at $s−120.0\ \text{m s}^{-1}$ . The driver’s reaction time is $s0.75\ \text{s}$ . Once the brakes are applied, the car decelerates uniformly at $s−26.0\ \text{m s}^{-2}$ . Find the stopping distance.

During the reaction time, the car continues at constant speed:

s_{\text{thinking}} = vt = 20.0 \times 0.75 = 15.0\ \text{m}

For braking, take forward as positive, so $20.0\ \text{m s}^{-1}$ , $v = 0$ , and $-6.0\ \text{m s}^{-2}$ :

0^2 = (20.0)^2 + 2(-6.0)s

s_{\text{braking}} = 33.3\ \text{m}

Add the two parts:

s_{\text{stopping}} = 15.0 + 33.3 = 48.3\ \text{m}

A sensible final answer is $m48\ \text{m}$ .

Common Mistake

Mixing up braking and stopping distance

Braking distance starts only when the brakes are applied. Stopping distance includes both the thinking distance and the braking distance.

Exam technique

In the exam

Define your positive direction before using the SUVAT equations, especially for vertical motion and braking.
Check that acceleration is constant before applying $v = u + a t$ , $\frac{1}{2}(u+v)t$ , $\frac{1}{2}at^2$ or $v^2 = u^2 + 2as$ .
For practical questions, mention repeats, electronic timing, uncertainty in distance/time measurements, and using a graph where possible.

Self review

Check yourself

If a ball is thrown upwards and upward is positive, what sign should the acceleration have?
Which constant-acceleration equation is most useful when time is not given?
What is the difference between thinking distance, braking distance and stopping distance?

What you'll learn

How to use the constant-acceleration equations for motion in a straight line.
How free-fall motion uses the acceleration of free fall, $g$ .
How motion and collisions can be investigated using light gates, ticker timers, data-loggers and video.
How reaction time, thinking distance, braking distance and stopping distance are linked.

Describing motion in one dimension

Linear motion means motion along a straight line. Because the motion is only along one line, you can choose one direction as positive and the opposite direction as negative.

Definition

Core motion quantities

Displacement, $s$ , is the change in position in a specified direction. Its unit is the metre, m.
Velocity, $v$ , is the rate of change of displacement. Its unit is m s^-1.
Acceleration, $a$ , is the rate of change of velocity. Its unit is m s^-2.
In the constant-acceleration equations, $u$ means initial velocity, $v$ means final velocity, and $t$ means time.

Key Idea

Use signs consistently

Constant acceleration and the SUVAT equations

Constant acceleration means the acceleration does not change with time. The velocity then changes by equal amounts in equal time intervals.

The equations are often called SUVAT equations because they connect displacement $s$ , initial velocity $u$ , final velocity $v$ , acceleration $a$ and time $t$ .

The velocity-time graph below shows why these equations work: the gradient gives acceleration, and the area under the graph gives displacement.

Velocity-time graph showing constant acceleration, gradient and area for displacement

Key Idea

The four constant-acceleration equations

Use these only for motion in a straight line with constant acceleration:

$v = u + a t$
$\frac{1}{2}(u + v)t$
$\frac{1}{2}at^2$
$v^2 = u^2 + 2as$

Tip

Choosing the right equation

Pick the equation that contains the quantity you want and the quantities you know, but avoids any unknown you do not need. For example, if time is not given, $v^2 = u^2 + 2as$ is often useful.

Common Mistake

Constant acceleration only

These equations are not valid if the acceleration changes significantly, such as for a falling object where air resistance becomes important.

Example

Thrown upwards under gravity

A ball is thrown vertically upwards with initial velocity $s−112.0\ \text{m s}^{-1}$ . Ignore air resistance. Find the time to reach its highest point and the height gained.

