Module 8 : Specialized Traffic Studies

Lecture 42 : Accident Studies

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	Energy theory Applying principle of conservation of energy or conservation of momentum also the initial speed of the vehicle can be computed if the skid marks are known. It is based on the concept that there is reduction in kinetic energy with the work done against the skid resistance. So if the vehicle of weight $W$ slow down from speed $v_1$ to $v_2$, then the loss in kinetic energy will be equal to the work done against skid resistance, where work done is weight of the vehicle multiplied by the skid distance and the skid resistance coefficient. $$\frac{W(v_1^2 - v_2^2)}{2g} = W.f.S$$ (1) where, $f$ is the skid resistance coefficient and $S$ is the skid distance. It also follows the law of conservation of momentum ($m_1$, $v_1$ are the mass and velocity of first vehicle colliding with another vehicle of mass and velocity $m_2$, $v_2$ respectively) $$m_1v_1 = m_2v_2$$ (2) Numerical example A vehicle of 2000 kg skids a distance of 36 m before colliding with a stationary vehicle of 1500 kg weight. After collision both vehicle skid a distance of 14 m. Assuming coefficient of friction 0.5, determine the initial speed of the vehicle. Solution:  Let the weight of the moving vehicle is $W_A$, let the weight of the stationary vehicle is $W_B$, skid distance before and after collision is $s_1$ and $s_2$ respectively, initial speed is $v_1$, speed after applying brakes before collision is $v_2$ and the speed of both the vehicles $A$ and $B$ after collision is $v_3$, and the final speed $v_4$ is 0. Then: After collision:  Loss in kinetic energy of both cars = Work done against skid resistance (can be obtained from equation mentioned below). Substituting the values we obtain $v_3$. $$\frac{(W_A + W_B) \times (v_3^2 - v_4^2)}{2g} = (W_A + W_B).f.s_2$$ $$\frac{(v_3)^2}{2g} = 0.5 \times 14 = 7$$ $$v_3 = 11.71 m/s$$ At collision:  Momentum before impact = momentum after impact (can be obtained from equation. 2) $$\frac{W_A.v_2}{g} = \frac{(W_A + W_B)v_3}{g}$$ $$v_2 = \frac{(W_A + W_B)v_3}{W_A}$$ $$v_2 = 20.5 m/s$$ Before collision (can be obtained from equation. 1): Loss in kinetic energy of moving vehicle = work done against braking force in reducing the speed $$\frac{(W_A) \times (v_1^2 - v_2^2)}{2g} = W_A.f.s_1$$ $$\frac{(v_1^2 - v_2^2)}{2g} = 0.5 \times 36$$ $$v_1~ = ~27.8~m/s~=~100~kmph$$ Ans: The pre-collision speed of the moving vehicle is 100 kmph.