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Car Collision Forensics Lab

Car Collision Forensics Lab. Ben O’Donnell, Robert Robinson, Jack O’Donnell. Introduction.

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Car Collision Forensics Lab

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  1. Car Collision Forensics Lab Ben O’Donnell, Robert Robinson, Jack O’Donnell

  2. Introduction • A traffic accident occurred in a 35 km/hr speed limit zone on Millway Street in which a 3000-kg Cadillac Escalade SUV rear-ended a 2000-kg Subaru Outback Wagon that was stopped at a stop sign. The entire police investigative division has gone on vacation to Bora Bora to relax, so the mayor has contracted with you and your team of experts to determine what happened and what traffic laws were broken.

  3. Accident Schematic

  4. Auto Expert In order to solve this portion of the lab we have to use a kinematics equation from an earlier chapter. Vf^(2 )=Vi^2+2a∆x We then plug in what we know from the information given to us in the problem. We know that the SUV has a mass of 3000kg, and the Wagon has a mass of 2000kg. We also know that the acceleration of the SUV with its brakes locked is -2m/s^2, and that the acceleration of the Wagon with its brakes locked is -3m/s^2. We then use this information to solve two kinematics problems, one for the final velocity of the wagon, the other for the final velocity of the SUV. Wagon: Vf^2=(0)^2+2(-3m/s^2 )(24) Vf^2= -144 Vf^2= √144 Vf=12 m/s SUV: Vf^2=(0)^2+2((-2m)/s^2 )(2) Vf^2= -8 Vf^2= √8 Vf=2.8 m/s So immediately after the collision the Subaru wagon had a velocity of 12 m/s. The SUV had a velocity of 2.8 m/s.

  5. Collision Expert • I had to find the initial velocity of the SUV using the equation I found the SUV was traveling at a speed of 10.8 meters per second when it rear ended the Wagon X= Wagon Y= SUV (𝑉_𝑖 𝑥)(𝑚𝑥)+(𝑉_𝑖 𝑦)(𝑚𝑦)=(𝑉_𝑓 𝑥)(𝑚𝑥)+(𝑉_𝑓 𝑦)(𝑚𝑦) (0)(2000)+(𝑉_𝑖 𝑦)(3000)=(12)(2000)+(2.8)(3000) 0+(𝑉_𝑖 𝑦)(3000)=24000+8400 (𝑉_𝑖 𝑦)(3000)=32400 (𝑉_𝑖 𝑦)=10.8 𝑚/𝑠

  6. Investigator • In the world of physics there are three types of collisions, these types are known as elastic, inelastic and perfectly inelastic. An Elastic Collision is when two objects collide and bounce off of each other with no deformation and have the same kinetic energy before and after the collision. An Inelastic Collision is exactly the opposite, with two objects colliding and a deformation occurring. A Perfect Inelastic collision occurs when two objects collide and move together as one object (such as throwing two pieces of clay together. • This crash exhibits characteristics of an inelastic collision. This is due to the fact that the two vehicles did dot stay joined together after the collision (the two vehicles had different final velocities). Using the kinetic energy formula (KE= ½ m*v^2) you can determine that the kinetic energies for the vehicles is different in the beginning than it is in the end. • Before: KEsuv= ½ (3000)x (10.8^2)= 174.96 KJ • KEwag= ½ (2000)x (0^2)= 0 KJ • After: KEsuv= ½ (3000)x (2.8^2)= 11.76 KJ • KEwag= ½ (2000)x (12^2)= 144 KJ • If added together the kinetic energies do not match (note: some energy is given off as heat) (174.96 does not = 155.76). So the only collision type left is an inelastic collision

  7. Conclusion • The cars collided with the SUV having a velocity of 10.8 m/s. The Subaru was at rest. After the accident the cars had a velocity of 12 m/s (Subaru) and 2.8 m/s (SUV). The type of collision that the cars experienced was a inelastic collision.

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