Choose upward as positive. At the highest point the final velocity is $v = 0$ , and the acceleration is $-9.81\ \text{m s}^{-2}$ .
Use $v = u + a t$ to find the time:

0 = 12.0 + (- 9.81) t

\frac{12.0}{9.81} = 1.22\ \text{s}

Use $v^2 = u^2 + 2as$ to find the height, because it avoids needing another time calculation:

0^2 = (12.0)^2 + 2(-9.81)s

\frac{(12.0)^2}{2 \times 9.81} = 7.34\ \text{m}

Common Mistake

Forgetting direction

A negative answer is not automatically wrong. It usually means the quantity is in the opposite direction to the positive direction you chose.

Free fall and $g$

Free fall means motion under gravity alone, with no air resistance. Near Earth’s surface, the gravitational field is approximately uniform, so the acceleration is constant.

Definition

Acceleration of free fall

The acceleration of free fall, $g$ , is the acceleration of an object falling freely in a gravitational field. Near Earth’s surface, $\approx 9.81\ \text{m s}^{-2}$ downward.

All objects have the same acceleration of free fall if air resistance is negligible. A feather and a steel ball would fall with the same acceleration in a vacuum.

Investigating motion and collisions

In practical work, the aim is to measure distances and times accurately enough to calculate velocities and accelerations. Electronic timing is useful because it avoids human reaction-time error.

Common methods include:

Ticker timers: dots are made at regular time intervals on a tape attached to a moving trolley. Increasing spacing means increasing speed.
Light gates: an interrupt card of length $L$ blocks a beam for time $Δt\Delta t$ , so the speed is $\frac{L}{\Delta t}$ .
Air tracks or low-friction runways: these reduce friction for trolley or glider experiments.
Data-loggers and video analysis: position can be recorded repeatedly, then velocities and accelerations can be calculated from the data.
Collision experiments: measure velocities just before and just after the collision, usually with light gates or video, while keeping the motion along one straight line.

Example

Finding acceleration using light gates

Calculate the speed at the first light gate:

\frac{0.050}{0.100} = 0.50\ \text{m s}^{-1}

Calculate the speed at the second light gate:

\frac{0.050}{0.0625} = 0.80\ \text{m s}^{-1}

Use $v = u + a t$ :

\frac{v-u}{t} = \frac{0.80 - 0.50}{0.80} = 0.375\ \text{m s}^{-2}

So the acceleration is about $s−20.38\ \text{m s}^{-2}$ .

Determining $g$ in the laboratory

Two standard arrangements are the trapdoor/electromagnet method and the light-gate method. Both are designed to remove timing by human reaction.

Laboratory arrangements for determining acceleration of free fall using an electromagnet and trapdoor, and using light gates

Electromagnet and trapdoor

A steel ball is held by an electromagnet. When the current is switched off, the timer starts and the ball falls. When it hits the trapdoor switch, the timer stops.

If the ball is released from rest and downward is positive:

\frac{1}{2}gt^2

A good method is to repeat the experiment for several heights and plot $s$ against $t^2$ . The gradient is $g2\frac{g}{2}$ , so $\times \text{gradient}$ .

Light gates and timer

An object falls through two light gates separated by a known distance $s$ . If its speed at the first gate is $u$ and at the second gate is $v$ , then:

v^2 = u^2 + 2gs

So:

\frac{v^2-u^2}{2s}

Example

Determining g with two light gates

A falling card has speed $s−11.00\ \text{m s}^{-1}$ at the first light gate and $s−11.72\ \text{m s}^{-1}$ at the second. The gates are separated by $m0.100\ \text{m}$ . Calculate $g$ .

Take downward as positive, so the acceleration is $g$ and the displacement between gates is positive.
Rearrange $v^2 = u^2 + 2gs$ :

\frac{v^2-u^2}{2s}

Substitute the data:

\frac{(1.72)^2 - (1.00)^2}{2(0.100)} = 9.6\ \text{m s}^{-2}

This is close to the accepted value, allowing for experimental uncertainty.

Common Mistake

Relying on one drop

One timing measurement gives a weak value for $g$ . Repeats, several heights, and a graph help reduce the effect of random uncertainty and make anomalies easier to spot.

