Skid numerical: bridge abutment impact at 20 kmph

The vehicle skidded 30 m on the pavement, left the pavement and skidded another 60 m on the gravel shoulder, and then hit the abutment at 20 km/h. Working backwards by the work-energy equation $v_1^2 = v_2^2 + 2gfd$ over each stretch:

 start -> [ 30 m pavement f=0.35 ] -> [ 60 m gravel f=0.50 ] -> abutment
 v0                                  v1                        20 km/h

Data: $v_{abutment} = 20$ km/h $= 5.556$ m/s.

Step 1: Speed at the start of the gravel shoulder

$$v_1 = \sqrt{v_a^2 + 2 g f_2 d_2} = \sqrt{(5.556)^2 + 2(9.81)(0.50)(60)} = 24.889\ \text{m/s} = 89.6\ \text{km/h}$$

Step 2: Initial speed at the start of the pavement skid

$$v_0 = \sqrt{v_1^2 + 2 g f_1 d_1} = \sqrt{(24.889)^2 + 2(9.81)(0.35)(30)} = 28.731\ \text{m/s}$$

Answer: initial speed of the vehicle = 28.73 m/s = 103.4 km/h.

(Check with the IRC/Indian field formula $v = \sqrt{254(f_1d_1+f_2d_2)+v_a^2} = \sqrt{254(0.35\times30 + 0.5\times60) + 20^2} = 103.4$ km/h.)