**Distances:**

A -> B 128ft 39.014m

B -> C 46ft 14.021m

C -> D 26ft 8.534m Tigra travels 29.41ft 8.9642m

D -> E 10ft 7.01m STRIKE RIGBY

E -> F 7ft 2.13m KERB

F -> G 33ft 10.05m Tigra travels 23.82 ft 11.6035m

**Gradient of the hill:**

From measuring the angle between the sloping base of the wall and its horizontal top, the gradient of the hill up Artillery Place has been reckoned by the author as being at 8.5%, or as being at an incline of 4.86 degrees from the horizontal.

**Acceleration of Tigra**:

According to the Government’s registration enquiry service, the Tigra with the plate N696 JWX, has an cylinder capacity of 1389cc and was manufactured in 1996. This makes it a Tigra 1.4i, capable of 0-62 mph in 11.5 seconds, according to specifications found online. This means that its ultimate rate of acceleration is

a = v-u/t = 27.716-0/11.5 = 2.41 ms²

**Minus Acceleration**** due to Incline:**

g, the acceleration due to gravity = 9.8m/s²

Incline = 4.86 degrees (calculated using imagery of the wall – the horizontal top line forms the height of a triangle; where the pavement cuts into it forms an hypotenuse). This incline is not necessarily uniform, but possibly represents the best estimation, and the greatest extent of the gradient.

Acceleration due to incline = -sin(4.86) x 9.8, or -9.8sin(4.86) = **-0.83m/s****²**

**Times at Positions after Commencement of Footage/Intervals:**

Lead car interval Tigra interval

A: 7.9s 16.3s

B: 10.8s 2.9s 21.6s 5.3s

C: 12.4s 1.6s 23.2s 1.6s

D: 13.0s 0.6s 24.0s 0.8s

E: 13.5s 0.5s

F: 13.65s 0.15s

G: 14.4s 0.65s

** A****verage Velocities:**

Lead Car: Tigra

A -> B 13.453m/s 30.09mph 7.361m/s 16.466mph

B -> C 8.763m/s 19.602mph 8.763m/s 19.602mph

C -> D 14.223m/s 31.81mph 10.667m/s 23.861mph

D -> E 14.02m/s 31.36mph

E -> F 14.2m/s 31.76mph

F -> G 15.461m/s 34.58mph

**Confirming Tigra’s Top Rate of Acceleration/Deceleration due to Incline:**

Using a = 2(s-ut/t²)

From B to C, assuming that Tigra is accelerating at top rate:

starting velocity, *u* = 7.361 (average across A->B)

distance, *s* = 14.021

time, *t* = 1.6

therefore, acceleration from B to C, *a* = 1.75m/s²

Now, theoretical top rate = 2.41m/s²

Notice that 2.41 + -0.83 (deceleration due to incline) = 1.58.

Let’s find a new starting velocity at B based on what we think the top rate of acceleration is:

Using u = s/t – at/2 = 14.021/1.6 – 1.58×1.6/2 = 7.499m/s

So this is supposed to be the starting speed at B, and it is actually quite close to what we thought the average speed across A->B would be. It makes a reasonable starting point.

**Minus Acceleration due to Braking:**

9.8 x μ x cos(4.86), with μ, friction co-efficient, reckoned at 0.7.

Therefore acceleration due to braking is reckoned at = **-6.835m/s²**

We can check this using this equation:

Stopping Distance = v²/2g(f+G)

g = 9.8; v = velocity; G = grade of road = 0.085; f = 0.7;

If travelling at 10m/s; stopping distance =

10²/2×9.8×0.785 = 100/15.386 = 6.49

a = v²-u²/2s

a = -7.7

and -6.835 + -0.83 = -7.665

**Velocity at B**

This was discovered above to be 7.499m/s, or **16.77mph**

**Velocity at C**

Using the initial velocity at B, and other numbers for the variables as above, final velocity, *v*, can be found using

v=u+at = 7.499+1.58×1.6

= 10.027m/s or **22.43mph**

**Velocity at D**

This assumes that the Tigra is still accelerating to maximum capability. Using the initial velocity at C, the final velocity at D can be found in the same way as for the previous case. The new numbers for the variables come from the data is the last chapter.

v = 10.027+1.58×0.8

= 11.291m/s or **25.26 mph**

**Velocity at E**

The official narrative makes a great point of the Tigra accelerating into Rigby. The interaction between the car and Rigby as has been described by witnesses, however, suggest that the car was decelerating. From now on, we need to calculate speeds for every case – acceleration at maximum capability, and deceleration due to the incline.

Using v²=u²+2as, with u = 11.291; a = 1.58, or -0.83; *s* = 7.01

Accelerating: *v* = 12.233m/s or 27.364mph (possible)

Decelerating : *v* = 10.763m/s or 24.076mph (preferred)

**Velocity at F**

This is at the kerb. Having now hit Rigby, we could also imagine at this stage what would happen if the Tigra had started to brake from E to F so as to be better able to control the coming manoeuvre up the kerb and to the road sign stanchion (denoted with a B).

Using v²=u²+2as, with u = 12.233 or 10.763; a = 1.58, or -0.83 or -7.665; *s* = 2.13

1. Continued acceleration: *v* = 12.505m/s or 27.97 mph

2. Decelerating (G) after accelerating at E: *v* = 12.088m/s or 27.04mph (possible)

3. Decelerating (B) after accelerating at E: *v* = 10.816m/s or 24.19mph

4. Decelerating (G) after decelerating at E: *v* = 10.597m/s or 23.71mph (preferred)

5. Decelerating (B) after decelerating at E: *v* = 9.121m/s or 20.4mph

**Velocity at G – Final Velocity at Stanchion**

This time, we’ll focus first on stopping due to gravity alone::

Using v²=u²+2as, with u = (1) 12.505 or (2) 12.088 or (3) 10.816 or (4) 10.597 or (5) 9.121; a = 1.58, or -0.83; *s* = 11.6035

Continued acceleration: *v* = 13.894m/s or 31.08mph

Decelerating (G) after case 1 at F: *v* = 11.709m/s or 26.19mph

Decelerating (G) after case 2 at F: *v* = 11.263m/s or 25.19mph (possible)

Decelerating (G) after case 3 at F: *v* = 9.885m/s or 22.11mph

Decelerating (G) after case 4 at F: *v* = 9.645m/s or 21.57mph (preferred)

Decelerating (G) after case 5 at F: *v* = 7.996m/s or 17.88mph

Now we’re going to see what happens when brakes are applied at F for the fastest (1) and slowest (5) cases. Remember, the distance between F and G is 11.6m

s=v²-u²/2a

so, when u = 12.505, s = 10.2m; when u = 9.121, s = 5.43m. For all cases, it would have been possible to brake so that the Tigra came to a stand still before hitting the stanchion. The slower the car was going at F, the more leisurely the slow down could have been.