5.5 Graphical analysis of braking distances, speed and kinetic energy of a moving road vehicle

What is the relationship between braking distance and kinetic energy?

Graph 1a assumes a uniform decrease in velocity i.e. uniform deceleration

See part 5.8 Some advanced calculations on braking force and removing vehicle kinetic energy

Diagram KEY:

KE = kinetic energy (J), m = mass (kg),   u = initial velocity (m/s)

v = final velocity (m/s),   s = speed (m/s)

a = acceleration or deceleration (m/s²)

W = work done (J),   F = force (N),   d = distance (m)

The graph 1b above takes the thinking distance, braking distance and stopping distance data and plots them against the typical speed of a road vehicle.

Obviously, all the distances increase with increase in speed, but note two other very important points.

You should notice ...

(i) two of the graphs curve upwards, so there is a sort of 'accelerating' effect of speed on the braking distance and overall stopping distance (the latter is due to the increase in braking distance).

Stopping distance and braking distance are not proportional to speed, and crucially,  the braking distance is proportional to speed squared. This means the stopping/braking distances increase faster than the increase in speed.

e.g. doubling speed quadruples the braking distance (2 ==> 2² = 4) and trebling speed increases the braking distance nine times (3 ==> 3² = 9).

The thinking distance is roughly proportional to speed, the graph is ~linear and does not curve upwards. This is because your response time, if fully alert, is pretty constant, so if your speed doubles, you just go twice as far in the same response time.

 

(ii) and if you examine the graph or data carefully, you can see that doubling the speed quadruples the braking distance.

This means by doubling your speed, approximately quadruples the stopping distance, obviously something you need to bear in mind the faster you drive.

Doubling speed quadruples braking distance and trebling speed increases it nine times! (see the REMINDER below)

This is discussed further and is related to the formula for kinetic energy

KE = ½mv².

 

By doubling the speed, you quadruple the kinetic energy of the car, hence you have quadrupled the kinetic energy to be removed by braking (because KE v²). See graphs 2 and 3 and notes below.

Therefore, on doubling the speed, for a constant braking force, you have four times as much KE to remove and will need four times the distance to remove it.

For more on kinetic energy calculations see Kinetic energy store calculations

and 5.8 Advanced calculations on braking force, removal of kinetic energy

• REMINDER: The kinetic energy of a moving object can be calculated using the equation:

  • kinetic energy (KE) = 0.5 × mass × (speed)²

  • E_ke = ¹/₂ m v²

  • kinetic energy, E_ke, in joules, J; mass, m, in kilograms, kg; speed/velocity, v, in metres per second, m/s

  • Note that by doubling the speed you quadruple the kinetic energy! (e.g. 2² : 4² is 4 : 16 or 1 : 4)

 

Question to illustrate some of the ideas above and using the chart below.

When travelling at 20 mph a driver's thinking distance is 6.0 m and the braking distance is 6.0 m.

(a) What is the stopping distance?

stopping distance = thinking distance + braking distance = 6.0 + 6.0 = 12.0 m

 

(b) Estimate the total stopping distance at 40 mph (scaling up factor of 2).

If the thinking distance is 6 m at 20 mph, it will be double that at 40 mph, 6 x 40 / 20 = 12 m.

From the KE argument and KE v2, the braking distance increases by the square of the scale factor.

So the braking distance 6 x 2² = 24 m

Therefore the stopping distance is 12 + 24 = 36 m (check on the chart)

 

(c) Estimate the total stopping distance at 80 mph (scale factor 4).

If the thinking distance is 6 m at 20 mph, it will be quadruple that at 40 mph, 6 x 80 / 20 = 24 m

The braking distance increases by the square of the scale factor.

So the braking distance 6 x 2⁴ = 96 m

Therefore the stopping distance is 24 + 96 = 120 m (not on the chart)