1.19-1.21 - Stopping distances and terminal velocity

Stopping distance

The stopping distance of a vehicle is the total distance it travels from the moment the driver first needs to stop to the moment the vehicle is at rest.

It has two parts:

• thinking distance: the distance travelled during the driver's reaction time, before the brakes are applied
• braking distance: the distance travelled after the brakes are applied, while the vehicle slows to rest

For example, if a car travels 12 m while the driver reacts and then 28 m while braking, its stopping distance is:

A common exam mistake is to describe only the braking distance. The stopping distance starts earlier, when the driver first notices the need to stop.

Thinking distance and reaction time

Thinking distance depends on how far the vehicle travels during the driver's reaction time. Reaction time is the time between noticing the hazard and starting to brake.

At a steady speed during the reaction time:

This means thinking distance increases if the vehicle is moving faster, because it covers more metres each second. It also increases if the driver's reaction time is longer.

Reaction time can be increased by factors such as tiredness, distraction, alcohol, drugs, or anything else that delays the driver noticing the hazard or moving to the brake. At the same speed, doubling reaction time doubles thinking distance.

Worked example: a car is travelling at 18 m/s. The driver's reaction time is 0.60 s.

The car travels 10.8 m before the brakes even begin to act.

Braking distance factors

Braking distance is affected by how quickly the vehicle can decelerate after the brakes are applied. The braking force must reduce the vehicle's speed to zero.

Several factors increase the braking distance:

For the same constant braking deceleration, braking distance rises with the square of the initial speed. From $0=u^2+2as$, $s=u^2/(2|a|)$. At $20\,\mathrm{m/s}$ and a deceleration of $5.0\,\mathrm{m/s^2}$, braking distance is $20^2/(2\times5.0)=40\,\mathrm{m}$. At twice the speed with the same deceleration it is $40^2/10=160\,\mathrm{m}$: four times as far. Thinking distance only doubles if reaction time stays the same.

Speed is especially important because it affects both parts of stopping distance. A faster vehicle covers more distance during the driver's reaction time and also needs a longer distance to slow down once braking starts.

The UK Highway Code's typical stopping distances show this pattern: at $30\,\mathrm{mph}$ a car needs about $9\,\mathrm{m}$ thinking distance and $14\,\mathrm{m}$ braking distance ($23\,\mathrm{m}$ in total); at $60\,\mathrm{mph}$ it needs about $18\,\mathrm{m}$ and $55\,\mathrm{m}$ ($73\,\mathrm{m}$ in total). The thinking distance doubles, but the braking distance is about four times larger.

When explaining road condition, connect the condition to friction. A phrase such as "the road is icy" is not enough by itself; the physics is that reduced friction gives a smaller braking force, so the vehicle decelerates less quickly.

Forces on falling objects

For the falling objects considered here, air upthrust is negligible and downward motion is taken as positive on the graph. The two important forces are:

• weight, acting downwards because of gravity
• air resistance or drag, acting upwards against the motion

When the object is first released, its speed is small, so the air resistance is small. Weight is larger than air resistance, so the resultant force is downwards and the object accelerates downwards. At the very start there is almost no air resistance, so the acceleration is close to $g$, about $10\,\mathrm{m/s^2}$.

As the object gets faster, air resistance increases. Weight stays almost constant, but the upward drag force grows. The resultant downward force becomes smaller, so the object still speeds up, but its acceleration decreases. Since $a = F/m$, a smaller resultant force on the same mass gives a smaller acceleration.

[DIAGRAM: terminal_velocity_velocity_time_graph: Stopping Distances and Terminal Velocity - diagram 01; asset_slug: p08_stopping_distances_and_terminal_velocity__diagram_01; recommended_method: matplotlib; description: NovaLearn pale-palette falling-object velocity-time graph, rising steeply and levelling at terminal velocity, with leaders explaining decreasing resultant force and acceleration and the final balanced forces.]

On this velocity-time graph the gradient is the acceleration. The curve starts steep (large acceleration), becomes less steep as air resistance grows (acceleration decreasing) and ends flat (zero acceleration at terminal velocity).

Terminal velocity

An object reaches terminal velocity when its weight and air resistance are equal in size and opposite in direction. The forces are balanced, so the resultant force is zero.

Zero resultant force does not mean the object stops. It means there is no acceleration. If the object is already moving downwards, it continues moving downwards at a constant velocity.

The sequence is:

1. At the start, weight is greater than air resistance, so the object accelerates downwards.
2. As speed increases, air resistance increases, so the resultant force and acceleration decrease.
3. At terminal velocity, air resistance equals weight, so resultant force is zero and velocity is constant.

In exam answers, use the chain of reasoning. Do not only say "the object reaches terminal velocity". Explain why: increasing speed increases air resistance until air resistance balances weight.

A skydiver shows the whole sequence. After jumping, the skydiver speeds up until air resistance balances weight, then falls at a steady terminal velocity. Opening the parachute greatly increases the surface area, so air resistance suddenly becomes larger than weight. The resultant force is now upwards, so the skydiver decelerates. As the speed falls, air resistance falls, until it balances weight again at a new, lower terminal velocity that is safe for landing.

Quick recap

• Stopping distance is thinking distance plus braking distance. At steady speed, thinking distance is speed multiplied by reaction time.
• Longer reaction time increases thinking distance; reduced braking force increases braking distance. Greater mass increases braking distance when braking force is unchanged.
• At the same constant braking deceleration, doubling initial speed makes braking distance four times larger.
• As a falling object speeds up, drag increases and acceleration decreases. At terminal velocity, drag balances weight: acceleration is zero but downward velocity remains constant.