Reaction time and stopping distances

When a driver sees a hazard, the car does not stop instantly. The total stopping distance is split into thinking distance and braking distance.

Car stopping distance split into thinking distance and braking distance

Definition

Stopping-distance terms

Reaction time is the time between seeing a hazard and starting to brake.
Thinking distance is the distance travelled during the reaction time.
Braking distance is the distance travelled while the brakes slow the vehicle to rest.
Stopping distance is thinking distance plus braking distance.

During the thinking distance, the car is still moving at its original speed:

s_{\text{thinking}} = vt_{\text{reaction}}

If the braking acceleration is constant and the vehicle slows from speed $u$ to rest, then:

0 = u^2 + 2as

Since braking acceleration is opposite to the motion, $a$ is negative if forward is positive.

Key Idea

Speed matters twice

For a fixed reaction time, thinking distance is proportional to speed. For constant braking deceleration, braking distance is proportional to speed squared, so higher speed has a very large effect.

Example

Calculating stopping distance

During the reaction time, the car continues at constant speed:

s_{\text{thinking}} = vt = 20.0 \times 0.75 = 15.0\ \text{m}

For braking, take forward as positive, so $20.0\ \text{m s}^{-1}$ , $v = 0$ , and $-6.0\ \text{m s}^{-2}$ :

0^2 = (20.0)^2 + 2(-6.0)s

s_{\text{braking}} = 33.3\ \text{m}

Add the two parts:

s_{\text{stopping}} = 15.0 + 33.3 = 48.3\ \text{m}

A sensible final answer is $m48\ \text{m}$ .

Common Mistake

Mixing up braking and stopping distance

Braking distance starts only when the brakes are applied. Stopping distance includes both the thinking distance and the braking distance.

Exam technique

In the exam

Define your positive direction before using the SUVAT equations, especially for vertical motion and braking.
Check that acceleration is constant before applying $v = u + a t$ , $\frac{1}{2}(u+v)t$ , $\frac{1}{2}at^2$ or $v^2 = u^2 + 2as$ .
For practical questions, mention repeats, electronic timing, uncertainty in distance/time measurements, and using a graph where possible.

Self review

Check yourself

If a ball is thrown upwards and upward is positive, what sign should the acceleration have?
Which constant-acceleration equation is most useful when time is not given?
What is the difference between thinking distance, braking distance and stopping distance?

What you'll learn

Describing motion in one dimension

Constant acceleration and the SUVAT equations

Free fall and $g$

Investigating motion and collisions

Determining $g$ in the laboratory

Electromagnet and trapdoor

Light gates and timer

Reaction time and stopping distances

Review Linear motion by teaching Genie

Remember key concepts with flashcards

Linear motion

What you'll learn

Describing motion in one dimension

Constant acceleration and the SUVAT equations

Free fall and $g$

Investigating motion and collisions

Determining $g$ in the laboratory

Electromagnet and trapdoor

Light gates and timer

Reaction time and stopping distances

Review Linear motion by teaching Genie

Remember key concepts with flashcards

Linear motion

What you'll learn

Describing motion in one dimension

Constant acceleration and the SUVAT equations

Free fall and ggg

Investigating motion and collisions

Determining ggg in the laboratory

Electromagnet and trapdoor

Light gates and timer

Reaction time and stopping distances

Review Linear motion by teaching Genie

Remember key concepts with flashcards

Linear motion Revision Guide

Linear motion

What you'll learn

Describing motion in one dimension

Constant acceleration and the SUVAT equations

Free fall and ggg

Investigating motion and collisions

Determining ggg in the laboratory

Electromagnet and trapdoor

Light gates and timer

Reaction time and stopping distances

Review Linear motion by teaching Genie

Remember key concepts with flashcards

Linear motion Revision Guide

Free fall and $g$

Determining $g$ in the laboratory

Free fall and $g$

Determining $g$ in the laboratory