You have been hired as an expert witness in a case involving an automobile accid

Question

You have been hired as an expert witness in a case involving an automobile accident. The accident involved car A of mass 1500 kg, which crashed into stationary car B of mass 1100 kg. The driver of car A applied his brakes 15 m before he skidded into car B. After the collision, car A slid 18 m while Car B slid 30 m before coming to rest. The coefficient of kinetic friction between the locked wheels and the road is 0.50. Show that the driver of car A was exceeding the 55 mi/h (90 km/h) speed limit before applying the brakes.

Explanation / Answer

You have to work it backwards.
• friction work B = Ff * d = µmgd = 0.5 * 1100kg * 9.8m/s² * 30m = 161.7 kJ
Therefore B's post-impact velocity can be found from
KE = 161.7 kJ = ½mv² = ½ * 1100kg * Vb² Vb = 17.15 m/s
Then the post-collision momentum for B was p_b = mVb = 18865 kg·m/s

• friction work A = 0.5 * 1500kg * 9.8m/s² * 18m = 132.3 kJ
So A's post-impact velocity was Va = (2E/m) = 13.28 m/s
Then the post-collision momentum for A was p_a = 19922.35 kg·m/s
• Then the pre-collision momentum p = 18865 kg·m/s + 19922.35 kg·m/s = 38787.35 kg·m/s
Then A's pre-collision velocity was V = 38787.35 kg·m/s / 1500kg = 25.86 m/s
meaning that his pre-collision KE = ½mv² = 501.5 kJ
• Energy lost in pre-collision skid: W = 0.5 * 1500kg * 9.8m/s² * 15m = 110.25 kJ
So before the skid, KE = 501.5 kJ + 110.25 kJ = 611.75 kJ

This translates to a velocity of
v = (2E/m) = 28.56 m/s * 2.237mph / 1m/s = 63.89 mph

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